TS Inter 2nd Year English Study Material Chapter 6 How to Avoid Foolish Opinions

Telangana TSBIE TS Inter 2nd Year English Study Material 6th Lesson How to Avoid Foolish Opinions Textbook Questions and Answers.

TS Inter 2nd Year English Study Material 6th Lesson How to Avoid Foolish Opinions

Annotations (Section A, Q.No. 1, Marks: 4)
Annotate the following in about 100 words each.

a) If someone maintains that two and two are five or that Iceland is on the equator, you feel pity rather than anger…….. (Revision Test – I)

Introduction: These beautiful lines are taken from the thought-provoking essay, “How to Avoid Foolish Opinions” written by Bertrand Russell. His clarity of thought and fluency of expression lend beauty to his style.

Context and Meanings: Russell gives us tips on how to avoid foolish opinions. He says that there are many ways to avoid being dramatic. To avoid foolish opinions, no super human or genius is required. Many matters are less easily brought to the test of experience. If we hear views opposite to our opinions, it makes us angry. It is a sign that we actually have no good reason for our opinion. If someone has very stupid and wrong opinions, we feel pity rather than anger. So, we must carefully reconsider our ideas.

Critical Comment: Though the article discusses many mistakes mankind is prove to make, it ends with a lively ray of hope.

కవి పరిచయం : బెర్ట్రాండ్ రస్సెల్ చే రచింపబడిన “మూర్ఖపు అభిప్రాయాలను ఎలా నివారించాలి” అను ఆలోచనను రేకెత్తించే వ్యాసం నుండి ఈ అందమైన పంక్తులు తీసుకొనబడినవి. ఇతని యొక్క స్పష్టమైన ఆలోచన మరియు ధారాళమైన అభిప్రాయం ఇతని శైలికి అందాన్ని ఇస్తున్నాయి.

సందర్భ౦ మరియు అర్థం : మూర్ఖపు అభిప్రాయాలను ఎలా నివారించాలి అనే దాని మీద రస్సెల్ మనకు సూచనలు ఇస్తున్నారు. మూర్ఖంగా వ్యవహరించడాన్ని నిరోధించుటకు అనేక మార్గాలున్నాయి. అనేక విషయాలు అనుభవం యొక్క పీక్షకు చాలా సులువుగా తీసుకురాబడతాయి. మన అభిప్రాయంకు వ్యతిరేఖంగా మనం అభిప్రాయాలు వింటే, అది మనకు కోపం తెప్పిస్తుంటే, అది మన అభిప్రాయానికి వాస్తవంగా సరైన కారణం లేదనే ఒక సూచిక. ఎవరైనా చాలా తెలివి తక్కువ మరియు తప్పుడు అభిప్రాయాలను కలిగిఉంటే, కోపం కంటే మనం జాలి చూపుతాము. కావున, మనం జాగ్రత్తగా మన అభిప్రాయాలను నిర్ధారించుకోవాలి.

విమర్శ : మానవాళి అనేక తప్పులు చేయటానికి అవకాశం ఉందని ఈ వ్యాసం చర్చించినప్పటికీ ఇది చక్కటి ఆశీకిరణంతో ముగుస్తుంది.

b) Persecution is used in the theology not in arithmetic because, there is knowledge, but in theology, there is only opinion. (Revision Test – I)

Introduction: These lines are taken from the thought provoking essay, “How to Avoid Foolish Opinion” written by Bertrand Russell. His clarity of thought and fluency of expression lend beauty to his style.

Context and Meanings: Russell gives us tips on how to avoid foolish opinion. Here, he tells us about controversies. The worst controversies are about matters which have no gcoa ev de ne either way. If you can not observe an issue, think about any biases you might have about it. This is because belief can go beyond facts. Persecution means annoying others deliberately all the time, the theology is required belief where as arithmetic is require facts and figures. Hence, persecution is not used in arithmetic but in theology – the study of God and religion.

Critical Comment: The author says that belief can go beyond facts.

కవి పరిచయం : బెర్ట్రాండ్ రస్సెల్ చే రచింపబడిన “మూర్ఖపు అభిప్రాయాలను ఎలా నివారించాలి” అనే ఆలోచనను రేకెత్తించే వ్యాసం నుండి ఈ పంక్తులు తీసుకొనబడినవి. ఇతని యొక్క స్పష్టమైన ఆలోచన మరియు ధారాళమైన అభిప్రాయాలు ఇతని శైలికి అందాన్ని ఇస్తున్నాయి.

సందర్భ౦ మరియు అర్థం : మూర్ఖపు అభిప్రాయాలను ఎలా నివారించాలి అనే దాని గురించి రస్సెల్ మనకు సూచనలు ఇస్తున్నారు. ఇతను మనకు వివాదాలు గురించి చెప్తున్నాడు. మంచి సాక్ష్యాలు లేని విషయాలు గురించి చెత్త వివాదాలు ఉన్నాయి. ఒక విషయాన్ని పరిశీలించలేకపోతే దాని గురించి మీకు ఏదైనా పక్షపాతాలు ఉన్నాయా లేదా ఆలోచించండి. ఎందుకంటే విశ్వాసం వాస్తవాలను అధిగమించగలదు. పీడించడం, చికాకు పెట్టడం అంటే ఇతరులను ఉద్దేశ్య పూర్వకంగా అన్ని వేళలా చూపించడం, చికాకు పెట్టడం. వేదాంతానికి నమ్మకం అవసరం, ఇక్కడ అంకగణితానికి వాస్తవాలు మరియు గణాంకాలు అవసరం. అందువల్ల, వేధింపు, చికాకు అనేది అంకగణితంలో ఉపయోగించబడదు కానీ వేదాంతశాస్త్రంతలో – దేవుని అధ్యయనంలో ఉపయోగిస్తారు.

విమర్శ: నమ్మకం వాస్తవాలకు అతీతంగా ఉంటుందని రచయిత చెప్పారు.

TS Inter 2nd Year English Study Material Chapter 6 How to Avoid Foolish Opinions

c) I have frequently found myself growing less dogmatic and cocksure through realising the possible reasonableness of a hypothetical opponent.

Introduction: These lines are taken from the thought provoking essay, “How to Avoid Foolish Opinions” written by Bertrand Russell. His clarity of thought and fluency of expression lend beauty to his style.

Context and Meanings: The writer advises us to argue with an imaginary character who has a different point of view. It is a good plan to test our arguments. Imaginary dialogues through psychological imagination serve a great purpose. Continue Conversation with that person hypothetically. You end up with reversing your ideas wherever necessary. He says that he has done the some experiment.

Critical Comment: Here, the author offers us hypothetical opponent to avoid foolish opinions and to become a less dogmatic.

కవి పరిచయం : బెర్ట్రాండ్ రస్సెల్ చే రచింపబడిన “మూర్ఖపు ఆలోచనలను ఎలా నివారించాలి” అనే ఆలోచనను రేకెత్తించే వ్యాసం నుండి ఈ పంక్తులు తీసుకొనబడినవి. ఇతని యొక్క స్పష్టమైన ఆలోచన మరియు అబిప్రాయ ధారాళం ఇతని శైలికి అందాన్ని చేకూర్చుతున్నాయి.

సందర్భ౦ మరియు అర్థం : మూర్ఖపు ఆలోచనలను ఎలా నివారించాలనే దానిపై తస్సెల్ మునకు సూచనలు ఇస్తున్నారు. ఇక్కడ అభిప్రాయ భేదం ఉన్న ఊహాజనిత పాత్రతో వాదించమని రచయిత మనకు సలహా ఇస్తున్నాడు. మన సంభాషణను పరీక్షించుకొనుటకు ఇది మంచి ప్రణాళిక. మన ఊహాశక్తితో కల్పిత సంభాషణల సృష్టి గొప్ప ఫలితాలను ఇస్తుంది. మీ అభిప్రాయాలకు వ్యతిరేఖ ఆలోచనలు గల వ్యక్తిని మీ ఎదురుగా ఉన్నట్లు భావించండి. వారితో కల్పిత సంభాషణలను కొనసాగించండి. దీని తుది ఫలితం అవసరమైన చోటల్లా మీ అభిప్రాయాలను సరిచేసుకొని మెరుగుపరుచుకొంటారు. రచయిత ఇదే ప్రయోగం చేశానని చెప్తున్నారు.

విమర్శ: ఇక్కడ, మూర్ఖపు అభిప్రాయాలను నివారించుటకు మరియు మొండిగా ఉండకుఁ డా ఉండుటకు, రచయిత ఒక ఊహాజనితపాత్రను చూస్తున్నాడు.

d) Be very way of opinions that flatter your self esteem.

Introduction: These lines are taken from the thought provoking essay, “How to Avoid Foolish Opinions” written by Bertrand Russell. This clarity of thought and fluency of expression lend beauty to his style.

Context and Meaning: In this essay, Russell give us tips on how to avoid foolish opinions. He cautions us against comments that boost one’s ego. This problem is tough to solve because everyone is sure of his/her supremacy. He advises a healthy dose of modesty and common sense. Our principles should not be dramatic. Other’s standards and ideals should equally be respected. He adds that dealing with a man’s self-esteem is challenging the only solution is to remind people of the episodic human life on a small planet.

Critical Comment: Though the essay discusses many mistakes mankind is prone to make, it ends with a lively ray of hope.

కవి పరిచయం : బెర్ట్రాండ్ రస్సెల్ చే రచింపబడిన “మూర్ఖపు అభిప్రాయాలను ఎలా నివారించాలి” అనే ఆలోచనను రేకెత్తించే వ్యాసం నుండి ఈ పంక్తులు తీసుకొనబడినవి. ఇతని ఆలోచన స్పష్టత మరియు వ్యక్తీకరణ యొక్క పటిమ అతని శైలికి అందాన్ని ఇస్తుంది.

సందర్భ౦ మరియు అర్థం : ఈ వ్యాసంలో రస్సెల్ తెలివి తక్కువ అభిప్రాయాలను ఎలా నివారించలో మనకు చిట్కాలు ఇస్తున్నారు. ఒకరి అహాన్ని పెంచే వ్యాఖ్యలకు వ్యతిరేఖంగా తనకు మనల్ని హెచ్చరిస్తున్నాడు. ప్రతి ఒక్కరూ తమ ఆధిపత్యాన్ని ఖచ్చితంగా కలిగిఉంటారు కాబట్టి ఈ సమస్యను పరిష్కరించడం చాలా కష్టం. అతను నమ్రత మరియు ఇంగితజ్ఞానం యొక్క ఆరోగ్యకరమైన మోతాదును సూచించాడు. మన సూత్రాలు పిడివాదంగా ఉండకూడదు. ఇతరుల యొక్క ప్రమాణాలు మరియు ఆదర్శాలను సముద్రంగా గౌరవించాలి. మనిషి యొక్క ఆత్మగౌరవంతో వ్యవహరించడం సవాలుతో కూడుకున్నదని అతను చెప్పాడు. ఒక చిన్న గ్రహం మీద ఎపిసోడ్ మాపైన జీవితాన్ని ప్రజలకు గుర్తు చేయడము ఏకైక పరిష్కారం.

విమర్శ : మానవజాతి చేయవలసిన అనేక తప్పులను వ్యాసం చర్చిస్తున్నప్పటికీ, ఇది సజీవమైన ఆశతో ముగుస్తుంది.

Paragraph Questions & Answers (Section A, Q.No.3, Marks: 4)
Answer the following Questions in about 100 words

a) How can we prevent developing a dogmatic attitude as per Russell’s suggestion ? (Revision Test – I)
Answer:
The thought provoking essay, “How to Avoid Foolish Opinion’s” is written by Bertrand Russell. In this essay, he says that there are many ways to avoid being dogmatic. Making a keen observation where it can settle the bias is the first way. Next to know what other people think.

One has to be aware of what they think. This can be done by going on vacation and talking to people with different ideas. The third is arguing with an imaginary character. The fourth one is to deal with one’s sense of self worth. To overcome conceit, we must remember that we live for a short while on a small planet in vast cosmos.

ఆలోచనను రేకెత్తించు వ్యాసం ‘మూర్ఖపు అభిప్రాయాలను ఎలా నివారించాలి’ అను వ్యాసం బెర్ట్రాండ్ రస్సెల్ చే రచింపబడింది. ఈ వ్యాసంలో, మొండిగా మాట్లాడాన్ని నివారించుటకు అనేక మార్గాలు ఉన్నాయని ఇతను చెప్తున్నారు. అందులో మొదటి మార్గం స్పష్టమైన పరిశీలన. రెండవది, ఇతరులు ఏమి ఆలోచిస్తున్నారో తెలుసుకోవటానికి, ముందు వారు ఏమనుకుంటున్నారో అవగాహన ఉండాలి. భిన్న అభిప్రాయాలు గల వ్యక్తులతో మాట్లాడటం మరియు సెలవుల యాత్రకు వెళ్ళటం వలన ఇది సాధ్యమౌతుంది. కల్పితపాత్రతో వాదించటం మరియు సంచరించటం మూడవ మార్గం. ఒకరి యొక్క ఆత్మ విలువ గౌరవం పరీక్షించుట నాల్గవ మార్గం. అహంను జయించుటకు, సువిశాల ప్రపంచంలో అణుమాత్రం కూడా ఉండని మన స్థానాన్ని గుర్తించటం. మన జీవితం క్షణకాలం మాత్రమే అని తెలుసుకోవటం.

TS Inter 2nd Year English Study Material Chapter 6 How to Avoid Foolish Opinions

b) Travel is an excellent educator. Explain with reference to Russell’s essay,”How to Avoid Foolish Opinions”.
Answer:
The thougnt provoking essay, How to Avoid Foolish Opinions” is written by Bertrand Russell. In this essay, he gives us tips on how to avoid foolish opinions and being dogmatic. Great and eminent people believe that fraud is an excellent educator. It gives travelers a lot of knowledge. It gives after the information about culture, custom, costumes, crops, climate and food habit of the people or area concerned.

It can explain all the details in elaborate manner. In the present essay, Russel also explains the importance of visiting, foreign countries. One can become aware of opinions held in different people. Talking to different people with different ideas, makes one know what other people think. Such type of observation teaches us to avoid foolish opinions. The writer personally experienced the benefits of living outside his own country.

‘మూర్ఖపు అభిప్రాయాలను ఎలా నివారించాలి’ అనే ఆలోచనను రేకెత్తించు వ్యాసం బెర్ట్రాండ్ రస్సెల్ చే రచింపబడింది. ఈ వ్యాసంలో అతను మూర్ఖపు అభిప్రాయాలను మరియు పిడివాదాన్ని ఎలా నివారించాలో మనకు చిట్కాలు ఇచ్చాడు. గొప్ప మరియు ప్రముఖ వ్యక్తులు ప్రయాణం ఒక అద్భుతమైన విద్యావేత్త అని నమ్ముతారు. ఇది ప్రయాణికులకు చాలా జ్ఞానాన్ని ఇస్తుంది. ఇది వారికి సంబంధించిన సంస్కృతి, ఆచారం, దుస్తులు పంటలు, వాతావరణం మరియు ఆహారపు అలవాట్ల గురించి సమాచారాన్ని అందిస్తుంది. ఇది అన్ని వివరాలను విస్తృతమైన పద్ధతిలో వివరించగలదు.

ప్రస్తుత వ్యాసంలో రీస్సెల్ విదేశాలను సందర్శించడం యొక్క ప్రాముఖ్యతను కూడా వివరించుట. వివిధ వ్యక్తులలో ఉన్న అభిప్రాయాల గురించి తెలుసుకోవచ్చు. విభిన్న ఆలోచనలతో ఉన్న విభిన్న వ్యక్తులతో మాట్లాడటం వలన ఇతర వ్యక్తులు ఏమనుకుంటున్నారో తెలుస్తుంది అటువంటి పరిశీలన మూర్ఖపు అభిప్రాయాలను నివారించేందుకు మనకు బోధిస్తుంది. రచయిత అని సొంత దేశం వెలుపల నివసించడం వల్ల కలిగే ప్రయోజనాలను వ్యక్తిగతంగా వివరించాడు.

c) According to Russell, what is the only way to tackle self pride?
Answer:
The thought provoking essay, How to Avoid Foolish Opinions” is written by Bertrand Russell. In this essay, he gives us tips on how to avoid foolish opinions and being dogmatic. Great and eminent people believe that travel is our excellent educator. It gives travellors a lot of knowledge. It gives them the information about culture, custom, costumes, crops, climate and food habit of the people or are concerned.

It can explain all the details in elaborate manner. In the present essay, Russell also explains the importance of visiting. Foreign countries once can become aware off opinions held in different people. Talking to different people with different ideas makes one know what other people think. Such type of observation teaches us to avoid foolish opinions. The wroter personally expressing

‘మూర్ఖపు అభిప్రాయాలను ఎలా నివారించాలి’ అనే ఆలోచనను రేకెత్తించు వ్యాసం బెర్ట్రాండ్ రస్సెల్ చే రచింపబడింది. ఈ వ్యాసంలో అతను మూర్ఖపు అభిప్రాయాలను ఎలా నివారించాలో మనకు చిట్కాలు మరియు ఇంగితజ్ఞానం యొక్క ఆరోగ్యకరమైన మోతాదును సూచించాడు. ఎందుకంటే అహంను వ్యవహరించడం చాలా కష్టం. కాబట్టి, మన సూత్రాలు పిడివాదంగా ఉండకూడదు.

ఇతరుల యొక్క ఆత్మగౌరవంతో వ్యతిరేకించడం సవాలుతో కూడుకున్నరని రచయిత చెప్పాడు. ఒక చిన్న గ్రహం మీద ఎపిసోడ్ జీవితాన్ని వారికి గుర్తు చేయటమే ఏకైక పరిష్కారం. అప్పుడు, అహం చితికిపోతుంది మరియు మూర్ఖపు అభిప్రాయాలు అంతరించిపోతాయి.

d) what does bertrand Russell say about a person getting angry about a difference of opinion.
Answer:
The thought provoking essay,”How to Avoid Foolish Opinions” is written by Bertrand Russell. In this essay, he gives us tips on how to avoid foolish opinions and being dogmatic. He advises us to identify our weak points and reconsider our opinions. when we hear views opposite to our opinions. It makes us angry. It is a clear sign of something wrong with our beliefs. then we must carefully reconsider our ideas. We have to be aware of what other people think. Thus, we can avoid such problem.

‘మూర్ఖపు అభిప్రాయాలను ఎలా నివారించాలి’ అనే ఆలోచనను రేకెత్తించు వ్యాసం బెర్ట్రాండ్ రస్సెల్ చే రచింపబడింది. ఈ వ్యాసంలో అతను మూర్ఖపు అభిప్రాయాలను ఎలా నివారించాలో మనకు చిట్కాలు ఇచ్చాడు. సున్నిత పరిశీలన ద్వారా విషయాలను ధృవీకరించుకోవడం మంచిదని సలహాఇస్తున్నాడు. మన బలహీనతలను గుర్తించి మన ఆలోచనలను పునఃపరిశీలించడం మంచిదని మనకు సలహా ఇస్తున్నాడు.

మన నమ్మకాలకు వ్యతిరేక అభిప్రాయాలు విన్నప్పుడు మనకు కోపం వచ్చిందంటే అది మన లోపభూయిష్ట ఆలోచనకు స్పష్ట సంకేతం. అప్పుడు మన అభిప్రాయాలను మనం జాగ్రత్తగా తిరిగి పరీక్షించుకోవాలి. ఇతరులు ఏమి ఆలోచిస్తున్నారో మనం అవగాహన కలిగి ఉండాలి. అలా అలాంటి సమస్యను మాత్రం నివారించగలము.

How to Avoid Foolish Opinions Summary in English

About Author

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Bertrand Arthur William Russell, 3rd Earl Russell OM FRS[66] (18 May 1872 February 1970) was a British philosopher, logician, and public intellectual.

Few of his notable works in English:

1896. German Social Democracy. London: Longmans, Green
1903. A Free man’s worship, and other essays.
1905. On Denoting, Mind, Vol. 14. ISSN 0026-4423. Basil Blackwell
1910. Philosophical Essays. London: Longmans, Green
1916. Why Men Fight. New York: The Century Co
1927. The Analysis of Matter. London: Kegan Paul, Trench, Trubner
1969. Dear Bertrand Russell… A Selection of his Correspondence with the General Public
1950-1968, edited by Barry Feinberg and Ronald Kasrils. London: George Allen and Unwin

Bertrand Russell’s essay, “How to Avoid Foolish Opinions” is a thought-provoking essay the author talks about people’s opinions and how we might avoid the foolish ones. He tells us to keep our beliefs in check, and elaborates about controversies and customs. He also mentions things that affect people’s opinions, such as self-esteem and Foolish opinions and how we might get past them.

To avoid foolish opinions, no superhuman genius is required. A few simple rules will keep us from silly error. If the matter can be settled by observation, we must make the observation ourselves. Aristotle could have avoided thinking, women have fever teeth than man. If he told Mrs. Aristotle to keep her mouth open while he counted, thinking that we know something when in fact we don’t is a big mistake. Many matters are less easily brought to the test of experience. If an opinion opposite to our own makes us angry, it is sign that we actually have no good reason for our opinion. If someone has very stupid and wrong opinions, we feel pity rather than anger.

The worst controversies are about matters which have no good evidence either way. A good way of ridding ourselves of certain kinds of rigid beliefs is to become aware of opinions held in social circles different from our own. If we cannot travel, we should find people with whom we disagree and read newspapers belonging to a party that is not ours. But becoming aware of foreign customs does not always have benefits.

For those with good imagination, it is a good plan to imagine an argument with a person having a different opinion. Mahatma Gandhi disliked railways and stem boats and machinery and would have liked to undo the industrial revolution. In western countries most people take the advantage of modern technology for granted. So it is rare to find someone with this opinion. But to make sure we are right, disagreeing with this opinion, it is a good plan to test the arguments by considering what Gandhi might have said to them.

We should be very wary of opinions that halter our self-steem. Both man and women are very convinced of the excellence of their own “Sex”. there is abundant evidence on both sides. The question actually can not be solved, but self-esteem hides this from most people. All people are convinced that their own nation is the best. Again there is no right answer to the question about the best nation. The only way of dealing with this general human vanity is to remind ourselves that man is a short episode in the life of small planet in a little corner of universe, and other parts of cosmos may contain beings much superior.

We must avoid blindly believing anything without concrete reason. We should confirm the things that we can with observation and hear out the opinions of other people regarding controversial topics. Self-esteem greatly affects opinion but it can be checked by remaining ourselves of our place in the universe.

TS Inter 2nd Year English Study Material Chapter 6 How to Avoid Foolish Opinions

How to Avoid Foolish Opinions Summary in Telugu

Note: This summary is only meant for Lesson Reference, not for examination purpose

బెర్ట్రాండ్ రస్సెల్ యొక్క వ్యాసం “మూర్ఖపు అభిప్రాయాలను ఎలా నివారించాలి” అనేది ఆలోచనను రేకెత్తించే వ్యాసం. రచయిత ప్రజల అభిప్రాయాల గురించి మరియు మనం తెలివితక్కువ వాటిని ఎలా నివారించవచ్చో మాట్లాడతాడు. అతను మన నమ్మకాలను అదుపులో ఉంచుకోమని చెప్పాడు మరియు వివాదాల గురించి మరియు ఆచారాల గురించి వివరిస్తాడు. ఆత్మగౌరవం మరియు మూర్ఖపు అభిప్రాయాలు మరియు మనం వాటిని ఎలా అధిగమించగలం వాటి వ్యక్తుల అభిప్రాయాలను ప్రభావితం చేసే విషయాలను కూడా అతను పేర్కొన్నాడు.

మూర్ఖపు అభిప్రాయాలను నివారించడానికి, మానవాతీత మేధాశక్తి అవసరం లేదు. కొన్ని సాధారణ నియమాలు చిన్న వెర్రి లోపం నుండి మనల్ని కాపాడతాయి. పరిశీలన ద్వారా సమస్యను పరిష్కరించగల్గితే మనమే పరిశీలన చేసుకోవాలి. అరిస్టాటిల్ తాను లెక్కపెట్టే సమయంలో నోరు తెరచి ఉంచమని మిసెస్ అరిస్టాటిల్కు చెబితే పురుషులకంటే స్త్రీలకు దంతాలు తక్కువగా ఉంటాయని భావించకుండా ఉంటాడు. వాస్తవానికి మనకు తెలియనప్పుడు మనకు తెలుసు అని ఆలోచించడం పెద్ద తప్పు. చాలా విషయాలు అనుభవం యొక్క పరీక్షకూ చాలా సులభంగా తీసుకోబడతాయి. మన అభిప్రాయానిక వ్యతిరేఖమైన అభిప్రాయం మనకు కోపం తెప్పిస్తే అది మన అభిప్రాయానికి సరైన కారణం లేదని సూచిస్తుంది. ఎవరైనా చాలా తెలివితక్కువ మరియు తప్పుడు అభిప్రాయాలను కలిగి ఉంటే, కోపం కంటే మనం జాలి చూపుతాము.

మంచి సాక్ష్యాలు లేని విషయాలు గురించి చెత్త వివాదాలు ఉన్నాయి. కొన్ని రకాల మొండి నమ్మకాలను వదిలించుకోవడానికి ఒక మంచి మార్గం, మన స్వంత అభిప్రాయాల కంటే భిన్నమైన సామాజిక వర్గాల్లో ఉన్న అభిప్రాయాలు గురించి తెలుసుకోవడం. మనం ప్రయాణం చేయలేకపోతే, మనం విభేదించే వ్యక్తులను కనుగొని, మనది కాని పార్టీకి సంబంధించిన వార్తాపత్రికలను చదవాలి. కానీ విదేశీ ఆచారాల గురించి తెలుసుకోవడం ఎల్లప్పుడూ ప్రయోజనాలను కలిగిఉండదు.

మంచి ఊహ ఉన్న వారికి భిన్నమైన అభిప్రాయాన్ని కలిగిఉన్న వ్యక్తితో వాదనను ఊహించుకోవడం మంచి ప్రణాళిక. మహాత్మాగాంధీ రైల్వేలుస్టీమ్ బోట్లను మరియు యంత్రాలను ఇష్టపడలేదు. మరియు పారిశ్రామిక విప్లవాన్ని రద్దుచేయడానికి ఇష్టపడేవారు. పాశ్చాత్య దేశాలలో చాలా మంది ప్రజలు ఆధునిక సాంకేతిక పరిజ్ఞానాన్ని సద్వినియోగం చేసుకుంటారు. కాబట్టి ఈ అభిప్రాయం ఉన్నవారు చాలా అరుదు. కానీ, మనము ఈ అభిప్రాయంతో ఏకీభవిస్తున్నామని నిర్ధారించుకోవడానికి, గాంధీ వారితో ఏమి చెప్పారో పరిశీలించడం ద్వారా వాదనలను పరీక్షించడం మంచి ప్రణాళిక.

మన ఆత్మగౌరవాన్ని దెబ్బతీసే అభిప్రాయాల పట్ల మనం చాలా జాగ్రత్తగా ఉండాలి. స్త్రీ, పురుషులు ఇరువురు వారివారి అహంపట్ల అనుకూలంగా ఉంటారు. ఇరువురి పట్ల ధారళమైన సాక్ష్యం ఉంది. వాస్తవానికి ఈ ప్రశ్న పరిష్కరించలేము. కానీ, ఆత్మగౌరవం ఈ విషయాన్ని చాలా మంది నుండి దాస్తుంది. అందరూ వారి జాతే గొప్ప అని ఒప్పించబడతారు. గొప్ప జాతి గురించి సరైన సమాధానం లేదు. ఇందులో నుంచి బయటపడే మార్గం విశాలవిశ్వం యొక్క విస్తృతితో పోల్చుకున్నప్పుడు ఆణుమాత్రం కూడా ఉండదని మన స్థానాన్ని గుర్తించడం అహం చితికిపోతుంది. మూర్ఖపు అభిప్రాయాలు అంతరించిపోతాయి.

సరైన కారణం లేకుండా ఏదైనా గుడ్డిగా నమ్మటాన్ని మనం నివారించాలి. మన పరిశీలన ద్వారా మనం చేయు విషయాలను ఋజువు చేసుకోవాలి మరియు ఎదైనా సమస్యలకు సంబంధించి ఇతరుల అభిప్రాయాల్ని తెలుసుకోవాలి. ఆత్మగౌరవం దీన్ని చాలా ప్రభావితం చేస్తుంది. కానీ మనకు ఈ విశ్యంలో మన పాత్రను గుర్తుచేసుకొని వీటి పరిశీలన చేసుకోగలగాలి.

TS Inter 2nd Year English Study Material Chapter 6 How to Avoid Foolish Opinions

How to Avoid Foolish Opinions Summary in Hindi

Note: This summary is only meant for Lesson Reference, not for examination purpose

बट्रेंड रसेल निबंध ‘मूर्खतापूर्ण विचारों से कैसे बचे एक विचारोत्तेजक निबंध है । हम मुर्खों से कैसे बच सकते हैं। वे हमें अपने विश्वासों को नियंत्रण में रखने के लिए कहते हैं और विवादों – रीति-रिवाजों के बारे में विस्तार से बताते हैं । वे उन चीजों का भी उल्लेख करते हैं जो लोगों की राय को प्रभावित करती हैं, जैसे कि आत्मसम्मान और मूर्खतापूर्ण राय और हम उनसे कैसे पार सकते हैं ।

मूर्खतापूर्ण विचारों से बचने के लिए किसी अतिमानवीय प्रतिभा की आवश्यकता नहीं है । कुछ सरल नियम हमें मूर्खतापूर्ण त्रुटि से बचाते हैं । यदि मामले को अवलोकन से सुलभाया जा सकता है, तो हमें स्वयं अवलोकन करना चाहिए । अरस्तू ने यह नहीं सोचा होगा कि महिलाओं के दाँत पुरुषों की तुलना में कम होते हैं यदि उन्होंने श्रीमती अरस्तू से कहा कि जब वह गिनों तो अपना मुँह खुला रखें । यह सोचना एक बड़ी भूल है कि हम जानते हैं कि वास्तव में हम कब नहीं जानते | कई चीजें बहुत आसानी से अनुभव की परीक्षा के अधीन होती हैं, अगर वृभारी विरोधी राय हमें गुस्सा दिलाती है, तो इसका मतलब है कि हमारी राय जायज नहीं है । अगर किसी के पास बहुत बेवकूफ और गलत विचार है । हम क्रोध के बजाय दया दिखाते हैं ।

सबसे खराब विवाद उन मामलों के बारे में हैं, जिन के पास किसी भी तरह से कोई अच्छा सबूत नहीं है । कुछ प्रकार के कठोर विश्वासों से खुद को मुक्त कनने का एक अच्छा तरीका सामाजिक मंडल में हमारे अपने से अलग राय के बारे में जागरूक होना है । अगर हम यात्रा नहीं कर सकते हैं, तो हम ऐसे लोगों को ढूँढना चाहिए जिन से हमें असमत है और उस पार्टी से संबंधित समाचार पत्र पढ़ना चाहिए, जो हमारी नहीं है । लेकिन विदेशी रीति-रिवाजों से अवगत होने से हमेशा लाभ नहीं होता है ।

अच्छी कल्पना वाले लोगों के लिए, एक अलग राय रखनेवाले व्यक्ति के साथ तर्क की कल्पना करना एक अच्छी योजना है। महात्मा गाँधी रेलवे और स्टीम बोट और मशीनरी को नापसंद करते थे और औद्योगिक क्रांति को पूर्ववत् करना पसंद करते थे । पश्चिमी देशों में ज्यादावर लोग आधुनिक तकनीक का फायदा उठा लेते हैं । तो इस राय होनेवाले व्यक्ति बहुत कम हैं । लेकिन यह सुनिश्चित करने के लिए कि हम इस राय से सही सहमत हैं, गाँधीजी ने दूसरों से क्या कहा होगा, इसपर विचार करके तर्कों का परीक्षण करता एक अच्छी योजना है ।

हमें उन विचारों से बहुत सावधान रहना चाहिए जो खयं की उत्कृष्टता के प्रति अश्वस्त हों, सेक्स के दोनों पक्षों में प्रचुर प्रमाण हैं । प्रश्न वास्तव में हल नहीं किया जा सकता है । लेकिन आत्मसम्मान इसे ज्यादावर लोगों से छुपाता है । सभी लोग आश्वस्त हैं कि उनका अपना राष्ट्र सर्वश्रेष्ठ है । फिर से सर्वश्रेष्ठ राष्ट्र रै प्रश्न का कोई सही उत्तर नहीं है । इस सामान्य मानवीय घमड से निपटाने का एकमात्र तरीका यह है कि हम खुद को याद दिलाएँ कि मनुष्य ब्रह्मांड के एक छोटे से कोने में छोटे ग्रह के जीवन में एक छोटी सी घटना है और लागत के अन्य हिस्सों में प्राणी बहुत बेहतर हो सकते हैं ।

हमें बिना ठोस कारण के किसी भी बात पर आँख मूँधकर विश्वास करने से बरना चाहिए । हमें उन चीजों की पहचान करनी चाहिए जो हम अवलोकन के साथ कर सकते हैं और विवादास्पद् विषयों के बारे में उनके लोगों की राय सुन सकते हैं, आत्म सम्मान राय को बहुत प्रभावित करता है । लेकिन ब्रह्मांड में अपने स्थान से खुद को बचाकर इसकी जाँच की जा सकती है ।

Meanings and Explanations

fatal (adj) / (ఫె ఇట్ ల్)/ ‘feɪ.təl/ : causing failure or disaster, వైఫల్యాన్ని, వైపరీత్యాన్ని కలిగించగల, विफलता या आपदा के कारण होना

prone (adj)/(ప్రఉన్)/prəʊn : liable; likely to do something bad; తప్పు చేసే లక్షణము, అవకాశము కల, करने की भावना

hedgehogs (n,pl) /(హెడ్జ్ హొగ్ జ్)/ ‘hedʒ.hɒgz : small brown nocturnal animals with needle like spikes on their backs,
చిన్న గోధుమ ఛాయలో ఉండే నిశాచర జంతువులు, వీపుపై సూదుల వంటి తీగలు ఉంటాయి
छोटे भूरे रंग के निशाचर जानवर जिनकी पीठ पर सुई की तरह कीले होती हैं ।

beetles (n, pl) / (బీట్ ల్ స్) /’bi.təls : large, black insects with hard case over their backs, వీపుపై పెంకులాంటి కప్పు ఉండే పెద్ద, నల్లటి కీటకములు, पीठ पर कठोर डिब्बेनाले बड़े, काले कीड़े निश्चित रूप से कहना

commit (v) / (కమిట్)//kə’mɪt : say definitely, give an opinion, ఖచ్చితమని పెప్పుట, అభిప్రాయము ప్రకటించుట, राय देना

unicorn (n,pl) / (యూనికో(ర్)న్)// ‘ju:.nɪ.kɔ:n : an imaginary white horse like animal with a long horn, పొడవైన కొమ్ము కల గుర్రంలాంటి తెల్ల కల్పిత జంతువు, एक लंबे सींग वाला एक काल्पनिक सफेद घोड़ा जैसा जानवर; लंबे सींगवाले घोड़े की तरह एक सफेद पौराणिक जानवर

salamanders (n,pl) / (స్యాలమండ(ర్)జ్) / ‘sæl.ə.mæn.dər : animals like lizards with short legs and long tails,
పొట్టి కాళ్ళు, పొడుగుతోక ఉండే బల్లుల లాంటి జంతువులు, छोटे पैर और लंबी पूँछ वाले छिपकली जैसे जानवर

dogmatic (n) /(డొ గ్ మ్యాటిక్)/ dɒg’mæt.ɪk : being certain that one’s beliefs are right and others should accept them without considering other opinions or evidence ఇతరుల అభిప్రాయాలు, ఆధారాలు పట్టించుకోకుండా తమ నమ్మకమే సరియైనది, ఇతరులు దానిని ఆమోదించాలి అని ఖచ్చితంగా అనుకునే
दूसरों की राय या आधार को ध्यान में रखे बिना, निश्चित होना कि किसी के विश्वास सही हैं और दूसरों को उन्हें स्वीकार करना चाहिए ।

passionate (adj) / (ప్యాషనట్) / ‘pæʃ.ən.ət /: strong opinions; మంచి ఉత్సాహము కల, उत्साह की तीव्र भावनाओं का होना या दिखाना

savage (adj) / (స్యావిజ్)/’sæv.ɪdʒ / : causing great harm, అత్యంత హానికర, बहुत नुकसान पहुँचा रहा हैं

persecutions(n) /(ప (ర్)సి క్యూషన్) /p3:.sɪ’kju:ʃən : annoying others deliberately all the time: అన్ని వేళలా ఉద్దేశ్య పూర్వకంగా చికాకు పెట్టటం, हर समय जान बूझकर परेशान करना

theology (n)/(థియోలజి)/ θi’ɒl.ə.dʒi : the study of religion, మత, దైవ అధ్యయన శాస్త్రము, धर्म का अध्ययन

warrants (v) /(వొరంట్ స్) / ‘wɒr.ənt : Justifies, సమర్థిస్తున్న, సరియైనదని తెలుపు, समर्थ करना, सही है कहना

diminishing (v+ing-adj)/(డిమినిషింగ్)/dɪ’mɪn.ɪʃın : decreasing, తగ్గుతున్న, हासमान / घटना

intensity (n)/(ఇంటెన్స టి)/ ɪn’ten.sə.ti : huge quantity; great degree, పెద్ద మొత్తం, ఎక్కువ స్థాయి, तीव्रता / बड़ीमात्रा

TS Inter 2nd Year English Study Material Chapter 6 How to Avoid Foolish Opinions

insular (adj)/(ఇన్ స్య ల(ర్)) /in.sjə.lər : only interested in one’s own opinions, తమ అభిప్రాయాలపట్ల మాత్రమే ఆసక్తిగల, केवल अपनी राय में दिल चस्पी है।

prejudice (n)/(ప్రెజుడిస్)/ ‘predʒ.ə.dɪs : an unreasonable liking or dislike without sensible base, దురభిప్రాయములు, ఆధార రహిత ఆసక్తి లేదా వ్యతిరేకత, युक्तियुक्त आधार के बिना अकारण और अनुचित प संद या नापसंद

perverse (adj)/(ప(ర్)వ(ర్)స్) / pə’v3:s : deliberately and determinedly doing what others think is wrong, ఇతరులు తప్పు అని భావించేదానిని కావాలని గట్టి పట్టుదలతో చేసే, जानबूझकर और निर्णायक रूप से वही करना जो दूसरे गलत समझते हैं ।

reflection (n) /(రిఫ్లెక్షన్)/ ri’flek.ʃən : careful thought, పరిశీలించి చేసిన ఆలోచన, सावधान विचार

Manchus (n prop); (మ్యాంసూచ్)/ mæn’tʃʊ : people who are originally living in Manchuria, who formed the last imperial dynasty in China (1644-1912) ; స్వతహాగా మంచూరియ దేశీయులు, 1644 నుండి 1912 వరకు చైనాను పాలించిన చివరి సామ్రాజ్యా ధినేతలు, सावधान विचार लोग जो मूल रूप मे मंचूरिया में रह रहे हैं, जिन्होंने चीन में अंतिम राजवंश का गठन किया (1644-1912)

pigtail (n)/(పిగ్ టె ఇల్) / ‘pɪg.teɪl : : hair bunched into a plait, జడ, बालों को एक चोटी के रूप में बाँधा गया

dominion (డమినిఅన్)/ də’mɪn.jən : authority to rule, పాలనాధికారము, शासन करने का अधिकार

deplores (v) /(డిప్లో(ర్)జ్)/ dɪ’plɔ:r : criticises strongly and publicly, బహిరంగంగా, తీవ్రంగా విమర్శించుట, कड़ी और खुलेआम आलोचना करना

refutation (n) /(రెఫ్యు టె ఇసన్) / ref.ju’teɪ.ʃən : proof that something is wrong, ఇతరులకు విసుగు కలిగేంచేంతగా అతి విశ్వాసముతో, ( पुनरावृत्ति) प्रमाण है कि यह गलत है

cocksure (adj)/(కొక్ షు ఆ(ర్))/kɒk’∫ɔ:r : confident that annoys, ఇతరులకు విసుగు కలిగేంచేంతగా అతి విశ్వాసముతో, दूसरों को परेशान करने की हद तक अति आविश्वास से

abundant (adj) /(అబండంట్)/o’bʌn.dənt : plentiful, పెద్ద మొత్తాలలో ఉన్న, बड़ी मात्रा में

inherently (adv)/(ఇన్ హి అరంట్లి)/ ɪn’her.ənt.li : in a way as a natural part of something, అంతర్లీనంగా, సహజంగా భాగమై ఉన్న, एक तरह से किसी चीज के प्राकृतिक भाग के रूप में

conceals (v)/(కన్ సీల్ జ్)/ kən’si:l : hides, wepe, छुपाना

persuaded (v,pt)/(ప(ర్)స్వె ఇడిడ్) / pə’sweɪd : made some one believe something is correct సరియైనది అని నమ్మేలా చేసేను,
किसी को विश्वास दिलाया कि कुछ सही है

trivial (adj) /(ట్రివిఆల్)/ ‘trɪv.i.əl : not important, ప్రాధాన్యత లేని, तुच्छ, महत्वपूर्ण नहीं

demonstrably (డిమొన్ స్ట్రబ్ లి)/ dɪ’mon.strə.bli : in a way that can be proved, నిరూపణకు వీలైన రీతిలో, एक तरह से जिसे साबित किया जा सकता है, एक सत्यापन योग्य तरीके से

aught (n)/(ఓట్)/ɔ:t/ : the smallest part, అతి చిన్న భాగము, सबसे छोटा हिस्सा

reflection (n) /(రిఫ్లెక్షన్) / rɪflek.ʃən : careful thought, పరిశీలించి చేసిన ఆలోచన, सावधान विचार

Manchus (n-prop)/(మ్యాంచూస్)/mæn’tʃʊ : people who are originally living in Manchuria, who formed the last imperial dynasty in China, (1644 – 1912) స్వతహాగా మంచూరియ దేశీయులు, 1644 నుండి 1912 వరకు చైనాను పాలించిన చివరి సామ్రాజ్యా ధినేతలు
सावधान विचार लोग जो मूल रूप मे मंचूरिया में रह रहे हैं, जिन्होंने चीन में अंतिम राजवंश का गठन किया (1644-1912)

pigtail (n) ⁄(పిగ్ టె ఇల్) / ‘pɪg.teɪl : hair bunched into a plait, జడ, बालों को एक चोटी के रूप में बाँधा गया

dominion (n) /(డమినిఅన్)/ də’mɪn.jən : authority to rule, పాలనాధికారము, शासन करने का अधिकार

deplores (v) /(డిప్లో(ర్)జ్)/dɪ’plɔ:r : criticises strongly and publicly, బహిరంగంగా, తీవ్రంగా విమర్శించుట, कड़ी और खुलेआम आलोचना करना

refutation (n) /(రెఫ్యు టెఇసన్)/ ref.ju’teı.ʃən : proof that something is wrong, ఇతరులకు విసుగు కలిగేంచేంతగా అతి విశ్వాసముతో, ( पुनरावृत्ति) प्रमाण है कि यह गलत है

cocksure (adj)/(కొక్ షు (అర్))/kɒk’ʃɔ:r : confident that annoys, ఇతరులకు విసుగు కలిగేంచేంతగా అతి విశ్వాసముతో, दूसरों को परेशान करने की हद तक अति आविश्वास से

abundant (adj) (అబండంట్)/ ɔ’bʌn.dənt : plentiful, పెద్ద మొత్తాలలో ఉన్న, बड़ी मात्रा में

inherently (adv)/(ఇన్ హిఅరంట్లి)/ ɪn’her.ənt.li : in a way as a natural part of something, అంతర్లీనంగా, సహజంగా భాగమై ఉన్న, एक तरह से किसी चीज के प्राकृतिक भाग के रूप में

conceals (v)/(కన్ సీల్ జ్)/ kən’si:l : hides, దాచిపెట్టుట, छुपाना

TS Inter 2nd Year English Study Material Chapter 6 How to Avoid Foolish Opinions

persuaded (v,pt)/(ప(ర్)స్వెఇడిడ్) / pə’sweɪd : made some one believe something is correct సరియైనది అని నమ్మేలా చేసేను, किसी को विश्वास दिलाया कि कुछ सही है

trivial (adj) (ట్రివిఆల్)/’trɪv.i.əl : not important, ప్రాధాన్యత లేని, तुच्छ, महत्वपूर्ण नहीं

(demonstrably /(డిమొన్ స్ట్రబ్ లి)/ dɪ’mɒn.strə.bli : in a way that can be proved, నిరూపణకు వీలైన రీతిలో, एक तरह से जिसे साबित किया जा सकता है, एक सत्यापन योग्य तरीके से

aught (n) : (ఓట్) : ɔ:t : the smallest part, అతి చిన్న భాగము, सबसे छोटा हिस्सा

TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

Telangana TSBIE TS Inter 1st Year Physics Study Material 14th Lesson Kinetic Theory Textbook Questions and Answers.

TS Inter 1st Year Physics Study Material 14th Lesson Kinetic Theory

Very Short Answer Type Questions

Question 1.
State the law of equipartition of energy. [TS Mar. ’18, ’16]
Answer:
Law of equipartition of energy :
The total energy of a gas is equally distributed in all possible energy modes, with each mode having an average energy equal to \(\frac{1}{2}\)KBT. This is known as law of “equipartition of energy”.

Question 2.
Define mean free path. [AP Mar. ’19, ’18, ’17, ’15, May ’18; TS May ’18, Mar. ’17, ’15]
Answer:
Mean free path :
The average distance by can travel an atom or a molecule without colliding is called “mean free path”.
l = \(\frac{1}{\sqrt{2} \pi n d^2}\)
Where n is the number of molecules per unit volume and d is the diameter of the molecule.
(or)
Mean tee path of gas molecules is the aver-age distance covered by a molecule between two successive collisions.

Question 3.
How does kinetic theory justify Avogadro’s hypothesis and show that Avogadro Number in different gases is same?
Answer:
Avogadro’s law or Avogadro’s hypothesis:
It states that “equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules

Consider two gases distinguished by subscripts 1 and 2.

Let
m1 and m2 = masses of molecules of 1 and 2 respectively
N1 and N2 = no. of molecules of 1 and 2 respectively
P1 and P2 = pressures of 1 and 2 respectively
V1 and V2 = volumes of 1 and 2 respectively
T1 and T2 = absolute temperatures of 1 and 2 respectively
\(\overline{\mathrm{V}_1^2}\) and \(\overline{\mathrm{V}_2^2}\) = mean square velocities of molecules of 1 and 2 respectively.
According to kinetic theory of gases,
P1V1 = \(\frac{1}{3}\) m1N1 \(\overline{\mathrm{V}_1^2}\) and P2V2 = \(\frac{1}{3}\) m2N2 \(\overline{\mathrm{V}_2^2}\)
For equal volumes at the same pressure,
P1V1 = P2 V2
∴ \(\frac{1}{3}\) m1N1 \(\overline{\mathrm{V}_1^2}\) = \(\frac{1}{3}\) m2N2 \(\overline{\mathrm{V}_2^2}\) ……………. (1)

If the gases are at the same temperature, the average kinetic energies of their molecules are equal.
∴ equation (1) becomes : N1 = N2

Hence, equal volumes of all gases under similar conditions of temperature and pressure have the same number of molecules.

TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

Question 4.
What are the units and dimensions of a specific gas constant? [AP Mar. ’14]
Answer:
For specific gas constant unit: Joule/Kelvin
Dimensional formula: r = \(\frac{PV}{T}\) =ML² T-2 K-1.

Question 5.
When does a real gas behave like an ideal gas? [AP Mar. ’19, ’14; TS Mar. ’19, ’16, May ’18, ’16; June ’15]
Answer:
No real gas is perfect or ideal. At extremely low pressures and high temperatures, some real gases (like H2, O2, N2, He, etc.) obey the gas laws to a fair degree of accuracy and hence, behave as nearly ideal gas.

Question 6.
State Boyle’s Law and Charles Law. [AP Mar. 18, June 15; TS Mar. 15, May. 16]
Answer:
Boyle’s Law :
At constant temperature, the volume (V) of a given mass of a gas is inversely proportional to its pressure (P).
∴ V ∝ \(\frac{1}{P}\) or PV = constant = K.

Charles Law :
At constant pressure, the volume (V) of a given mass of a gas is directly proportional to its absolute temperature (T).
∴ V ∝ T or \(\frac{V}{T}\) = K (constant)

Question 7.
State Dalton’s law of partial pressures. [TS Mar. ’18, ’17; AP Mar. ’16, ’14]
Answer:
Dalton’s law of partial pressures :
For a mixture of non interacting ideal gases at same temperature and volume total pressure in the vessel is the sum of partial pressures of individual gases.
i.e., P = P1 + P2 + ………… where P is total pressure
P1, P2 ……….. etc, are individual pressures of each gas.

Question 8.
Define absorptive power of a body. What is the absorptive power of a perfect black body? [AP May ’14]
Answer:
Absorptive power of a body is defined as the ratio of energy absorbed by the body within the wave length range of λ and λ + dλ to the total energy flux following on the body.
Absorptive power,
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 1

Question 9.
Pressure of an ideal gas in container is independent of shape of the container – explain. [AP June ’15]
Answer:
Pressure exerted by a gas is due to continuous bombardment of gaseous molecules with the walls of the container. During each collision, certain momentum is transferred to the walls of the container and this transfer is independent of its shapes because area A and time interval ∆t do not appear in the final formula. Hence, pressure of an ideal gas in a container is independent of the shape of the container.

TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

Question 10.
Explain the concept of degrees of freedom for molecules of a gas.
Answer:
The number of degrees of freedom of a dynamical system is defined as the total number of co-ordinates or independent quantities required to describe completely the position and configuration of the system.

Question 11.
What is the expression between the pressure and kinetic energy of a gas molecule?
Answer:
The pressure exerted by an ideal gas is numerically equal to \(\frac{2}{3}\) rd of the mean kinetic energy of translation per unit volume of gas.
P = \(\frac{2}{3}\)E

Question 12.
The absolute temperature of a gas is increased 3 times. What will be the increase in rms velocity of the gas molecule? [TS Mar. ’19, June ’15]
Answer:
The relation between r.m.s. velocity and absolute temperature of a gas is C ∝ √T.
Therefore, the r.m.s. velocity becomes √3 C.
Hence, increase in r.m.s. velocity
√3 C – C = 0.732 C = 73.2 %

Short Answer Questions

Question 1.
Explain the kinetic interpretation of Temperature.
Answer:
According to kinetic theory of gases, the pressure P exerted by one mole of an ideal gas is given by
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 2
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 3

Thus, the square root of the absolute temperature of an ideal gas is directly proportional to root mean square velocity of its molecules.
Also, from eqn. (1)
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 4
But, \(\frac{1}{2}\)mc² is average translational K.E per molecule of a gas.

Therefore, average kinetic energy of translation per molecule of a gas is directly proportional to the absolute temperature of the gas.

TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

Question 2.
How can specific heat capacity of monoatomic, diatomic, and polyatomic gases be explained on the basis of Law of equipartition of Energy? [AP May ’17. ’16; Mar. ’13]
Answer:
From law of equipartition of energy, energy per each degree of freedom is \(\frac{1}{2}\) KBT Per atom or molecule.

1) Monoatomic gas has three degrees of freedom.
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 5
But molar specific heat at constant volume Cv = \(\frac{dU}{dT}\)
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 6

2) A diatomic gas has 3 translational and two rotational degrees of freedom.
∴ Kinetic energy per molecule U1 = 5.\(\frac{1}{2}\)KB.T
For one gram mole total energy U = \(\frac{5}{2}\)KB.T.NA
Molar-specific heat at constant volume
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 7

3) A polyatomic gas has three translational, three rotational, and at least one vibrational degrees of freedom.
∴ Kinetic energy per molecule
U1 = 3.\(\frac{1}{2}\)KB.T + 3.\(\frac{1}{2}\)KB.T + f = (3 + f)KBT
f = Number of vibrational degrees of freedom

Kinetic energy
= per gram mole of molecules
= U1NA = U = (3 + f)KB.NA.T = (3 + f)RT
Molar specific heat Cv = \(\frac{dU}{dT}\) = (3 + f) R
∴ CP = (u + f)R
∴ For polyatomic gases Cv = (3 + f) R & CP = (4 + f)R

Question 3.
Explain the concept of absolute zero of temperature on the basis of kinetic theory.
Answer:
According to kinetic theory of gases, the pressure ‘P* exerted by one mole of an ideal gas is,
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 8
Absolute zero:
When T = 0, from equation (1), C = 0.

Hence, absolute zero of temperature may be defined as that temperature at which the root mean square velocity (C) of the gas molecules reduces to zero.

It means, molecular motion ceases at absolute zero.

This definition holds in cases of an ideal gas or perfect gas.

Question 4.
Prove that the average kinetic energy of a molecule of an ideal gas is directly proportional to the absolute temperature of the gas.
Answer:
According to kinetic theory of gases, the pressure ‘P’ exerted by one mole of an ideal gas is
P = \(\frac{1}{3}\) ρC² where p is density of the gas.
∴ P = \(\frac{1}{3}\frac{M}{V}\)C² ⇒ PV = \(\frac{1}{3}\)MC²
But PV = RT for one mole of ideal gas
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 9

But, \(\frac{1}{2}\) mC² is averaSe translational kinetic energy per molecule of a gas.

Therefore, average kinetic energy of molecule of an ideal gas is directly proportional to the absolute temperature of the gas.

TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

Question 5.
Two thermally insulated vessels 1 and 2 of volumes V1 and V2 are joined with a valve and filled with air at temperatures (T1, T2) and pressures (P1, P2) respectively. If the valve joining the two vessels is opened, what will be the temperature inside the vessels at equilibrium.
Answer:
According to Standard gas equation
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 10

As the vessels are thermally insulated and no work is done, total energy remains conserved.
Therefore, \(\frac{3}{2}\) (P1V1 + P2V2) = \(\frac{3}{2}\) P(V1 + V2)
where P is resultant pressure
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 11

For mixture of two gases,
1 + µ2)RT = P(V1 + V2)
where T is resultant temperature.

Substitute equation (1) and equation (2) in the above equation, we have
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 12

This is the final equilibrium temperature of the air in the vessels.

Question 6.
What is the ratio of r.m.s speed of Oxygen and Hydrogen molecules at the same temperature?
Answer:
The r.m.s. speed of oxygen molecules at temperature T is
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 13

The r.m.s. speed of Hydrogen molecules at same temperature (T) is
CH = \(\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{H}}}}\) But molecular weight of hydrogen = 2
Ratio of r.m.s. speed of Oxygen and Hy-drogen molecules is
∴ CH = \(\sqrt{\frac{3 \mathrm{RT}}{2}}\)
Ratio of r.m.s. speed of Oxygen and Hydrogen molecules is
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 14
∴ Cghy : CH = 1 : 4

Question 7.
Four molecules of a gas have speeds 1,2,3 and 4 km/s. Find the rms speed of the gas molecule.
Answer:
Given, C1 = 1 km/s ; C2 = 2 km/s
C3 = 3 km/s ; C4 = 4 km/s

The root mean square speed of molecules is
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 15
Crms = 2.74 km/s

Question 8.
If a gas has ‘f degrees of freedom, find the ratio of Cp and Cv.
Answer:
Suppose a polyatomic gas molecule has ‘f degrees degrees of freedom (0 and ratio of Cp and of freedom.
∴ Internal energy of one gram mole of the gas is
U = f × \(\frac{1}{2}\)kBT × NA = \(\frac{f}{2}\)RT
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 16

This is the relation between number of degrees of freedom (f) and ratio of Cp and Cv & (γ)

Question 9.
Calculate the molecular K.E of 1 gram of Helium (Molecular weight 4) at 127°C. Given R = 8.31 J mol-1C-1.
Answer:
n = number of moles of Helium = \(\frac{1gm}{4gm mol^{-1}}\) = 0.25 mole
R = 8.314 J mol-1 KT-1,
T = 127°C = 127 + 273 = 400 K
∴ Kinetic energy = \(\frac{3}{2}\) nRT = \(\frac{3}{2}\) × 0.25 mol × 8.314 J mol-1 K-1 × 400K = 1247.1 J

Question 10.
When pressure increases by 2%, what is the percentage decrease in the volume of a gas, assuming Boyle’s law is obeyed?
Answer:
Let ‘P’ be the initial pressure and ‘V’ be the volume.
When pressure is increased by 2%, new pressure, P¹ = P + \(\frac{2}{100}\)P ⇒ P¹ = \(\frac{102}{100}\)P
According to Boyle’s law, PV = constant ⇒ PV = P’V’
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 17
% decrease in volume = \(\frac{2}{100}\) × 100 = 1.96%

∴ When pressure increases by 2%, the volume decreases by 1.96%.

Long Answer Questions

Question 1.
Derive an expression for the pressure of an ideal gas in a container from Kinetic Theory and hence give Kinetic Interpretation of Temperature.
Answer:
Consider a cube of side ‘l’ and an ideal gas is filled in it. Let the co-ordinates X, Y and Z will coincide with the sides of the cube. Let velocities of gas molecules along these directions are Vx, Vy and Vz. Consider a gas molecule moving along X – axis with a velocity Vx. Its motion is perpendicular to Y – Z plane. Let the gas molecule suffered elastic collision with Y – Z plane and bounced back. In this case the velocities Vy and Vz are not considered because collision is along X – direction only.
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 18

Change in momentum along X-direction is final momentum (-mVx) – Initial momentum (mVx) = -2mVx ………… (1)

Let area of one side of the cube is A. Then during the time ∆t only the molecules at a distance of Vx ∆t will collide with the walls. Let number of gas molecules in the volume AV∆t are say ‘n’. In these molecules half of the molecules will move towards the wall and remaining will move away from the wall.
∴ Amount of momentum transferred to the wall = Q
= 2mVx × number of molecules collided with wall
∴ Q = 2mVx(\(\frac{1}{2}\)nAVx∆t)

Pressure on Y – Z plane =
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 19

If behaviour of gas is isotropic that is equal in all directions then Vx = Vy = Vz
∴ Root mean square velocity of gas
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 20
∴ Pressure of gas P = \(\frac{1}{3}\)nm\(\overline{\mathrm{V^2}}\)
From Pascal’s law pressure is same throughout the container so pressure in any direction (say x, y or z) is same.

Kinetic interpretation of temperature :
From gas equation
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 21

For Ideal gas internal energy is purely kinetic energy.
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 22
But \(\frac{E}{N}=\frac{1}{2}\)mV² and \(\frac{E}{N}=\frac{3}{2}\) KBT

∴ \(\frac{E}{N}\) ∝ T is the average kinetic energy of gas molecule is proportional to absolute temperature of gas. But it does not depend on volume of container.
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 23

Additional Problems

Question 1.
Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP, Take the diameter of an oxy-gen molecule to be 3 Å.
Solution:
Here, diameter, d – 3Å,
r = \(\frac{d}{2}=\frac{3}{2}\)Å = \(\frac{3}{2}\) × 10-8
Molecular volume, V = \(\frac{4}{3}\) πr³ . N, where N is
Avogadro’s number = \(\frac{4}{3}\times\frac{22}{7}\) (1.5 × 10-8)³ × (6.023 × 1023) = 8.52 cc.
Actual volume occupied by 1 mole of oxygen at STP, V’ = 22400 cc
∴ \(\frac{V}{V’}=\frac{8.52}{22400}\) = 3.8 × 10-4 ≈ 4 × 10-4

TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

Question 2.
Molar volume is the volume occupied by 1 mole of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0°Q. Show that it is 22.4 litres.
Solution:
For one mole of an ideal gas, PV = RT
∴ V = \(\frac{RT}{P}\)
Put R = 8.31 .1 mole-1 K-1, T = 273 K,
P = 1 atmosphere = 1.013 × 105 Nm-2
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 24
= 0.0224 × 106 cc = 22400 cc = 22.4 litre.

Question 3.
An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12°C. To what volume does it grow when it reaches the surface, which is at a temperature of 35°C?
Solution:
V1 = 1.0 cm³ = 1.0 × 10-6m³;
T1 = 12°C = 12 + 273 = 285 K;
P1 = 1 atm. + h1 p g = 1.01 × 105 + 40 × 10³ × 9.8 = 493000 Pa.
When the air bubble reaches at the surface of lake, then
V2 = ? ; T2 = 35°C = 35 + 273 = 308 K ;
P2 = 1 atm. = 1.01 × 105 Pa
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 25

Question 4.
At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a he-lium gas atom at – 20°C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).
Solution:
Let C and C’ be the r.m.s. velocities of argon and a helium gas atoms at temperature TK and T’ K respectively.
Here, M = 39.9 ; M’ = 4.0 ; T = ?
T’ = -20 + 273 = 253 K
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 26
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 27

TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

Question 5.
From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm³ s-1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm³ s-1. Identify the gas.
[Hint: Use Graham’s law of diffusion: R1/R2 = (M2/ M1)1/2, where R1, R2 are diffusion rates of gases 1 and 2, and M1 and M2 their respective molecular masses. The law is a simple consequence of kinetic theory.]
Solution:
According to Graham’s law of diffusion, = \(\frac{r_1}{r_2}=\sqrt{\frac{\mathrm{M}_2}{\mathrm{M}_1}}\)
Where, r1 = diffusion rate of hydrogen = 28.7 cm³ s-1
r2 = diffusion rate of unknown gas = 7.2 cm³ s-1
M1 = molecular mass of hydrogen = 2 u
M2 = ?
TS Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 28

TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics

Telangana TSBIE TS Inter 1st Year Physics Study Material 13th Lesson Thermodynamics Textbook Questions and Answers.

TS Inter 1st Year Physics Study Material 13th Lesson Thermodynamics

Very Short Answer Type Questions

Question 1.
Define Thermal equilibrium. State the zeroth law of thermodynamics (or) How does it lead to Zeroth Law of Thermodynamics? [AP May ’13]
Answer:
Thermal Equilibrium:
Two systems are said to be in thermal equilibrium with each other if they are at same temperature.

Explanation :
If two different temperature of systems are kept in contact. Heat transfer from hot to cold body. If they are at same, then they are said to be in thermo equilibrium.

Generally at thermal equilibrium, the temperature of two systems is same. This concept leads to Zeroth law of thermodynamics.

Zeroth law of thermodynamics :
It states that if two systems say A & B are in thermal equilibrium with a third system ‘C’ separately then the two systems A and Bare also in thermal equilibrium with each other.

Question 2.
Define Calorie. What is the relation between calorie and mechanical equivalent of heat?
Answer:
Calorie:
The amount of heat required to rise the temperature of 1g of water by 1°C is called calorie.

Relation between calorie and Joule, 1 calorie = 4.2 J.

Question 3.
What thermodynamic variables can be defined by a) Zeroth Law b) First Law?
Answer:
Zeroth law refers temperature and first law refers internal energy.

Question 4.
Define specific heat capacity of the substance. On what factors does it depend?
Answer:
Specific heat capacity:
The quantity of heat required to rise the temperature of unit mass of the substance through 1 °C or IK is called the “specific heat capacity of the substance.”
S = \(\frac{dQ}{m.dT}\)
The specific heat capacity depends upon the factors like temperature and nature of the substance.

Question 5.
Define molar specific heat capacity. [AP May ’13]
Answer:
Molar specific heat capacity:
It is defined as the amount of heat required to rise the temperature of one mole of a gas through 1 °C or IK.

TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics

Question 6.
For a solid, what is the total energy of an oscillator?
Answer:
For a solid, the total energy of an oscillator can be expressed as the sum of its potential and kinetic energies.

Question 7.
Indicate the graph showing the variation of specific heat of water with temperature. What does it signify?
Answer:
The variation of specific heat of water with the temperature is as shown in the graph.
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 1

From the graph, we find that at T = 15°C, specific heat of water, S = 1 calg-1 °C-1

At T = 0°C, S = 1.008 Calg-1 °C-1, the highest and at T = 30°C, s = 0.9976 Calg-1 °C-1, the lowest and beyond 30°C, specific heat of water increases slightly with rise in temperature.

At T = 100°C, specific heat of water is 1.0057 calg-1 °C-1.

It signifies that the specific heat of water decreases with increase in temperature from 0° to 30°C and increases from 30c – 100°C,

Question 8.
Define state variables and equation of state.
Answer:
State variables:
The variables which deter mine the thermodynamic behaviour of a sys tern are called “state variables.”

If the system is a gas, then P, V, and T (for a given mass) are called state variables.

Equation of state:
The general relationship between pressure, volume, and temperature for a given mass of the system (eg., gas) is called “equation of the state.”

For n moles of an ideal gas, the equation of state is, PV = nRT.

Question 9.
Why a heat engine with 100% efficiency can never be realised in practise?
Answer:
The efficiency of heat engine, η = 1 – \(\frac{T_2}{T_1}\)

The efficiency will be 100%, or 1, if T2 = OK or T1 = ∞.

Since, both these conditions cannot be attained practically, a heat engine cannot have 100% efficiency.

Question 10.
In summer, when the valve of a bicycle tube is opened, the escaping air appears cold. Why?
Answer:
This happens due to adiabatic expansion of the air in the tube of the bicycle. Hence the air cools.

TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics

Question 11.
Why does the brake drum of an automobile get heated up while moving down at constant speed?
Answer:
When an automobile moving down with constant speed its potential energy decreases. This decrease in potential energy is converted in the form of heat energy. As a result, the break drum of an automobile get heated.

Question 12.
Can a room be cooled by leaving the door of an electric refrigerator open?
Answer:
No. When a refrigerator is working in a closed room with its door closed, it is rejecting heat from inside to the air in the room. So, temperature of room increases gradually.

When the door of refrigerator is kept open, heat rejected by the refrigerator to the room will be more than the heat taken by the refrigerator from the room. Therefore, temperature of room will increase at a slower rate compared to the first case.

Hence, a room cannot be cooled by leaving the door of an electric refrigerator open.

Question 13.
Which of the two will increase the pressure more, an adiabatic or an isothermal process, in reducing the volume to 50%?
Answer:
In an adiabatic process, no exchange of heat is allowed between the system and surroundings. Hence, the work done during reducing the volume to 50% results in the increase in the temperature of the system thereby further increase in the pressure (∵ PV = RT). In case of isothermal compression, the excess heat is exchanged with the surroundings, maintaining constant temperature. Hence the increase in pressure is only due to decrease in volume obeying Boyle’s law.

Hence adiabatic compression increases the pressure more than isothermal compression.

Question 14.
A thermos flask containing a liquid is shaken vigorously. What happens to its temperature?
Answer:
Temperature of the liquid increases, because work is done in shaking the liquid. W ∝ Q.

Question 15.
A sound wave is sent into a gas pipe. Does its internal energy change?
Answer:
Yes, the internal energy changes when a sound wave is sent into a gas pipe. Because the sum of all the energies contained in the system in equilibrium is called its internal energy.

Question 16.
How much will be the internal energy change in
i)isothermal process ii) adiabatic process
Answer:
i) In an isothermal process,
T = constant i.e., dT = 0 ∴ dU = 0
So, in a isothermal process, the internal energy does not change.

ii) In an adiabatic process,
Q = constant i.e., dQ = 0 ∴ dU = -dW ≠ 0

So, in an adiabatic process, the change in internal energy is equal to the amount of work done.

Question 17.
The coolant in a chemical or a nuclear plant should have high specific heat. Why?
Answer:
“Specific heat of a substance is the amount of heat required to raise the temperature of unit mass of the substance through 1 °C or IK”. The coolant in a chemical or nuclear plant should be able to absorb more amount of heat released from the plant. Hence, the coolant should have high specific heat.

TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics

Question 18.
Explain the following processes i) Isochoric process ii) Isobaric process
Answer:
i) Isochoric process:
The process that occurs at constant volume is called “Isochoric process” or “Isovolumic process”.
∴ dV = 0.

ii) Isobaric process :
The process that occurs at constant pressure is called “Isobaric process.”
∴ dP = 0

Short Answer Questions

Question 1.
State and explain first law of thermodynamics.
Answer:
First law of thermodynamics :
The heat energy (dQ) supplied to a system is equal to the sum of the increase in the internal energy (dU) of the system and external work done (dW) by it
i.e., dQ = dU + dW ……….. (1)

If dV is the increase in the volume under constant pressure (P) then dW = PdV.
∴ dQ = dU + PdV ………….. (2)

The importance of this law is that it defines, the thermodynamic quantity, internal energy which has a fixed value in a state.

Increase in internal energy dU = nCvdT

Where n is the number of moles of the gas. This equation helps to calculate change of internal energy of the system when the temperature change by ∆T.

Limitations of 1st law of thermody namics:

  1. It does not tell about the direction of heat flow. That is it does not specify the conditions under which a body can use the heat energy to produce the work.
  2. It does not give any information about the efficiency with which heat can be converted into work.

Question 2.
Define two principal specific heats of a gas. Which is greater and why? [TS June ’15]
Answer:
i) Specific heat of a gas at constant vok ume (Cv) :
It is defined as the amount of heat energy required to raise the temperature of one gram of gas through 1°C of 1K, when volume of the gas is kept constant.

It is measured in cal.g-1.K-1 or J.g-1. K-1,

ii) Specific heat of a gas at constant pressure (Cp) :
It is defined as the amount of heat energy required to raise the temperature of one gram of gas through 1 °C or IK, when pressure of the gas is kept constant.
It is also measured in cal. g-1.K-1 or J.g-1.K-1.

Out of the two principal specific heats of a gas, Cp > Cv. This can be justified as follows:

a) When heat is given to a gas at constant volume, it is only used in increasing the internal energy of the gas, i.e., in raising the temperature of the gas, and no heat is spent in the expansion of the gas.

b) When heat is given to a gas at constant pressure, it is spent in two ways :

  1. Part of the heat is increasing the internal energy of the gas and hence the temperature of the gas.
  2. Remaining amount of heat is used in doing work i.e., in the expansion of the gas against the external pressure.

Therefore to raise the temperature of 1 mole of a gas through 1°C or 1K, more heat is required at constant pressure (Cp) than at constant volume (Cp).

Hence, Cp > Cv.

TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics

Question 3.
Derive a relation between the two specific heat capacities of gas on the basis of first law of thermodynamics. [TS June ’15]
Answer:
Relationship between the two specific heat capacities of gas:
To derive Cp – Cv = R:
Let one gram mole of given mass of gas is enclosed within a cylinder with a frictionless air tight-piston. Let P, V be the pressure and volume of the gas at a temperature T.

i) In specific heat at constant volume :
The volume of the given mass of gas must remain constant. Hence, the piston is fixed in position AB.

Let Cv be the amount of heat energy supplied. It is utilised only to raise the temperature of the gas by 1°C.
∴ dU = CvdT
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 2

ii) In specific heat at constant pressure, the pressure must remain constant. Hence the piston is allowed to move freely. The amount of heat energy supplied is used not only to do external work but also to increase the temperature of the gas by 1°C.

In this case the piston moves forward and work is done against external pressure. Suppose by the time the piston moves from AB to CD, the temperature increases by 1 °C.

Let the work done in moving the piston through a distance ‘dl’ be dW = P dV.

The energy supplied has to increase the internal energy and to do external work.
∴ CpdT = dQ = dU + dW

From first law of thermodynamics
dQ = dU + dW
∴ CPdT = CvdT + dW
(CP – Cv) dT = dW = PdV
But work done, dW = F × S
= P × A × dl (where A is area of cross-section of the piston) = PdV
But PV = RT (for one mole of gas).

∴ dW = PdV = RdT OR (CP – Cv) dT = RdT
But change of temperature = dT = 1°C (from definition of specific heat)
∴ Cp – Cv = R

So difference of molar specific heats of the gas is equals to universal gas constant R.

Question 4.
Obtain an expression for the work done by an ideal gas during isothermal change.
Answer:
Work done during isothermal process:
Consider n mole of a perfect gas contained in a cylinder. When the piston moves through a small distance dx, then small work dW will be done by it
∴ dW = P A dx = P dV,
where ‘A’ = the area of cross-section of the piston.
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 3

Therefore, when the system goes from initial state A (P1, V1) to the final state B (P2, V2), the amount of work done,
W = \(\int_{v_1}^{v_2} P d V\) ………….. (1)
But PV = nRT (or) P = \(\frac{nRT}{V}\)
Substitute P in equation (1),
W = \(\int_{v_1}^{v_2} \frac{n R T}{V} d V\)
During an isothermal process, temperature remains constant.
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 4

All the heat supplied to the gas is used only to do work since temperature remains constant, internal energy does not change.
∴ dQ = PdV

Question 5.
Obtain an expression for the work done by an ideal gas during adiabatic change and explain.
Answer:
Work done during Adiabatic process :
Consider n mole of perfect gas contained in a cylinder having insulating walls. When piston moves through a small distance dx, then small work (dW) will be done.
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 6
∴ dW = (Pa) dx = PdV,
where a is area of cross-section of the piston. Therefore, when the system goes from initial state A (P1, V1) to the final state B ( P2, V2) the amount of work done,
W = \(\int_{v_1}^{v_2} P d V\) ………….. (1)
For an adiabatic change,
PVγ = K (a constant ) or P = KV

Substituting for P in equation (1), the work done in an adiabatic process
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 7

∴ Work done in adiabatic process,
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 8

∴ Work done in adiabatic change
W = \(\frac{nR}{(\gamma -1)}\)(T1 – T2)

Since heat is not supplied to the gas, it can do work only by expanding its internal energy. dQ = 0 = dU + PdV or PdV = – dU.
So the gas cools in adiabatic expansion.

Question 6.
Compare isothermal and an adiabatic process.
Answer:

Isothermal changesAdiabatic changes
1. Temperature (T) remains constant,
i.e., ∆T = 0
1. Heat content (Q) remains constant,
i.e., ∆Q = 0.
2. System is thermally conducting to the surroundings.2. System is thermally insulated from the surroundings.
3. The changes occur slowly.3. The changes occur suddenly.
4. Internal energy (U) remains constant,
i.e., ∆U = 0.
4. Internal energy changes, i.e., U ≠ constant
∴ ∆U ≠ 0.
5. Specific heat becomes infinite.5. Specific heat becomes zero.
6. Equation of isothermal changes is PV = constant.6. Equation of adiabatic changes is PVγ = constant
7. Slope of isothermal curve, \(\frac{dP}{dV}\) = -(P/V)7. Slope of adiabatic curve, \(\frac{dP}{dV}\) = -γ(P/V)
8. Coefficient of Isothermal elasticity; Ei = P8. Coefficient of adiabatic elasticity; Ea = γP

TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics

Question 7.
Explain the following processes
i) Cyclic process with example
ii) Non-cyclic process with example
Answer:
i) Cyclic process :
A process in which the system after passing through various stages such as change in pressure, volume and temperature etc. returns to its initial state is defined as “cyclic process”.

For a thermodynamic system the internal energy of the system depends on thermodynamic variables such as pressure, volume, temperature, etc. In cyclic process the system finally returns to the initial state and it is in thermal equilibrium with surroundings. So change in internal energy of the system dU = 0.

Hence, in a cyclic process work done is equal to energy absorbed in the cyclic process.

So for cyclic process dU = 0 and dQ = dW.

Example:
Generally, all heat engines (or) refrigerators are operated in cyclic process.

ii) Non-cyclic process:
A non-cyclic process consists of a series of changes involved do not return the system back to its initial state.

Example:
Suppose a gas with variables P1, V1, T1 is taken through a series of different states subjecting to a number of changes including isothermal expansions and compressions. In the final state, if the system does not come back to P1V1T1, then the gas is said to be undergo a non-cyclic process.

Work done in a non-cyclic process depends upon the path chosen or the series of changes involved.

Question 8.
Write a short note on Quasistatic process.
Answer:
Quasi-static process:
A Quasistatic process can be defined as an infinitesimally show process in which at each and every intermediate stage the system remains in thermal and mechanical (thermodynamic) equilibrium with the surroundings through out the entire process.
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 9

Explanation:
A non-equilibrium thermodynamic system can be treated as an idealized process in which at every stage the system is in equilibrium state.

In a thermodynamic system let the piston moves in a frictionless manner. Instead of sudden compression of piston imagine the piston moves very very slowly i.e., it appears almost static. Then the pressure inside the cylinder P + ∆P and temperature T + ∆T are almost equal to external pressure P and temperature T.

Since for extremely slow process the values of ∆P and ∆T are so small that we can treat P + ∆P = P and T + ∆T = T. Such type of process is called Quasi static process.

For Example, to take a gas from the state (P, T) to another state (P1, T1), via a quasistatic process, we change the external pressure / temperature by a very small amount and allow the system to equalise its pressure / temperature with the surroundings. Continue the process infinitely slowly till the final state (P1, T1) is attained.

A quasi-static process is a hypothetical construct. The process must be infinitely slow, should not involve large temperature differences or accelerated motion of the piston of the container.

Question 9.
Explain qualitatively the working of a heat engine.
Answer:
Heat engine :
A heat engine is a device used to convert heat energy into mechanical work.

Generally heat engines will work in a cyclic process. Heat engine consists of three important units.

1) Source :
Which is an object or system at high temperature. A heat engine will absorb heat energy Q1 from source.

2) Working substance :
Every heat engine requires a working substance to do work Generally the working substance is like steam or fuel vapour and air mixture etc. A part of heat energy of working substance is converted into mechanica work.

3) Sink :
In every heat engine heat, energy content of working substance is not converted into work totally. So some energy (Q2) is wasted or rejected by the engine. This rejected energy (Q2) is delivered to some other body or system at low temperature. This body with low temperature is called “sink”.

Efficiency of heat engine,
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 10

where
Q1 = heat energy supplied by source
Q2 = heat energy delivered to sink
Block diagram of heat engine is as shown.
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 11

Long Answer Questions

Question 1.
Explain reversible and irreversible processes. Describe the working of Carnot engine. Obtain an expression for the efficiency. [AP Mar. ’18, ’17, ’16, ’14; May 18. 17, ’16; TS Mar. ’19, ’17, 15, May ’17]
Answer:
Reversible process :
In reversible process, a thermodynamic system can be retraced back in opposite direction to the changes that take place in the direct process or in forward process.

A reversible process is only an ideal concept.
Examples for reversible process :

  1. Peltier effect and Seebeck effect.
  2. Fusion of ice and vapourisation of water.

Irreversible process:
A thermodynamic process that cannot be taken back in opposite direction is called an “irreversible process.”

Examples:

  1. Work done against friction.
  2. Magnetization of materials.

Carnot’s Engine :
Carnot’s engine works on the principle of reversible process within the temperatures T1 and T2.

It consists of four continuous processes. The total process is known as Carnot Cycle.

Step 1 :
In Carnot cycle, the 1st step consists of isothermal expansion of gases. So temperature T is constant, P, V changes are
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 12

Step 2 :
In this stage gases will expand adiabatically. So energy to the system Q is constant.
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 13

Step 3 :
In this stage gases will be compressed isothermally. So P1V change are
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 14

Step 4 :
In the fourth stage the gas suffers adiabatic compression and returns to original stage.
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 15

The total work done W = Q1 – Q2 i.e., the difference to heat energy absorbed from source and heat energy given to sink
Efficiency of Carnot engine
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 16

Question 2.
State second law of thermodynamics. How is heat engine different from a refrigerator? Explain. [TS Mar. ’18, ’16, May ’18, ’16; AP Mar. ’19, ’16, ’15, ’13; June ’15; May ’14]
Answer:
Second Law of Thermodynamics :
First law of thermodynamics is based on “Law of conservation of energy”, while second law of thermodynamics gives “information about the transformation of heat energy”.

So, there are two conventional statements of second law depending on common experience.

1) Kelvin-Plank statement:
It is impossible for an engine working in a cyclic process to extract heat from a hot body and to convert it completely into work.

2) Clausius Statement:
It is impossible for a self-acting machine, unaided by any external agency to transfer heat from a cold body to a hot reservoir. In other words, heat cannot by itself flow from a colder body to a hotter body.

Heat engine:
A heat engine is a device used to convert heat energy into mechanical work.

Generally heat engines will work in a cyclic process. Heat engine consists of three important units.

1) Source :
Which is an object or system at high temperature. A heat engine will absorb heat energy Q1 from source.

2) Working substance :
Every heat engine requires a working substance to do work. Generally, the working substance is like steam or fuel vapour and air mixture etc. A part of heat energy of working substance is converted into mechanical work.

3) Sink :
In every heat engine heat energy content of working substance is not converted into work totally. So some energy (Q2) is wasted or rejected by the engine. This rejected energy (Q2) is delivered to some other body or system at low temperature. This body with low temperature is called sink.

Efficiency of heat engine,
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 10

where
Q1 = heat energy supplied by source
Q2 = heat energy delivered to sink
Block diagram ot heat engine is as shown.
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 11

Refrigerator :
A refrigerator works in the reverse process of heat engine.

It extracts heat energy Q2 from sink i.e., from low temperature body with the help of external work and delivers heat energy Q1 to high temperature body called source.
Work done W = Q1 – Q2.
In refrigerators external work is done on working substance.
A block diagram of refrigerator is as shown.
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 17

Difference between heat engine and refrigerator :
Refrigerator extracts heat energy from sink i.e., from low temperature body with the help of external work and delivers heat energy to high temperature body called source. A heat engine will absorb heat energy from source and reject heat energy to the sink. Heat engine will work in a reversible process but the refrigerator works in the reverse process of heat engine.

TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics

Question 3.
State second Saw of thermodynamics. Describe the working of Carnot engine. Obtain an expression for the efficiency. [AP Mar. ’16]
Answer:
Second Law of Thermodynamics :
First law of thermodynamics is based on Law of conservation of energy, while second law of thermodynamics gives information about the transformation of heat energy. So, there are two conventional statements of second law depending on common experience.

1) Kelvin-Plank statement:
It is impossible for an engine working in a cyclic process to extract heat from a hot body and to convert it completely into work.

2) Clausius Statement:
It is impossible for a self-acting machine, unaided by any external agency to transfer heat from a cold body to a hot reservoir. In other words, heat cannot by itself flow from a colder body to a hotter body.

Carnot’s Engine :
Carnot’s engine works on the principle of reversible process within the temperatures T1 and T2.

It consists of four continuous processes. The total process is known as Carnot Cycle.

Step 1 :
In Carnot cycle, the 1st step consists of isothermal expansion of gases. So temperature T is constant, P, V changes are
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 18
Work done in isothermal process
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 19

Step 2 :
In this stage gases will expand adiabatically. So energy to the system Q is constant.
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 20

Step 3 :
In this stage gases will be compressed isothermally. So PjV changes are
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 21

Step 4 :
In the fourth stage the gas suffers adiabatic compression and returns to original stage.
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 22
Total work done in Carnot Cycle
W = W1,2 + W2, 3 + W3, 4 + W4, 1

TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 23
The total work done W = Q1 – Q2 i.e., the difference to heat energy absorbed from source and heat energy given to sink Efficiency of Carnot engine
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 24

Question 4.
What is the difference between heat engine and refrigerator. [TS May. ’16]
Answer:
Differences between heat engine and refrigerator:
Refrigerator extracts heat energy from sink i.e., from low temperature body with the help of external work and delivers heat energy to high temperature body called source.

A heat engine will absorb heat energy from source and reject heat energy to the sink. Heat engine will work in a reversible process but the refrigerator works in the reverse process of heat engine.

Problems

Question 1.
If a monoatomic ideal gas of volume 1 litre at N.T.P. is compressed (I) adiabatically to half of its volume, find the work done on the gas. Also find (ii) the work done if the compression is isothermal. (γ = 5/3)
Solution:
i) During an adiabatic process T1V1γ-1 = T2V2γ-1

ii) Work done during isothermal compression is
W = 2.3026 nRT log10 \(\frac{V_2}{V_1}\)
n = number of moles = \(\frac{1}{22.4}\)
T = 273 K; R = 8.314 J mol-1K-1
TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 26

TS Inter 1st Year Physics Study Material Chapter 13 Thermodynamics

Question 2.
Five moles of hydrogen when heated through 20 K expand by an amount of 8.3 × 10-3m³ under a constant pressure of 105 N/m². If Cv = 20 J/mole K, find Cp.
Solution:
We know that Cp – Cv = R.
Multiplying throughout by n ∆ T
nCp ∆T – nCv ∆T = nR ∆T
n ∆ T (Cp – Cv) = P ∆ V
5 × 20 (Cp – 20) = 105 × 8.3 × 10-3 (∵ nR∆T = P∆V)
Cp – 20 = 8.3
Cp = 28.3 J/mole K.
(∵ n = 5, ∆T = 20 K, P = 1 × 105N/m² & Cv = 20 J/mole K and ∆V = 8.3 × 10³ m³)

TS Inter 2nd Year English Study Material Chapter 5 Fear

Telangana TSBIE TS Inter 2nd Year English Study Material 5th Lesson Fear Textbook Questions and Answers.

TS Inter 2nd Year English Study Material 5th Lesson Fear

Annotations (Section A, Q.No. 2, Marks: 4)
Annotate the following in about 100 words each.

a) It is said that before entering the sea
a river themselves with fear.

Introduction: These are the opening lines of the poem, “Fear”, written by Khalil Gibran, a Lebanese-American writer. He became famous for his book, “The Prophet”, a collection of philosophical essays. His writings deal with Spiritual Love and Life Issues.

Context and Meaning: In the poem, the poet expresses his philosophical under-standing of overcoming fear. The speaker thinks of the image of a river flowing into the sea. The poet’s point of view is not wholly his own. He may have heard of the river’s fear and chose to give it some strength through the poetry. The river may have traversed difficult paths before entering the ocean. Yet it themselves with fear at the sight of the vastness of the ocean.

Critical comment: The poet compares humanity to rivers. Even for a man there is always the fear of the unknown and being lost in it.

కవి పరిచయం : ఇవి లెబనీస్ – అమెరికన్ రచయిత ఖలీల్ గిబ్రాన్ రాసిన పద్యం, “భయం” యొక్క ప్రారంభ పంక్తులు. అతను తన పుస్తకం. ప్రవక్త, తాత్విక వ్యాసాల సేకరణకు ప్రసిద్ధి చెందాడు. అతని రచనలు ఆధ్యాత్మిక ప్రేమ మరియు జీవిత విషయాలను తెలియజేస్తుంది.

సందర్భ౦ మరియు వివరణ: ఈ పద్యంలో, కవి భయాన్ని అధిగమించడానికి తన తాత్విక అవగాహనను వ్యక్తపరుస్తాడు. వక్త సముద్రంలోకి ప్రవహించే నది చిత్రం గురించి ఆలోచిస్తాడు. కవి యొక్క దృక్కోణం పూర్తిగా అతనిదికాదు. అతను నది భయం గురించి విని, కవిత్వం ద్వారా దానికి కొంత బలాన్ని ఇచ్చేందుకు ఎంచుకున్నాడు. సముద్రంలోకి ప్రవేశించే ముందు నది కష్టతరమైన మార్గాలను దాటి ఉండవచ్చు. అయినప్పటికీ, అది సముద్రపు విశాలతను చూసి భయంతో వణికిపోతుంది.

విమర్శ : కవి మానవత్వాన్ని నదులతో పోల్చాడు. మనిషికి కూడా తెలియని భయం మరియు దానిలో కనుమరుగవుతున్న భయం ఎప్పుడూ ఉంటుంది.

b) And in front of her, she sees an ocean
so vast, that to enter there seems
nothing more than to disappear

Introduction: These are the opening lines of the poem, “Fear”, written by Khalil Gibran, a Lebanese-American writer. He became famous for his book, “The Prophet”, a collection of philosophical essays. His writings deal with Spiritual Love and Life Issues.

Context and Meaning: The poet conveys his philosophical insight about overcoming fear in the poem. The speaker imagines a river that flows into the sea. The river may have traversed difficult paths before entering the ocean. Yet it trembles with fear at the sight of the vastness of the ocean. She looks back at the path she has travelled. But, there is no other option but to move forward and enter the vast ocean. The river must realize that the fear of disappearing forever must be done away with and embrace the truth.

Critical comment: The poet compares humanity to the river. Even for a man there is always the fear of the unknown and being lost in it.

కవి పరిచయం : ఇవి లెబనీస్ – అమెరికన్ రచయిత ఖలీల్ గిబ్రాన్ చే రచించబడిన “భయం” అను పద్యం నుండి తీసుకొనబడినవి. ఇతని తాత్విక కవితా సంపుటి, ప్రవక్తతో ప్రసిద్ధి చెందాడు. అతని రచనలు ఆధ్యాత్మిక ప్రేమ మరియు జీవితసత్యాలను తెలియజేస్తాయి.

సందర్భ౦ మరియు వివరణ : ఈ పద్యంలో భయాన్ని అధిగమించడం గురించి కవి తన తాత్విక అంతర దృష్టిని తెలియజేస్తాడు. వక్త సముద్రంలోకి ప్రవహించే నదిని ఊహించాడు. సముద్రంలోకి ప్రవేశించే ముందు నది కష్టమైన మార్గాలను దాటి ఉండవచ్చు. అయినప్పటికీ సముద్రపు విశాలతను చూసి భయంతో వణికిపోతుంది. ఆమె ప్రయాణించి వచ్చిన మార్గం వైపు తిరిగిచూస్తుంది. కానీ ముందుకు సాగడం మరియు విశాలమైన సముద్రంలోకి ప్రవేశించటం తప్ప వేరే మార్గం లేదు. నది శాశ్వతంగా మాయమైపోతుందనే భయాన్ని వదిలి మరియు సత్యాన్ని స్వీకరించాలి.

విమర్శ : కవి మానవాలిని నదులతో పోల్చాడు. మనిషికి కూడా తెలియని భయం ఉంటుంది మరియు దానిలో కనుమరుగపు తీరని భయం ఎప్పుడూ ఉంటుంది.

TS Inter 2nd Year English Study Material Chapter 5 Fear

c) The river cannot go back. Nobody can go back.
To go back is impossible in existence. (Revision Test – V)

Introduction: These spiritual lines are taken from the “Fear”, written by Khalil Gibran, a Lebanese-American writer. He is famous for his book, “The Prophet”, a collection of philosophical essays. His writings, deal with Spiritual Love and Life Issues.

Context and Meaning: The river travels through mountains and plains to merge with an ocean. The poet talks about her fear directly. He compares humanity to rivers. He discusses the year that human beings encounter too. There is a desire to go back. But, that is impossible in existence. People as well as ‘the river need to accept the fact that there is no other option but to move forward. Thus, people must take risks and believe in themselves.

Critical comment: The poem shows a variety of themes. The error of moving forward, the anxiety of losing oneself, and the journey life till death are some of the major themes.

కవి పరిచయం : ఈ ఆధ్యాత్మిక పంక్తులు లెబనీస్ – అమెరికన్ రచయిత ఖలీల్ గిబ్రాన్చే రచించబడిన “భయం” అను పద్యం నుండి తీసుకొనబడినవి. ఇతని తాత్విక కవితా సంపుటి, ప్రవక్త తో ప్రసిద్ధి చెందాడు. అతని రచనలు ఆధ్యాత్మిక ప్రేమ మరియు జీవితసత్యాల గురించి తెలియజేస్తాయి.

సందర్భ౦ మరియు వివరణ : నది పర్వతశిఖరాలను దాటి వచ్చి సముద్రంలో కలుస్తుంది. కవి తిన్నగా నది భయాన్ని గురించి తెలియజేస్తున్నాడు. మానవాళిని నదితో పోల్చాడు. మానవాళి ఎదుర్కొనే భయాన్ని కూడా ఇతని చర్చిస్తున్నాడు. వెనక్కి వెళ్ళాలనిపిస్తుంది. కానీ, జీవితంలో ఇది సాధ్యం కాదు. మానవాళి అదేవిధంగా నది కూడా ముందుకు సాగటం తప్ప మరో అవకాశం, మార్గం లేదు. అలా ప్రజలు తప్పనిసరిగ్గా రిస్క్ తీసుకోవాలి. మరియు వారిలో వారికి నమ్మకం ఉండాలి.

విమర్శ : పద్యం వివిధ ఇతి వృత్తాలను చూపుతుంది. ముందుకు సాగుటకు భయం, తనను తాను కోల్పోయే ఆందోళన మరియు మరణం వరకు జీవితప్రయాణం కొన్ని ప్రధాన ఇతివృత్తాలు.

d) It’s not about disappearing into the ocean,
but of becoming the ocean.

Introduction: These are the concluding lines of the poem, “Fear”, written by Khalil Gibran, a Lebanese-American writer. He became famous for his book, “The Prophet”, a collection of philosophical essays. His writings deal with Spiritual Love and Life Issues.

Context and Meaning: The poem gives us glimpse of how a river feels when it travels through mountains and plains to merge with an ocean. There is a desire to go back. But, that is impossible in existence. There is no other option but to more forward and enter the ocean. The poet says that it is not disappearing into the ocean. But, it is becoming the ocean by overcoming fear. Just like the river, people must take risks and overcome fear.

Critical comment: The poet says that the fear of disappearing forever must be done away with and embrace the truth.

కవి పరిచయం : లెబనీస్ – అమెరికన్ రచయిత ఖలీల్ గిబ్రాన్చే రచించబడిన “భయం” అను పద్యంలోని ముగింపు పంక్తులు ఇవి. ఇతను తన ఆధ్యాత్మిక కవితా సంపుటి “ప్రవక్త”కి ప్రసిద్ధి. ఇతని రచనలు ఆధ్యాత్మిక, ప్రేమ మరియు జీవిత సత్యాలను భోదిస్తాయి.

సందర్భ౦ మరియు అర్థం : పర్వతాలు మరియు మైదానాల గుండా ప్రయాణించి సముద్రంలో కలిసిపోయినప్పుడు నది ఎలా ఉంటుందో ఈ పద్యం మనకు ఒక సంగ్రహనలోకనం ఇస్తుంది. తిరిగి వెళ్ళాలనే కోరిక ఉంది నదికి కానీ, అది ఉనికిలో అసాధ్యంముందుకు సాగడం మరియు సముద్రంలోకి ప్రవేశించడం తప్ప వేరే మార్గం లేదు. అది సాగరంలో కనుమరుగవడం లేదని కవి చెప్తాడు. కానీ, భయాన్ని జయించి, సముద్రంగా మారుతుంది. నది వలె ప్రజలు రిస్క్ తీసుకోవాలి మరియు భయాన్ని అధిగమించాలి.

విమర్శ: శాశ్వతంగా అదృశ్యమౌతుందనే భయాన్ని పోగొట్టి, సత్యాన్ని స్వీకరించాలని కవి చెప్పాడు.

Paragraph Questions & Answers (Section A, Q.No.4, Marks: 4)
Answer the following Questions in about 100 words

a) What is the central idea of the poem Fear?
Answer:
The poem, ‘Fear’, is written by Khalil Gibran, a Lebanese-American writer. The poem gives us a glimpse of how a river feels when it travels through mountains and plains to merge with an ocean. The poet refers to the river as ‘She’ to infuse life into the river. He talks about her fear directly. He discusses the fear that human beings encounter too. There is a desire to revisit the past, and ‘to go back’ that is ‘impossible in existence’.

The later poet suggests that people need to accept the fact that there is no other option. But to move forward and meet the world by relying on the distance already traveled. As a result, people must take risks and believe in themselves. Thus, the message of the poem is overcoming fear.

‘భయం’ అను కావ్యంను లెబనీస్ – అమెరికన్ రచయిత ఖలీల్ గిబ్రాన్ వ్రాశాడు. పర్వతాలు మరియు మైదానాల గుండా సముద్రంలో కలిసి పోయేటప్పుడు నది ఎలా ఉంటుందో ఈ పద్యం మనకు అందిస్తుంది. కవి నదిలోకి జీవం పోయడానికి నదిని ‘ఆమె’గా సూచించాడు. అలా ఆమె భయం గురించి నేరుగా మాట్లాడతాడు. అతను మానవులకు కూడా ఎదురయ్యే భయాన్ని చర్చిస్తాడు. తిరిగి వెళ్ళాలనే కోరిక ఉంది.

మరియు తిరిగి సందర్శించాలంటే ఉనికిలో అసాధ్యం. ఇప్పటకీ ప్రయాణించిన దూరం, అనుభవంపై ఆధారపడి ముందుకు సాగడం మరియు ప్రపంచాన్ని ఎదుర్కోవడం తప్ప మరో మార్గం లేదనే వాస్తవాన్ని ప్రజలు అంగీకరించాలని తర్వాత భాగం తెలియజేస్తుంది. దాని ఫలితంగా, ప్రజలు తప్పని సరిగా రిస్క్ తీసుకోవాలి మరియు వారిని వారు నమ్మాలి. అలా, భయాన్ని జయించటమే ఈ పద్యం యొక్క సందేశం.

TS Inter 2nd Year English Study Material Chapter 5 Fear

b) What does “nobody can go back” mean in the poem “Fear”?. Explain from your point of you.
Answer:
The poem,”Fear”, is written by Khalil Gibran, a Lebanese – American writer. He is famous for his book, “The Prophet”, a collection of Philosophical essays in his poem, he expresses his philosophical understanding of overcoming fear. He imagines a river that flows into the sea. The river may have traversed difficult paths before entering the ocean. Yet it trembles with fear at the sight of the vastness of the ocean.

He not only talks about her fear directly but discusses the fear that human beings encounter too. There is a desire to go back. But, that is impossible in existence. Everyone should accept the fact that there is no other option but to move forward and meet the world by relying on the distance already traveled. Hence, people must take risks to achieve success.

‘భయం’ అను కావ్యం లెబనీస్ అమెరికన్ రచయిత ఖలీల్ గిబ్రాన్చే రచించబడింది. ఇతను ఆధ్యాత్మిక కవితా సంపుటి, ప్రవక్తచే ప్రసిద్ధి చెందాడు. ఈ పద్యంలో భయాన్ని అదిగమించడం గురించి కవి తన తాత్విక అంతరదృష్టిని తెలియజేస్తాడు. ఇతను సముద్రంలోకి ప్రవహించే నదిని ఊహించాడు. సముద్రంలోకి ప్రవహించుటకు ముందు నది కష్టమైన మార్గాలను దాటి ఉండవచ్చు అయినప్పటికి, సువిశాలమైన సముద్రంను చూసి, నది భయంతో వణుకుతుంది. ఇతను కేవలం తిన్నగా నది భయం గురించే చెప్పటంలేదు.

మానవులు ఎదురుకునే భయాన్ని కూడా అంతర్లీనంగా చెప్తున్నాడు. వెనక్కి వెళ్ళాలి అనుకుంటాం. కానీ, జీవితంలో అది సాధ్యంకాదు. ఇప్పటకీ ప్రయాణించిన దూరం, అనుభవంపై ఆధారపడి ముందుకు సాగడం మరియు ప్రపంచాన్ని ఎదుర్కోవడం తప్ప మరో మార్గం లేదనే వాస్తవాన్ని అందరూ అంగీకరించాలి. కాబట్టి, విజయం సాధించడానికి ప్రజలు రిస్క్ తీసుకోవాలి, కష్టాలను, ప్రమాదాలను తట్టుకొని ముందుకు వెళ్ళాలి.

c) How can one overcome fear? Explain (or) What does the line ” The river needs to take the risk of entering the ocean” mean? Discuss. (Revision Test – V)
Answer:
The poem,”Fear”, is written by Khalil Gibran, a Lebanese – American writer. He is famous for his book, “The Prophet”, a collection of Philosophical essays. The poet conveys his philosophical insight about overcoming fear in the poem. He imagines a river that flows into the sea. He refers to the river as ‘She’ to infuse life into the river. He talks about her fear directly. He discusses the fear that human beings encounter too.

The river trembles with fear at the sight of the vastness of the ocean. She looks back at the path she has traversed. But, that is impossible in existence. There is no other option to her. She has to move forward and accept the truth. So, she takes the risk of entering the ocean so that she becomes an ocean.

The poet compares humanity to the river. Through the river’s emotions, the poet sends a powerful message to those who fear, losing their identity, death, change, being forgotten in this universe and so on. So, once can overcome fear by taking risks to achieve success.

‘భయం’ అను పద్యం లెబనీస్ – అమెరికన్ రచయిత ఖలీల్ గిబ్రాన్చే రచించబడింది. ఇతను ఆధ్యాత్మిక కవితా సంపుటి, ‘ప్రవక్త’ చే ప్రసిద్ధి చెందాడు. ఈ పద్యంలో భయాన్ని అధిగమించటం గురించి కవి తన తాత్విక అంతర దృష్టిని తెలియజేస్తాడు ఇతను సముద్రంలోకి ప్రవహించే నదిని ఊహించాడు. అతను నదిలోకి జీవనాన్ని నింపడానికి నదిని ఆమె అని సూచిస్తాడు. అతను ఆమె భయం గురించి నేరుగా మాట్లాడతాడు. అతను కూడా ఎదురయ్యే భయాన్ని చర్చిస్తాడు. సముద్రపు విశాలతను చూసి నది భయంతో వణికిపోయింది.

ఆమె తాను ప్రయాణించిన మార్గం వైపు తిరిగి చూస్తుంది. కానీ అది ఉనికిలో అసాధ్యం. ఆమెకు వేరే అవకాశం లేదు. ఆమె ముందుకు సాగాలి మరియు సత్యాన్ని అంగీకరించాలి. కాబట్టి, ఆమె సముద్రంలోకి ప్రవేశించే ప్రమాదాన్ని తీసుకుంటుంది తద్వారా ఆమె సముద్రం అవుతుంది. కవి మానవాళిని నదులతో పోల్చాడు. నది యొక్క భావోద్వేగాల ద్వారా, కవి భయపడే వారికి, ఆమె గుర్తింపును కోల్పోతున్న వారికి, మరణం, ఈ విశ్వంలో మరచిపోయిన మార్పు మొ|| వారికి శక్తివంతమైన సందేశాన్ని పంపాడు. కాబట్టి, విజయం సాధించడానికి రిస్క్లను తీసుకోవడం ద్వారా భయాన్ని అధిగమించవచ్చు.

Fear Summary in English

About Author

TS Inter 2nd Year English Study Material Chapter 5 Fear 1

Born in a village of the Ottoman-ruled Mount Lebanon Mutasarrifate to a Maronite family, the young Gibran immigrated with his mother and siblings to the United States in 1895. As his mother worked as a seamstress, he was enrolled at a school in Boston, where his creative abilities were quickly noticed by a teacher who presented him to photographer and publisher F. Holland Day. Gibran was sent back to his native land by his family at the age of fifteen to enroll at the Collège de la Sagesse in Beirut.

Returning to Boston upon his youngest sister’s death in 1902, he lost his older half-brother and his mother the following year, seemingly relying afterwards on his remaining sister’s income from her work at a dressmaker’s shop for some time.

The poem ‘Fear’, is written by Khalil Gibran, a Lebanese – American writer. He is famous for his book, the prophet, a collection of philosophical essays. In the poem, he expresses his philosophical understanding of over coming fear. He imagines a river that flows into the sea. The statement “it is said” that implies that his point of view is not wholly his own. He may love heard of the river’s fear and close to it some strength through the poetry. The poet refers to the river as ‘she’ to infuse life into the river. She may have braversed difficult paths before entering the ocean, yet it trembles with fear at the sight of the vastness of the ocean. He talks about her fear directly. He discusses the fear that human beings encounter too.

The river looks back at the path she has travelled through. She desires to go back. But, that is impossible in existence. She realizes that she has to accept and embrace the truth. It is because nobody can go back. The poet compares humanity to the rivers. The river as well as the people should accept the truth and go forward.

The final stanza suggests that the river needs to take the risk of entering the ocean. It is because fear will disappear due to the realization of the river. She comes to understand that she is not disappearing into the ocean. In-fact she is becoming the ocean. Through the emotions of the river, the poet sends a powerful message to those who fear losing their identity death, change being forgotten in this universe and so on so people need to accept the fact that there is no other option but to more forward. They must take risks to achieve success and believe in themselves. Thus, the message of the poem is “Overcoming Fear”.

Fear Summary in Telugu

Note: This summary is only meant for Lesson Reference, not for examination purpose

‘భయం’ అను పద్యం లెబనీస్ – అమెరికన్ రచయిత ఖలీల్ గిబ్రాన్ చే రచింపబడింది. ఇతని తాత్విక కవితా సంపుటి, ‘ప్రవక్త’తో ప్రసిద్ధి చెందాడు. ఈ పద్యంలో భయాన్ని అధిగమించడం గురించి కవి తన తాత్విక అంతర దృష్టిని తెలియజేస్తాడు. ఇతను సముద్రంలోకి ప్రవహించే నదిని ఊహించాడు. కవి యొక్క దృక్కోణం పూర్తిగా అతని సొంతం కాదు. అతను నది భయం గురించి విని, కవిత్వం ద్వారా దీనికి కొంత బలాన్ని ఇచ్చేందు ఎంచుకున్నాడు. కవి నదికి జీవంపోయడానికి నదిని ‘ఆమె’గా సూచించాడు. సముద్రంలోకి ప్రవహించుటకు ముందు నది కష్టమైన మార్గాలను దాటి ఉండవచ్చు. అయినప్పటికీ సువిశాలమైన సముద్రంను చూసి నది భయంతో వణికిపోతుంది. ఇతను నేరుగా నది భయాన్ని గురించి చెప్తున్నాడు. ఇతను మానవులకు కూడా ఎదురయ్యే భయాన్ని చర్చిస్తాడు.

(ఆమె) నది తాను ప్రవహించిన మార్గం వైపు తిరిగి చూస్తుంది. తిరిగి వెళ్ళాలనుకుంటుంది. కానీ, అది అసాధ్యం. ఆమె ముందుకు సాగాలి మరియు వాస్తవాన్ని అంగీకరించాలన్న సత్యాన్ని గ్రహిస్తుంది. ఎందుకంటే, జీవితంలో ఎవ్వరూ వెనక్కి వెళ్ళలేరు. కవి మానవాళిని నదితో పోల్చాడు. నది అదే విధంగా ప్రజలు వాస్తవాన్ని అంగీకరించాలి మరియు ముందుకు సాగాలి అంటున్నాడు కవి.

చివరి చరణంలో నది సముద్రంలోకి ప్రవేశించే రిస్క్ తీసుకోవాల్సివచ్చింది అని చెప్తుంది. నది వాస్తవాన్ని గ్రహించటం వల్లనే భయం తొలగిపోయింది. తాను సముద్రంలో కనుమరుగు అవ్వటంలేదన్న వాస్తవాన్ని గ్రహిస్తుంది. వాస్తవంగా, నదీ సముద్రమౌతుంది. నది భావోద్వేగాల ద్వారా, కవి భయపడేవారికి, తమ గుర్తింపును కోల్పోతున్న వారికి, మరణం ఈ విశ్వంలో మరిచిపోయిన మార్పు మొ॥న వారికి శక్తివంతమైన సందేశాన్ని కవి పంపాడు. కాబట్టి ముందుకు సాగడం తప్ప, మరొక మార్గం లేదు. అన్న వాస్తవాన్ని ప్రజలు అంగీకరించాలి. వారు రిస్క్ తీసుకోవాలి తప్పని సరిగా మరియు వారిని వారు నమ్మాలి. అలా భయాన్ని జయించుటకు ఈ పద్యం యొక్క సందేశం.

Fear Summary in Hindi

Note: This summary is only meant for Lesson Reference, not for examination purpose

कविता ‘फ़ियर’ एक लेबनानी पुस्तक, ‘द प्रोकेट’, दर्शनिक निबंध संग्रह द्वारा प्रसिद्ध हैं। इस कविता में, वे भच पर काबू पाने की उनकी दार्शनिक समझ को व्यक्त करते हैं । वे समुद्र की बहती हुई समुद्र में मिलनेवाली नदी की कल्पना करते हैं । कथन, ‘ऐसा कहा जाता है” का तात्पर्य है कि लेखक का दृष्टिकोण पूरी तरह से उनका अपना नहीं है । उन्होंने नदी के डर के बारे में सुना होगा और कविता के माध्याम से इसे कुछ बल देने का फैसला किया होगा । कवि नदी को नदी में जान डालने केलिए ‘वह’ के रूप में संदर्भित करते हैं । वह भले ही समुद्र में प्रवेश करने से पहले कठिन रास्तों को पार कर चुकी हो, फिर भी वह समुद्र की विशालता को देखकर भय से कांपती है । वे उसके डर के बारे में सीधे बताते हैं । वे उस डर पर चर्चा करते हैं, जिसका सामना मनुष्य भी करते हैं ।

नदी पीछे मुड़कर तकिए गए रास्ते की ओर देखती है । वह वापस जाने की इच्छा करती है । लेकिन वह अस्तित्व में असंभव है । उसे पता चलता है कि सच्चाई को स्वीकार करना और गले लगाना है । ऐसा इसलिए कि कोई भी वापस नहीं जा सकता । कवि मानवता की नदियों तुलना करते हैं। नदी और लोग सच्छाई को स्वीकार कर आगे बढ़ना चाहिए ।

अंतिम छंद बताता है कि नदी को समुद्र में प्रवेश करने का जोखिम उठाने की जरूरत हैं । क्योंकि नदी पूर्ण रूप से समझने के बाद उसका भय मिठ जाएगा । उसे समझ में आ जाता है कि वह सागर में विलीन नहीं हो रही है । वास्तव में वह सागर बन रही है। नदी की भावनाओं के माध्यम से कवि उन लोगों को एक शक्तिशाली संदेश भेजते हैं कि जो अपनी पहचान खोने से, मृत्यु से, | परिवर्तन से, इत्यादि से डरते हैं । इसलिए लोगों को इस तथ्य को स्वीकार करने की आवश्यकता है कि आगे बढ़ाने के बजाए कोई विकल्प नहीं है। सफलता हासिल करने केलिए लोगों को जोखिम उठाना चाहिए और खुद पर विश्वास करना चाहिए । इस प्रकार, इस कविता का संदेश है, ‘डर पर काबू पाना ।

TS Inter 2nd Year English Study Material Chapter 5 Fear

Meanings and Explanations

trembles/(ట్రెంబల్జ్)/ ‘trem.bəl / : shakes, quivers, vibrates (out of fear) వణుకుట (భయంతో) काँपना

path (n)/ (పాత్)/ pa:θ / : a way; a track; దారి మార్గము

traversed(v-past ten)/ (ట్రవ(ర్))/trə’v3:s / : travelled across; ద్వరా పయనించెను, के माध्यम से यात्रा की

winding(v+ing-ore.part.adj)/(వైండింగ్)/ waın.dŋ/ : with twists and turns, వంపులు, మెలికలు కల, एक घुमाया सर्पिल माला

vast(adj)/(వాస్ట్)/va:st/ : very large, చాలా విశాలమైన, विशाल

existence/ (ఇగ్జిస్టన్స్స్) /ɪg’zɪs.təns/ : the state of being, ఉనికి, जीवित होने की अवस्था

risk (n) / రిస్క్ /rɪsk/ : a possible adverse event; danger, కష్టం కలిగించే అవకాశం కల అంశము సవాలు ప్రమాదము, खतरा

TS Inter 1st Year Maths 1A Matrices Important Questions Long Answer Type

Students must practice these Maths 1A Important Questions TS Inter 1st Year Maths 1A Matrices Important Questions Long Answer Type to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Matrices Important Questions Long Answer Type

Question 1.
Without expanding the determinant show that \(\left|\begin{array}{lll}
\mathbf{b}+\mathbf{c} & \mathbf{c}+\mathbf{a} & \mathbf{a}+\mathbf{b} \\
\mathbf{c}+\mathbf{a} & \mathbf{a}+\mathbf{b} & \mathbf{b}+\mathbf{c} \\
\mathbf{a}+\mathbf{b} & \mathbf{b}+\mathbf{c} & \mathbf{c}+\mathbf{a}
\end{array}\right|\) = 2\(\left|\begin{array}{lll}
\mathbf{a} & \mathbf{b} & \mathbf{c} \\
\mathbf{b} & \mathbf{c} & \mathbf{a} \\
\mathbf{c} & \mathbf{a} & \mathbf{b}
\end{array}\right|\) [Mar. 15 (AP); May 98, 96, 91]
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 1

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 2.
Show that \(\left|\begin{array}{ccc}
1 & a^2 & a^3 \\
1 & b^2 & b^3 \\
1 & c^2 & c^3
\end{array}\right|\) = (a – b) (b – c) (c – a) (ab + bc + ca). [Mar. 17(AP), 09: May 15 (AP); 02]
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 2
= (a – b) (b – c) (c – a) [0 (c3 – c2 (a + b + c) – (a + b) (0 – a – b – c) + (a2 + ab + b2) (0 – 1)]
= (a – b) (b – c) (c – a) [0 + a2 + ab + ac + ab + b2 + bc – a2 – ab – b2]
= (a – b) (b – c) (c – a) (ab + bc + ca) = RHS.

Question 3.
Show that \(\left|\begin{array}{ccc}
\mathbf{a}-\mathbf{b}-\mathbf{c} & \mathbf{2 a} & \mathbf{2 a} \\
\mathbf{2 b} & \mathbf{b}-\mathbf{c}-\mathbf{a} & \mathbf{2 b} \\
\mathbf{2 c} & \mathbf{2 c} & \mathbf{c}-\mathbf{a}-\mathbf{b}
\end{array}\right|\) = (a + b + c)3. [Mar. 11; May 11]
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 3

Question 4.
Find the value of x if \(\left|\begin{array}{ccc}
x-2 & 2 x-3 & 3 x-4 \\
x-4 & 2 x-9 & 3 x-16 \\
x-8 & 2 x-27 & 3 x-64
\end{array}\right|\) = 0 [Mar 15 (TS); Mar. 06]
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 4
⇒ (x – 2) (30 – 24) – (2x – 3) (10 – 6) + (3x – 4) (4 – 3) = 0
⇒ (x – 2)6 – (2x – 3)4 + (3x – 4)(1)
⇒ 6x – 12 – 8x + 12 + 3x – 4 = 0
⇒ x – 4 = 0
⇒ x = 4.

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 5.
Show that \(\left|\begin{array}{ccc}
a+b+2 c & a & b \\
c & b+c+2 a & b \\
c & a & c+a+2 b
\end{array}\right|\) = 2 (a + b + c)3 [Mar. 18, 16 (AP); Mar. 16 (TS), 10 Mar.19 (TS), May 12, 10, 08, 03, 99]
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 5= 2 (a + b + c)2 [1{(c + a + 2b) – 0} – a (0 – 0) + b (0 – 1)]
= 2(a + b + c)2 [c + a + 2b – b]
= 2(a + b + c)2 (a + b + c) = 2(a + b + c)3

Question 6.
Show that \(\left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|^2=\left|\begin{array}{ccc}
2 b c-a^2 & c^2 & b^2 \\
c^2 & 2 a c-b^2 & a^2 \\
b^2 & a^2 & 2 a b-c^2
\end{array}\right|\) = (a3 + b3 + c3 – 3abc)2. [Mar. 19 (AP) Mar. 18 (TS): May 14. 09; Mar. 12, 01]
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 6

Question 7.
Show that \(\left|\begin{array}{ccc}
a^2+2 a & 2 a+1 & 1 \\
2 a+1 & a+2 & 1 \\
3 & 3 & 1
\end{array}\right|\) = (a – 1)3 [Mar. 13, 07]
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 7
= (a – 1)2 [(a + 1) (1 – 0) – 1 (2 – 0) + 0 (6 – 3)]
= (a – 1)2 [a + 1 – 2] = (a – 1)2 (a – 1) = (a – 1)3 = R.H.S.

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 8.
Show that \(\left|\begin{array}{ccc}
\mathbf{a} & \mathbf{b} & \mathbf{c} \\
\mathbf{a}^2 & \mathbf{b}^2 & \mathbf{c}^2 \\
\mathbf{a}^3 & \mathbf{b}^3 & \mathbf{c}^3
\end{array}\right|\) = abc (a – b) (b – c) (c – a). [May. 06]
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 8
= abc(a – b)(b – c) [0(c2 – bc – c2) – 0(c2 – ac – bc) + 1(b + c – a – b)]
= abc (a – b) (b – c) (c – a) = R.H.S

Question 9.
If A = \(\left[\begin{array}{lll}
\mathbf{a}_1 & \mathbf{b}_1 & \mathbf{c}_1 \\
\mathbf{a}_2 & \mathbf{b}_2 & \mathbf{c}_2 \\
\mathbf{a}_{\mathbf{3}} & \mathbf{b}_3 & \mathbf{c}_3
\end{array}\right]\) is a non-singular matrix, then show that A is invertiable and A-1 = \(\frac{{Adj} \mathbf{A}}{{det} \mathbf{A}}\). [Mar. 17 (AP). May 15 (AP). 13, 10, 07, 06, 02, Mar. 07, 02, 99, 94, 82, 80]
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 9

Question 10.
Solve the system of equations 3x + 4y + 5z = 18, 2x – y + 8z = 13, 5x – 2y + 7z = 20 by using Cramer’s rule. [Mar. 12, 03; May 09]
Answer:
Given system of linear equations are 3x + 4y + 5z = 18, 2x – y + 8z = 13, 5x – 2y + 7z = 20
Let A = \(\left[\begin{array}{rrr}
3 & 4 & 5 \\
2 & -1 & 8 \\
5 & -2 & 7
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and D = \(\left[\begin{array}{l}
18 \\
13 \\
20
\end{array}\right]\)
Then we can write the given equations in the form of matrix equation as AX = D.
Δ = det A = \(\left|\begin{array}{rrr}
3 & 4 & 5 \\
2 & -1 & 8 \\
5 & -2 & 7
\end{array}\right|\) = 3 (- 7 + 16) – 4 (14 – 40) + 5 (- 4 + 5)
= 3(9) – 4 (- 26) + 5 (1)
= 27 + 104 + 5 = 136 ≠ 0
Hence, we can solve the given equations by using Cramer’s rule.
Δ1 = \(\left|\begin{array}{rrr}
18 & 4 & 5 \\
13 & -1 & 8 \\
20 & -2 & 7
\end{array}\right|\) = 18(- 7 + 16) – 4(91 – 160) + 5(- 26 + 20) = 18(9) – 4(- 69) + 5(- 6) = 162 + 276 – 30 = 408
Δ2 = \(\left|\begin{array}{lll}
3 & 18 & 5 \\
2 & 13 & 8 \\
5 & 20 & 7
\end{array}\right|\) = 3(91 – 160) – 18(14 – 40) + 5(40 – 65) = 3(- 69) – 18(- 26) + 5(- 25) = – 207 + 468 – 125 = 136
Δ3 = \(\left|\begin{array}{rrr}
3 & 4 & 18 \\
2 & -1 & 13 \\
5 & -2 & 20
\end{array}\right|\) = 3(- 20 + 26) – 4(40 – 65) + 18(- 4 + 5) = 3(6) – 4(- 25) + 18(1)
= 18 + 100 + 18 = 136
Hence, by Cramer’s rule,
x = \(\frac{\Delta_1}{\Delta}=\frac{408}{136}\) = 3, y = \(\frac{\Delta_2}{\Delta}=\frac{136}{136}\) = 1, z = \(\frac{\Delta_3}{\Delta}=\frac{136}{136}\) = 1
∴ The solution of the giveñ system of equations is x = 3, y = 1, z = 1.

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Solve the system of equations 2x – y + 3z =9, x + y + z = 6, x – y + z = 2 by using Cramer’s rule. [Mar. 17 (TS), 16 (AP), 02; May 13]
Answer:
x = 1, y = 2, z = 3

Question 11.
Solve: 3x + 4y + 5z = 18, 2x – y + 8z = 13 and 5x – 2y + 7z = 20 by using the matrix inversion method. [Mar. ‘19 (TS): Mar. ‘15 (AP) ; Mar. ‘13. ‘08, ‘01, ‘00, 96]
Answer:
Given system of linear equations are 3x + 4y + 5z = 18, 2x – y + 8z = 13, 5x – 2y + 7z = 20
Let A = \(\left[\begin{array}{rrr}
3 & 4 & 5 \\
2 & -1 & 8 \\
5 & -2 & 7
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\), D = \(\left[\begin{array}{l}
18 \\
13 \\
20
\end{array}\right]\)
This can be represented as AX = B and X = A1B is a solution.
∆ = det A = \(\left|\begin{array}{rrr}
3 & 4 & 5 \\
2 & -1 & 8 \\
5 & -2 & 7
\end{array}\right|\) = 3(- 7 + 16) – 4(14 – 40) + 5(- 4 + 5) = 3(9) – 4(- 26) + 5(1) = 27 + 104 + 5 = 136
Cofactor of 3 is A1 = +(- 7 + 16) = 9
Cofactor of 5 is A3 = +(32 + 5) = 37
Cofactor of -1 is B2 = +(21 – 25) = – 4
Cofactor of 5 is C1 = +(- 4 + 5) = 1
Cofactor of 7 is C3 = +(- 3 – 8) = – 11
Cofactor of 2 is A2 = – (28 + 10) = – 38.
Cofactor of 4 is B1 = – (14 – 40) = 26
Cofactor of -2 is B3 = – (24 – 10) = – 14
Cofactor of 8 is C2 = – (- 6 – 20) = 26
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 10

Solve 2x – y + 3z = 8, – x + 2y + z = 4, 3x + y – 4z = 0 by using matrix Inversion method. [May 15 (AP); May. 12]
Answer:
x = 2, y = 2, z = 2

Solve x + y + z = 1, 2x + 2y + 3z = 6, x + 4y + 9z = 3 by using matrix Inversion method. [May 03, 93]
Answer:
x = 7, y = – 10, z = 4

Question 12.
Solve the equations 3x + 4y + 5z = 18, 2x – y + 8z = 13 and 5x – 2y + 7z = 20 by Gauss-Jordan method. [May 15(TS): May 06, 01: Mar. 01]
Answer:
Given system of Linear equations are 3x + 4y + 5z = 18, 2x – y + 8z = 13 and 5x – 2y + 7z = 20
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 11
In this case, system of equations have unique solution.
i.e., x = 3, y = 1, z = 1.
∴ The solution of the given system of equations is x = 3, y = 1, z = 1.

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 13.
Solve the equation 2x – y + 3z = 9, x + y + z = 6, x – y + z = 2 by Gauss-Jordan method. [Mar. 18(AP): Mar. 11, 10; May 11]
Answer:
The given system of linear equations are 2x – y + 3z = 9, x + y + z = 6, x – y + z = 2
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 12
In this case, system of equations has unique solution i.e., x = 1; y = 2; z = 3.
∴ The solution of given system of equations is x = 1; y = 2; z = 3.

Question 14.
Solve the following system of equations by Gauss – Jordan method : x + y + z = 9, 2x + 5y + 7z = 52, 2x + y – z = 0. [May 10, 07; Mar. 09, 1, 99]
Answer:
Given system of linear equations are x + y + z = 9, 2x + 5y + 7z = 52; 2x + y – z = 0. The matrix form is AX = D
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 13
In this case, the system of equations has unique solution i.e., x = 1; y = 3; z 5.
∴ The solution of given system of equations is x = 1; y = 3; z = 5.

Solve the equations 2x – y + 3z = 8, – x + 2y + z = 4, 3x + y – 4z = 0 by Gauss-Jordan method. [Mar. 07]
Answer:
x = 2, y = 2, z = 2

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Solve the equations 2x – y + 8z = 13, 3x + 4y + 5z = 18, 5x – 2y + 7z = 20 by Gauss-Jordan method. [May 09; Mar. 03]
Answer:
x = 3, y = 1, z = 1

Question 15.
Examine whether the system of equations x + y + z = 9, 2x + 5y + 7z = 52, 2x + y – z = 0 are consistent or Inconsistent and If consistent, find the complete solution. [May 15(TS); May 11]
Answer:
The given system of linear equations are x + y + z = 9, 2x + 5y + 7z = 52, 2x + y – z = 0.
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 14
∴ Rank [A] = 3
Now, Rank [AD] = 3
Since, the 3 × 3 sub matrix is \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) whose det is 1 ≠ 0.
∴ Rank [A] = Rank [AD] = 3.
In this case, system of equations has unique solution. i.e., x = 1, y = 3, z = 5.
Hence, the given system is consistent.

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 16.
Examine whether the system of equations x + y + z = 6, x – y + z = 2, 2x – y + 3z = 9 are consistent or inconsistent and if consistent, find the complete solution. [Mar.11, 05]
Answer:
Given system of linear equations are x + y + z = 6, x – y + z = 2, 2x – y + 3z = 9
The given system of equations can be written as AX = D
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 15
∴ Rank of AD = 3.
Rank [A] = Rank [AD] = 3.
In this case, system of equations has unique solution. i.e., x = 1, y = 2, z = 3.
Hence, given system is consistent.

Question 17.
Examine whether the system of equations x + y + z = 1, 2x + y + z = 2, x + 2y + 2z = 1 are consistent or Inconsistent and if consistent, find the complete solution. [Mar. 15 (TS); May 05]
Answer:
Given system of linear equations are x + y + z = 1; 2x + y + z = 2; x + 2y + 2z = 1
The system of equations can be written as AX = D
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 16
∴ Rank [A] = Rank [AD]
In this case, system of equations has infinitely many solutions. x = 1; y + z = 0
∴ The solution of the given system of equations is x = 1, y + z = 0.
Hence, the given system is consistent.

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 18.
Examine whether the following system of equations x + y + z = 6, x + 2y + 3z = 10, x + 2y + 4z = 1 are consistent or Inconsistent and if consistent, find the complete solution. [May. 02]
Answer:
The given system of linear equations are x + y + z = 6; x – 2y + 3z = 10; x + 2y – 4z = 1
The system of equations can be written as AX = D
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 17
∴ Rank [AD] = 3
∴ Rank [A] = Rank [AD] = 3
∴ In this case, system of equations has unique solution.
i.e., x = – 7; y = 22; z = – 9
Hence, the given system is consistent.

Question 19.
If A = \(\left[\begin{array}{ccc}
3 & 2 & -1 \\
2 & -2 & 0 \\
1 & 3 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
-3 & -1 & 0 \\
2 & 1 & 3 \\
4 & -1 & 2
\end{array}\right]\) and X = A + B then find X.
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 18

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 20.
If \(\left[\begin{array}{ccc}
x-1 & 2 & 5-y \\
0 & z-1 & 7 \\
1 & 0 & a-5
\end{array}\right]\) = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 4 & 7 \\
1 & 0 & 0
\end{array}\right]\), then find the values of x, y, z and a.
Answer:
Given \(\left[\begin{array}{ccc}
x-1 & 2 & 5-y \\
0 & z-1 & 7 \\
1 & 0 & a-5
\end{array}\right]\) = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 4 & 7 \\
1 & 0 & 0
\end{array}\right]\)
From equality of matrices
x – 1 = 1 ⇒ x = 2
5 – y = 3 ⇒ y = 2
z – 1 = 4 ⇒ z = 5
a – 5 = 0 ⇒ a = 5
∴ x = 2, y = 2, z = 5, a = 5

Question 21.
If \(\left[\begin{array}{ccc}
x-1 & 2 & y-5 \\
z & 0 & 2 \\
1 & -1 & 1+a
\end{array}\right]=\left[\begin{array}{ccc}
1-x & 2 & -y \\
2 & 0 & 2 \\
1 & -1 & 1
\end{array}\right]\) then find the values of x, y, z and a.
Answer:
Given
\(\left[\begin{array}{ccc}
x-1 & 2 & y-5 \\
z & 0 & 2 \\
1 & -1 & 1+a
\end{array}\right]=\left[\begin{array}{ccc}
1-x & 2 & -y \\
2 & 0 & 2 \\
1 & -1 & 1
\end{array}\right]\)
From the equality of matrices,
x – 1 = 1 – x ⇒ 2x = 2 ⇒ x = 1
y – 5 = – y ⇒ 2y = 5 ⇒ y = 5/2
z = 2
1 + a = 1 ⇒ a = 0
∴ x = 1, y = 5/2, z = 2, a = 0

Question 22.
find the trace of A
if A = \(\left[\begin{array}{ccc}
1 & 2 & -1 / 2 \\
0 & -1 & 2 \\
-1 / 2 & 2 & 1
\end{array}\right]\).
Answer:
Given A = \(\left[\begin{array}{ccc}
1 & 2 & -1 / 2 \\
0 & -1 & 2 \\
-1 / 2 & 2 & 1
\end{array}\right]\)
The elements of the principal diagonal of ‘A’ are 1, – 1, 1
Hence, the trace of A = 1 + (- 1) + 1
= 1 – 1 + 1 = 1

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 23.
If A = \(\left[\begin{array}{ccc}
0 & 1 & 2 \\
2 & 3 & 4 \\
4 & 5 & -6
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
-1 & 2 & 3 \\
0 & 1 & 0 \\
0 & 0 & -1
\end{array}\right]\) find B – A and 4A – 5B.
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 19

Question 24.
If A = \(\left[\begin{array}{lll}
0 & 1 & 2 \\
2 & 3 & 4 \\
4 & 5 & 6
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
1 & -2 & 0 \\
0 & 1 & -1 \\
-1 & 0 & 3
\end{array}\right]\) find A – B and 4B – 3A.
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 20

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 25.
If A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
-4 & 2 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 5 \\
2 & 1
\end{array}\right]\) do AB and BA exist? If they exist find them. Do A and B commute with respect to multiplication?
Answer:
Given A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
-4 & 2 & 5
\end{array}\right]\), B = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 5 \\
2 & 1
\end{array}\right]\)
The order of matrix A is 2 × 3
The order of matrix B is 3 × 2
The no.of columns in A The no.of rows in B.
∴ AB is defined
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 21
The no.of columns in B = The no.of rows in A.
∴ BA is defined.
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 22
∴ AB ≠ BA
∴ A and B is not commute with respect to multiplication.

Question 26.
Find A2 where A = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]\).
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 23

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 27.
If A = \(\left[\begin{array}{ll}
\mathrm{i} & 0 \\
0 & \mathrm{i}
\end{array}\right]\) find A2.
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 24

Question 28.
If A = \(\left[\begin{array}{ccc}
1 & 1 & 3 \\
5 & 2 & 6 \\
-2 & -1 & -3
\end{array}\right]\) then find A3.
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 25

Question 29.
If A = \(\left[\begin{array}{cc}
-1 & 2 \\
0 & 1
\end{array}\right]\) then find AA’. Do A and A’ commute with respect to multiplication of matrices?
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 26
∴ A and A’ do not commute with respect to multiplication of matrices.

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 30.
Find the determinant of the matrix \(\left[\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right]\).
Answer:
Let A = \(\left[\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right]\)
det A = a(bc – f2) – h(ch – gf) + g(hf – bg)
= abc – af2 – ch2 + fgh + fgh – bg2
= abc + 2fgh – af2 – bg2 – ch2

Question 31.
Find the determinant of the matrix \(\left[\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right]\).
Answer:
Let A = \(\left[\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right]\)
det A = a(bc – a2) – b(b2 – ac) + c(ab – c2)
= abc – a3 – b3 + abc + abc – c3
= 3abc – a3 – b3 – c3

Question 32.
Find the adjoint and the inverse of the matrix A = \(\).
Answer:
Given A = \(\left[\begin{array}{cc}
1 & 2 \\
3 & -5
\end{array}\right]\)
Cofactor of 1 is A1 = + (- 5) = – 5
Cofactor of 2 is B1 = – (3) = – 3
Cofactor of 3 is A2 = – (2) = – 2
Cofactor of – 5 is B2 = +(1) = 1
∴ The cofactor matrix of A is
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 27
Now det A = ad – bc = 1(- 5) – 2(3)
= – 5 – 6 = – 11 ≠ 0
Hence A is invertiable.
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 28

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 33.
If A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\) then show that A2 – 4A – 5I = 0. [Mar. 16(AP)]
Answer:
Given A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\)
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 29

Question 34.
Find the adjoint and the inverse of the matrix \(\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 1 & 0 \\
3 & 2 & 1
\end{array}\right]\).
Answer:
Let A = \(\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 1 & 0 \\
3 & 2 & 1
\end{array}\right]\)
Cofactor of 1, A1 =+(1 – 0) = 1
Cofactor of 0, B1 = – (2 – 0) = – 2
Cofactor of 2, C1 = + (4 – 3) = 1
Cofactor of 2, A2 = – (0 – 4) = 4
Cofactor of 1, B2 = + (1 – 6) = – 5
Cofactor of 0, C2 = – (2 – 0) = – 2
Cofactor of 3, A3 = + (0 – 2) = – 2
Cofactor of 2, B3 = – (0 – 4) = 4
Cofactor of 1, C3 = + (1 – 0) = 1
∴ Cofactor matrix of A = B
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 30

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 35.
Find the adjoint and the inverse of the matrix \(\left[\begin{array}{lll}
2 & 1 & 2 \\
1 & 0 & 1 \\
2 & 2 & 1
\end{array}\right]\).
Answer:
Let A = \(\left[\begin{array}{lll}
2 & 1 & 2 \\
1 & 0 & 1 \\
2 & 2 & 1
\end{array}\right]\)
Cofactor of 2, A1 = + (0 – 2) = – 2
Cofactor of 1, B1 = – (1 – 2) = 1
Cofactor of 2, C1 = + (2 – 0) = 2
Cofactor of 1, A2 = – (1 – 4) = 3
Cofactor of 0, B2 = + (2 – 4) = – 2
Cofactor of 1, C2 = – (4 – 2) = —2
Cofactor of 2, A3 = + (1 – 0) = 1
Cofactor of 2, B3 = – (2 – 2) = 0
Cofactor of 1, C3 = + (0 – 1) = – 1
∴ Cofactor matrix of A = B
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 31

Question 36.
Solve the system of equations
2x – y + 3z = 9, x + y + z = 6, x – y + z = 2 by using Cramer’s rule.
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 32

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 37.
Solve 2x – y + 3z = 8, – x + 2y + z = 4, 3x + y – 4z = 0 by using matrix inversion method.
Answer:
The given system of linear equations are
2x – y + 3z = 8, – x + 2y + z = 4, 3x + y – 4z = 0
Let A = \(\left[\begin{array}{crr}
2 & -1 & 3 \\
-1 & 2 & 1 \\
3 & 1 & -4
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\), D = \(\left[\begin{array}{l}
8 \\
4 \\
0
\end{array}\right]\)
Then we can write the given equations in the form of AX = D.
detA = 2(- 8 – 1) + 1(4 – 3) + 3(- 1 – 6)
= 2(- 9) + 1 (1) + 3(- 7)
= – 18 + 1 – 21 = – 38 ≠ 0
Hence, we can solve the given equations ¡n matrix inversion method.
Cofactor of 2 is A1 = + (- 8 – 1) = – 9
Cofactor of – 1 is B1 = – (4 – 3) = – 1
Cofactor of 3 is C1 = + (- 1 – 6) = – 7
Cofactor of – 1 is A2 = – (4 – 3) = – 1
Cofactor of 2 is B2 = + (- 8 – 9) = – 17
Cofactor of 1 is C2 = – (2 + 3) = – 5
Cot actor of 3 is A3 = + (- 1 – 6) = – 7
Cofactor of 1 is B3 = – (2 + 3) = – 5
Cofactor of – 4 is C3 = + (4 – 1) = 3
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 33

Question 38.
Solve x + y + z = 1, 2x + 2y + 3z = 6, x + 4y + 9z = 3 by using matrix inversion method.
Answer:
Given system of linear equations are
x + y + z = 1, 2x + 2y + 3z = 6, x + 4y + 9z = 3
Let A = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
2 & 2 & 3 \\
1 & 4 & 9
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\), D = \(\left[\begin{array}{l}
1 \\
6 \\
3
\end{array}\right]\)
Then we can write the given equations in the form of AX = D.
det A = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 2 & 3 \\
1 & 4 & 9
\end{array}\right|\) = 1 (18 – 12) – 1 (18 – 3) + 1 (8 – 2) = 1(6) – 1(15) + 1(6)
= 6 – 15 + 6 = – 3 ≠ 0
Hence we can solve the given equations using matrix inversion method.
Cofactor of 1 is A1 = + (18 – 12) = 6
Cofactor of 1 is B1 = – (18 – 3) = – 15
Col actor of 1 is C1 = + (8 – 2) = 6
Cofactor of 2 is A2 = – (9 – 4) = – 5
Cofactor of 2 is B2 = + (9 – 1) = 8
Cofactor of 3 is C2 = – (4 – 1) = – 3
Cofactor of 1 is A3 = + (3 – 2) = 1
Cofactor of 4 is B3 = – (3 – 2) = – 1
Cofactor of 9 is C3 = + (2 – 2) = O
∴ Cofactor matrix of A = B
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 34
∴ The solution of given system of equations is x = 7, y = – 10, z = 4.

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 39.
Solve the equations 2x – y + 3z = 8, – x + 2y + z = 4, 3x + y – 4z = 0 by Gauss – Jordan method.
Answer:
Given system of linear equations are 2x – y + 3z = 8; – x + 2y + z = 4; 3x + y – 4z = 0.
Matrix equation form is AX = D
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 35
In this case, the system of equations has unique solution. i.e., x = y = z = 2
∴ The solution of given system of equations is x = 2; y = 2; z = 2.

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 40.
Solve the equations 2x – y + 8z = 13, 3x + 4y + 5z = 18, 5x – 2y + 7z = 20 by Gauss – Jordan method.
Answer:
Given system of linear equations are 2x – y + 8z = 13, 3x + 4y + 5z = 18, 5x – 2y + 7z = 20.
Matrix equation form is AX = D
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 36
In this case, the system of equations has unique solution. i.e., x = 3, y = 1, z = 1.
∴ The solution of given system of equations is x = 3; y = 1; z = 1.

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Some More Maths 1A Matrices Important Questions

Question 1.
If A = \(\left[\begin{array}{ccc}
2 & 3 & 1 \\
6 & -1 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
1 & 2 & -1 \\
0 & -1 & 3
\end{array}\right]\) then find the matrix X such that A + B – X = 0. What is the order of the matrix X?
Answer:
Given A = \(\left[\begin{array}{ccc}
2 & 3 & 1 \\
6 & -1 & 5
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & 2 & -1 \\
0 & -1 & 3
\end{array}\right]\)
A and B are matrices of same order 2 × 3.
If A + B – X is to be defined the order of X also must also be 2 × 3.
Given A + B – X = 0 ⇒ X = A + B
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 37
∴ Order of X is 2 × 3.

Question 2.
Construct a 3 × 2 matrix whose elements are defined by aij = \(\frac{1}{2}\) |i – 3j|.
Answer:
In general a 3 × 2 matrix is given by
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 38

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 3.
If A = \(\left[\begin{array}{cc}
-1 & 3 \\
4 & 2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & 1 \\
3 & -5
\end{array}\right]\), X = \(\left[\begin{array}{ll}
\mathbf{x}_1 & \mathbf{x}_2 \\
\mathbf{x}_3 & \mathbf{x}_4
\end{array}\right]\) and A + B = X, then find the values of x1, x2, x3 and x4.
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 39

Question 4.
A certain book shop has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are Rs. 80, Rs. 60 and Rs. 40 each respectively. Using matrix algebra, find the total value of the books in the shop.
Answer:
Number of 3 types of books is expressed by the row matrix A
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 40
= [120 96 120]
Selling price of 3 types of books is expressed by the column matrix B
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 41
Total value of the books in the shop is given by AB
AB = [120 96 120] \(\left[\begin{array}{l}
80 \\
60 \\
40
\end{array}\right]\)
= [120 × 80 + 96 × 60 + 120 × 40]
= [9600 + 5760 + 4800]
= [20160]
∴ Total value of the books = Rs. 20160

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 5.
If A = \(\left[\begin{array}{ll}
2 & 1 \\
1 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
3 & 2 & 0 \\
1 & 0 & 4
\end{array}\right]\), find AB and BA, if it exists.
Answer:
Given A = \(\left[\begin{array}{ll}
2 & 1 \\
1 & 3
\end{array}\right]\), B = \(\left[\begin{array}{lll}
3 & 2 & 0 \\
1 & 0 & 4
\end{array}\right]\)
The order of matrix A is 2 × 2
The order of matrix 13 is 2 × 3
The no.of columns in A = The no.of rows in B
∴ AB is defined
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 42
The no.of columns in B ≠ The no.of rows in A
∴ BA is not defined.

Question 6.
Give examples of two square matrices A and B of the same order for which AB = 0. But BA ≠ 0.
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 43

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 7.
If A = \(\left[\begin{array}{rr}
7 & -2 \\
-1 & 2 \\
5 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
-2 & -1 \\
4 & 2 \\
-1 & 0
\end{array}\right]\) then find AB’ and BA’. [Mar. 18 (AP)]
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 44

Question 8.
Find the minors of – 1 and 3 in the matrix \(\left[\begin{array}{ccc}
2 & -1 & 4 \\
0 & -2 & 5 \\
-3 & 1 & 3
\end{array}\right]\).
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 45

Question 9.
Find the cofactors of the elements 2, – 5 in the matrix \(\left[\begin{array}{ccc}
-1 & 0 & 5 \\
1 & 2 & -2 \\
-4 & -5 & 3
\end{array}\right]\).
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 46

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 10.
Show that the determinant of skew – symmetric matrix of order three is always zero.
Answer:
Let A = \(\left[\begin{array}{ccc}
0 & -c & -b \\
c & 0 & -a \\
b & a & 0
\end{array}\right]\) is a skew – symmetric matrix of order ‘3’.
det A = 0(0 + a2) + c(0 + ab) – b(ac – 0)
= 0 + abc – abc = 0 + 0 = 0
∴ The determinant of skew symmetric matrix of order 3 is always zero.

Question 11.
Show that \(\left[\begin{array}{ccc}
\mathbf{y}+\mathbf{z} & \mathbf{x} & \mathbf{x} \\
\mathbf{y} & \mathbf{z}+\mathbf{x} & \mathbf{y} \\
\mathbf{z} & z & \mathrm{x}+\mathbf{y}
\end{array}\right]\) = 4xyz.
Answer:
LHS = \(\left[\begin{array}{ccc}
\mathbf{y}+\mathbf{z} & \mathbf{x} & \mathbf{x} \\
\mathbf{y} & \mathbf{z}+\mathbf{x} & \mathbf{y} \\
\mathbf{z} & z & \mathrm{x}+\mathbf{y}
\end{array}\right]\)
= (y + z) [(z + x) (x + y) – yz] – x[y(x + y) – yz] + x[yz – z(z + x)]
= (y + z) [zx + xy + zy + x2 – yz] – x[xy + y2 – yz] + x[yz – z2 – zx]
= xyz + xy2 + zy2 + x2y – y2z + z2 x + xyz + z2y + x2z – yz2 – x2y – xy2 + xyz + xyz – xz2 – zx2
= 4xyz
= RHS.

Question 12.
If Δ1 = \(\left|\begin{array}{ccc}
1 & \cos \alpha & \cos \beta \\
\cos \alpha & 1 & \cos \gamma \\
\cos \beta & \cos \gamma & 1
\end{array}\right|\), Δ2 = \(\left|\begin{array}{ccc}
0 & \cos \alpha & \cos \beta \\
\cos \alpha & 0 & \cos \gamma \\
\cos \beta & \cos \gamma & 0
\end{array}\right|\) and Δ1 = Δ2, then show that cos2α + cos2β + cos2γ = 1.
Answer:
Δ1 = \(\left|\begin{array}{ccc}
1 & \cos \alpha & \cos \beta \\
\cos \alpha & 1 & \cos \gamma \\
\cos \beta & \cos \gamma & 1
\end{array}\right|\)
= 1(1 – cos2γ) – cos α(cos α – cos β cos γ) + cos β (cos α cos γ – cos β)
= 1 – cos2 γ – cos2 α + cos α cos β cos γ + cos α cos β cos γ – cos2β
= 1 – cos2 γ – cos2α – cos2β + 2 cos α cos β cos γ

Δ2 = \(\left|\begin{array}{ccc}
0 & \cos \alpha & \cos \beta \\
\cos \alpha & 0 & \cos \gamma \\
\cos \beta & \cos \gamma & 0
\end{array}\right|\)
= 0(0 – cos2γ) – cos α (0 – cos γ cos β) + cos β (cos α cos γ – 0)
= cos α cos β cos γ + cos α cos β cos γ
= 2 cos α cos β cos γ

Given Δ1 = Δ2
1 – cos2 α – cos2β – cos2γ + 2 cos α cos β cos γ = 2 cos α cos β cos γ
1 – cos2α – cos2β – cos2γ = 0
cos2α + cos2 β + cos2 γ = 1.

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 13.
Show that \(\left|\begin{array}{ccc}
1 & a & a^2-b c \\
1 & b & b^2-c a \\
1 & c & c^2-a b
\end{array}\right|\) = 0.
Answer:
LHS = \(\left|\begin{array}{ccc}
1 & a & a^2-b c \\
1 & b & b^2-c a \\
1 & c & c^2-a b
\end{array}\right|\)
= 1(bc2 – ab2 – b2c + c2a) – a(c2 – ab – b2 + ac) + (a2 – bc) (c – b)
= bc2 – ab2 – b2c + c2a – ac2 + a2b + ab2 – a2c + a2c – a2b – bc2 . cb2 = 0
= RHS.

Question 14.
Solve the following system of equations by using Cramer’s rule. [Mar. 15 (TS)]
x – y + 3z = 5, 4x + 2y – z = 0, – x + 3y + z = 5
Answer:
Given system of equations can be written as:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 47

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 15.
If A = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\) and B = \(\frac{1}{2}\left[\begin{array}{lll}
\mathbf{b}+\mathbf{c} & \mathbf{c}-\mathbf{a} & \mathbf{b}-\mathbf{a} \\
\mathbf{c}-\mathbf{b} & \mathbf{c}+\mathbf{a} & \mathbf{a}-\mathbf{b} \\
\mathbf{b}-\mathbf{c} & \mathbf{a}-\mathbf{c} & \mathbf{a}+\mathbf{b}
\end{array}\right]\) then show that ABA-1 is a diagonal matrxi.
Answer:
Given A = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\),
B = \(\frac{1}{2}\left[\begin{array}{lll}
\mathbf{b}+\mathbf{c} & \mathbf{c}-\mathbf{a} & \mathbf{b}-\mathbf{a} \\
\mathbf{c}-\mathbf{b} & \mathbf{c}+\mathbf{a} & \mathbf{a}-\mathbf{b} \\
\mathbf{b}-\mathbf{c} & \mathbf{a}-\mathbf{c} & \mathbf{a}+\mathbf{b}
\end{array}\right]\)
Cot actor of 0 is A1 = + (0 – 1) = – 1
Cofactor of ‘1’ is B1 = – (0 – 1) = 1
Cofactor of 1 is C1 = + (1 – 0) = 1
Cofactor of 1 is A2 = – (0 – 1) = 1
Cofactor of 0 is B2 = + (0 – 1) = – 1
Cofactor of 1 is C2 = – (0 – 1) = 1
Cofactor of 1 is A3 = + (1 – 0) = 1
Cofactor of 1 is B3 = – (0 – 1) = 1
Cofactor of 0 is C3 = +(0 – 1) = – 1
∴ Cofactor matrix of
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 48
det A = 0(0 – 1) – 1 (0 – 1) + 1 (1 – 0)
= 0 + 1 + 1 = 2 ≠ 0
∴ A is invertiable.
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 44

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 16.
If A = \(\left[\begin{array}{rrr}
3 & -3 & 4 \\
2 & -3 & 4 \\
0 & -1 & 1
\end{array}\right]\) then show that A-1 = A3
Answer:
Given A = \(\left[\begin{array}{lll}
3 & -3 & 4 \\
2 & -3 & 4 \\
0 & -1 & 1
\end{array}\right]\)

Cofactor of 3 is A1 =+(- 3 + 4) = 1
Cofactor of – 3 is B1 = – (2 – 0) = – 2
Cofactor of 4 is C1 = (- 2 + 0) = – 2
Cofactor of 2 is A2 = – (- 3 + 4) = – 1
Cofactor of – 3 is B2 = + (3 – 0) = 3
Cofactor of 4 is C2 = – (- 3 + 0) =3
Cofactor of 0 is A3 = + (- 12 + 12) = 0
Cofactor of – 1 is B3 = – (12 – 8) = – 4
Cofactor of 1 is C3 = + (- 9 + 6) = – 3
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 50

Question 17.
For any square matrix A, show symmetric. [Mar. 15 (AP)]
Answer:
Let ‘A” be a square matrix
(AA’)’ = (A’)’ A’ = AA’
∴ (AA’)’ = AA’
⇒ AA’ is a symmetric matrix.

Question 18.
Find the rank of the matrix \(\left[\begin{array}{ccc}
1 & 4 & -1 \\
2 & 3 & 0 \\
0 & 1 & 2
\end{array}\right]\).
Answer:
Let A = \(\left[\begin{array}{ccc}
1 & 4 & -1 \\
2 & 3 & 0 \\
0 & 1 & 2
\end{array}\right]\)
det A = 1(6 – 0) – 4(4 – 0) – 1(2 – 0)
= 6 – 16 – 2 – 12 ≠ 0
∴ A is a non – singular.
Hence Rank (A) = 3.

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 19.
Find the rank of the matrix \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]\). [Mar. 19 (AP), Mar. 15 (TS)]
Answer:
Let A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]\)
det A = 1 (6 – 4) – 2 (4 – 0) + 3 (2 – 0) = 2 – 8 + 6 = 0
Since det A = 0, Rank (A) ≠ 3.
Now, \(\left[\begin{array}{ll}
1 & 2 \\
2 & 3
\end{array}\right]\) is a sub matrix of ‘A’ whose determinant is 3 – 4 = – 1 ≠ 0.
Hence Rank (A) = 2.

Question 20.
Solve the following system of homogeneous equations x – y + z = 0, x + 2y – z = 0, 2x + y + 3z = 0. [Mar.16 (TS)]
Answer:
The coefficient matrix is \(\left[\begin{array}{ccc}
1 & -1 & 1 \\
1 & 2 & -1 \\
2 & 1 & 3
\end{array}\right]\)
Its determinant is 1(6 + 1) + 1(3 + 2) + 1(1 – 4) = 1(7) + 1(5) + 1(- 3) = 7 + 5 – 3 = 9
Hence the system has the trivial solution x = y = z = 0 only.

Question 21.
Solve the following system of equations by using Matrix inversion method.
2x – y + 3z = 9, x + y + z = 6, x – y + z = 2. [Mar. 16 (TS)]
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 51

TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type

Question 22.
Solve x + y + z = 9, 2x + 5y + 7z = 52 and 2x + y – z = 0 by using matrix inversion method. [Mar. 17 (AP)]
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 52

Question 23.
Solve the following system of equations by Cramer’s rule: 2x – y + k = 8, – x + 2y + z = 4, 3x + y – 4z = 0 [Mar. 18 (TS)]
Answer:
Given equations are
2x – y + 3z = 8,
– x + 2y + z = 4,
3x + y – 4z = 0
TS Inter First Year Maths 1A Matrices Important Questions Long Answer Type 53

TS Inter 1st Year Maths 1A Matrices Important Questions Short Answer Type

Students must practice these Maths 1A Important Questions TS Inter 1st Year Maths 1A Matrices Important Questions Short Answer Type to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Matrices Important Questions Short Answer Type

Question 1.
If A = \(\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\), , then show that for all the positive integers n, An = \(\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\) [May 98, 91]
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type 1
∴ S(k + 1) is true.
∴ By the principle of mathematical induction, S(n) is true for all n ∈ N.
∴ An = \(\left[\begin{array}{cc}
\cos n \theta & \sin n \theta \\
-\sin n \theta & \cos n \theta
\end{array}\right]\), ∀ n ∈ N.

TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type

Question 2.
If A = \(\left[\begin{array}{rrr}
1 & -2 & 1 \\
0 & 1 & -1 \\
3 & -1 & 1
\end{array}\right]\), then find A3 – 3A2 – A – 3I, where I is unit matrix of order 3. [Mar. 19 (TS); Mar. 11, 98; May 98]
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type 2

Question 3.
If I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) and E = \(\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]\), then show that (aI + bE)3 = a3I + 3a2bE, where I is unit matrix of order 2. [Mar. 16 (TS), 15(AP), 10; May 05]
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type 3

TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type

Question 4.
If θ – Φ = \(\frac{\pi}{2}\), then show that \(\left[\begin{array}{cc}
\cos ^2 \theta & \cos \theta \sin \theta \\
\cos \theta \sin \theta & \sin ^2 \theta
\end{array}\right]\left[\begin{array}{cc}
\cos ^2 \phi & \cos \phi \sin \phi \\
\cos \phi \sin \phi & \sin ^2 \phi
\end{array}\right]\) = 0 [May 15 (TS); May 11, 09, 96; Mar. 04]
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type 4

Question 5.
If A = \(\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right]\), then show that An = \(\left[\begin{array}{cc}
1+2 n & -4 n \\
n & 1-2 n
\end{array}\right]\) for any integer n ≥ 1, by using mathematical induction. [May 08, 02]
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type 5
∴ S(k + 1) is true.
∴ By using the principle of mathematical Induction, S(n) is true for all n ∈ N.
∴ An = \(\left[\begin{array}{cc}
1+2 n & -4 n \\
n & 1-2 n
\end{array}\right]\) ∀ n ∈ N.

TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type

Question 6.
For any n × n matrix A, prove that A can be uniquely expressed as a sum of a symmetric matrix and a skew symmetric matrix. (Mar. ‘03)
Answer:
Let A be a square matrix.
TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type 6
∴ A can be expressed as a sum of a symmetric matrix and a skew symmetric matrix.

Question 7.
Show that \(\left|\begin{array}{ccc}
\mathbf{1} & \mathbf{a} & \mathbf{a}^2 \\
\mathbf{1} & \mathbf{b} & \mathbf{b}^2 \\
\mathbf{1} & \mathbf{c} & \mathbf{c}^2
\end{array}\right|\) = (a – b) (b – c) (c – a). [Mar. 17 (TS). 05]
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type 7
= (a – b) (b – c) (c – a) [0(c2 – c) – 1(0 – 1) + (a + b) (0 – 0)]
= (a – b) (b – c) (c – a) (1) = (a – b) (b – c) (c – a) = RHS

Question 8.
Show that \(\left|\begin{array}{lll}
b c & b+c & 1 \\
c a & c+a & 1 \\
a b & a+b & 1
\end{array}\right|\) = (a – b) (b – c) (c – a). [Board Paper]
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type 8
= (a – b) (c – a) [b (1 – 0) – 1 (c – 0) + 0 (ac + bc – ab)]
= (a – b) (c – a) [b – c] = (a – b) (b – c) (c – a) = R.H.S.

TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type

Question 9.
Show that \(\left|\begin{array}{ccc}
b+c & c+a & a+b \\
a+b & b+c & c+a \\
a & b & c
\end{array}\right|\) = a3 + b3 + c3 – 3abc. [May 13, 07; Mar. 08]
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type 9
= (a + b + c) [c2 – bc – ac + ab + a2 – 2ab + b2]
= (a + b + c) [a2 + b2 + c2 – ab – bc – ca]
= a3 + b3 + c3 – 3abc

Question 10.
If \(\left|\begin{array}{ccc}
a & a^2 & 1+a^3 \\
b & b^2 & 1+b^3 \\
c & c^2 & 1+c^3
\end{array}\right|\) = 0 and \(\left|\begin{array}{lll}
a & a^2 & 1 \\
b & b^2 & 1 \\
c & c^2 & 1
\end{array}\right|\) ≠ 0 then show that abc = – 1. [Mar. 14, 04; May. 98, 95]
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type 10

TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type

Question 11.
Show that \(\left|\begin{array}{lll}
a-b & b-c & c-a \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\) = 0. [May. 08]
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type 11

Question 12.
Let A and B be invertiable matrices then show that (AB)-1 = B-1 A-1. [May. 03]
Answer:
A is invertible matrix then A-1 exists and AA-1 = A-1 A = I
B is an invertible matrix then B-1 exists and BB-1 = B-1B = I
Now (AB) (B-1 A-1) = A(BB-1)A-1 = A(T)A-1 = AA-1 = I
∴ (AB) (B-1 A-1) = I ………………… (1)
(B-1 A-1) (AB) = B-1 (A-1 A) B = B-1 (I) B = B-1 B = 1
∴ (B-1 A-1) (AB) = I ………………….. (2)
From (1) & (2)
(AB) (B-1 A-1) = (B-1 A-1) (AB) = I
AB is invertiable and (AB)-1 = B-1 A-1.

Question 13.
Find the adjoint and the inverse of the matrix A = \(\left[\begin{array}{lll}
1 & 3 & 3 \\
1 & 4 & 3 \\
1 & 3 & 4
\end{array}\right]\) [May 14; Mar. 08]
Answer:
Given A = \(\left[\begin{array}{lll}
1 & 3 & 3 \\
1 & 4 & 3 \\
1 & 3 & 4
\end{array}\right]\)
Cofactor of 1 is A11 = +(16 – 9) = 7
Cofactor of 3 is A12 = – (4 – 3) = – 1
Cofactor of 3 is A13 = (3 – 4) = – 1
Cofactor of 1 is B11 = – (12 – 9) = -3
Cofactor of 4 is B12 = (4 – 3) = 1
Cofactor of 3 is B13 = – (3 – 3) = 0
Cofactor of 1 is C11 = (9 – 12) = – 3
Cofactor of 3 is C12 = – (3 – 3) = 0
Cot actor of 4 is C13 = (4 – 3) = 1
∴ Cofactor matrix,
TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type 12
Now, det A = 1(16 – 9) – 3(4 – 3) + 3 (3 – 4)
= 1(7) – 3(1) + 3(- 1) = 7 – 3 – 3 = 1
Hence A is invertiable.
A-1
TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type 13

TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type

Question 14.
Show that A = \(\left[\begin{array}{lll}
1 & 2 & 1 \\
3 & 2 & 3 \\
1 & 1 & 2
\end{array}\right]\) is non-singular and find A-1. [Mar. 17 (TS). 12, 98; May 89]
Answer:
Given A = \(\left[\begin{array}{lll}
1 & 2 & 1 \\
3 & 2 & 3 \\
1 & 1 & 2
\end{array}\right]\)
det A = 1 (4 – 3) – 2 (6 – 3) + 1 (3 – 2)
= 1 – 6 + 1 = – 4 ≠ 0
∴ A is a non-singular matrix.
Cofactor of 1 is A1 = + (4 – 3) = 1
Cofactor of 2 is B1 = – (6 – 3) = – 3
Cofactor of 1 is C1 =+(3 – 2) = 1
Cofactor of 3 is A2 = – (4 – 1) = – 3
Cofactor of 2 is B2 = + (2 – 1) = 1
Cofactor of 3 is C2 = – (1 – 2) = + 1
Cofactor of 1 is A3 = + (6 – 2) = 4
Cofactor of 1 is B3 = – (3 – 3) = 0
Cofactor of 2 is C3 = + (2 – 6) = – 4
∴ Cofactor matrix of A is B
TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type 14

Find the adjoint and the inverse of the matrix \(\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 1 & 0 \\
3 & 2 & 1
\end{array}\right]\). [Mar. 05; May 98]
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type 15

Find the adjoint and the inverse of the matrix \(\left[\begin{array}{lll}
2 & 1 & 2 \\
1 & 0 & 1 \\
2 & 2 & 1
\end{array}\right]\). [Mar. 08, 89]
Answer:
\(\left[\begin{array}{rrr}
-2 & 3 & 1 \\
1 & -2 & 0 \\
2 & -2 & -1
\end{array}\right],\left[\begin{array}{rrr}
-2 & 3 & 1 \\
1 & -2 & 0 \\
2 & -2 & -1
\end{array}\right]\)

TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type

Question 15.
If A = \(\left[\begin{array}{rrr}
-1 & -2 & -2 \\
2 & 1 & -2 \\
2 & -2 & 1
\end{array}\right]\), then show that the adjoint of A is 3A’. Find A-1. [Mar. 19 (AP), May 08]
Answer:
Given A = \(\left[\begin{array}{rrr}
-1 & -2 & -2 \\
2 & 1 & -2 \\
2 & -2 & 1
\end{array}\right]\)

Cofactor of – 1 is A1 = + (1 – 4) = – 3
Cofactor of – 2 is B1 = – (2 + 4) = – 6
Cofactor of – 2 is C1 = + (- 4 – 2) = – 6
Cofactor of 2 is A2 = – (- 2 – 4) = 6
Cofactor of 1 is B2 = + (- 1 + 4) = 3
Cofactor of – 2 is C2 = – (2 + 4) = – 6
Cofactor of 2 is A3 = + (4 + 2) = 6
Cofactor of – 2 is B3 = – (2 + 4) = – 6
Cofactor of 1 is C3 = + (- 1 + 4) = 3
∴ Cofactor matrix of
TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type 16
∴ Adj A = 3A’
det A = – 1 (1 – 4) + 2 (2 + 4) – 2 (- 4 – 2)
= – 1 (- 3) + 2(6) – 2 (- 6) = + 3 + 12 + 12 = 27 ≠ 0
∴ A is invertiable.
A-1 = \(\frac{{adj} A}{{det} A}=\frac{1}{27}\left[\begin{array}{ccc}
-3 & 6 & 6 \\
-6 & 3 & -6 \\
-6 & -6 & 3
\end{array}\right]\)

Question 16.
If abc ≠ 0, find the inverse of \(\left[\begin{array}{lll}
a & 0 & 0 \\
0 & b & 0 \\
0 & 0 & c
\end{array}\right]\). [Mar. 06; Oct. 96]
Answer:
Let A = \(\left[\begin{array}{lll}
a & 0 & 0 \\
0 & b & 0 \\
0 & 0 & c
\end{array}\right]\)
Cofactor of a is A1 + (bc – 0) = bc
Cofactor of 0 is B1 = – (0 – 0) = 0
Cofactor of 0 is C1 = + (0 – 0) = 0
Cofactor of 0 is A2 = – (0 – 0) = 0
Cofactor of b is B2 = + (ac – 0) = ac
Cofactor of 0 is C2 = – (0 – 0) = 0
Cofactor of 0 is A3 = + (0 – 0) = 0
Cofactor of 0 is B3 = – (0 – 0) = 0
Cofactor of c is C3 = + (ab – 0) = ab
∴ Cofactor matrix of
TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type 17

TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type

Question 17.
If 3A = \(\left[\begin{array}{rrr}
1 & 2 & 2 \\
2 & 1 & -2 \\
-2 & 2 & -1
\end{array}\right]\), then show that A-1 = A’. [Mar. 14, 09; May. 12]
Answer:
TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type 18

Question 18.
If A = \(\left[\begin{array}{ccc}
1 & -2 & 3 \\
0 & -1 & 4 \\
-2 & 2 & 1
\end{array}\right]\), then find (A’)-1. [Board Paper]
Answer:
Given A = \(\left[\begin{array}{ccc}
1 & -2 & 3 \\
0 & -1 & 4 \\
-2 & 2 & 1
\end{array}\right]\)
Cofactor of 1 is A1 = + (- 1 – 8) = – 9
Cofactor of 0 is A2 = – (- 2 – 6) = 8
Cofactor of – 2 is A3 = (- 8 + 3) = – 5
Cofactor of – 2 is B1 = – (0 + 8) = – 8
Cot actor of – 1 is B2 = + (1 + 6) = 7
Cofactor of 2 is B3 = – (4 – 0) = – 4
Cofactor of 3 is C1 = + (0 – 2) = – 2
Cofactor of 4 is C2 = – (2 – 4) = 2
Cofactor of 1 is C3 = (- 1 + 0) = – 1
TS Inter First Year Maths 1A Matrices Important Questions Short Answer Type 19

TS Inter 2nd Year English Study Material Chapter 4 Awake

Telangana TSBIE TS Inter 2nd Year English Study Material 4th Lesson Awake Textbook Questions and Answers.

TS Inter 2nd Year English Study Material 4th Lesson Awake

Annotations (Section A, Q.No. 2, Marks: 4)
Annotate the following in about 100 words each.

a) Who kneel in thy presence to serve and adore thee!

Introduction: This beautiful line is taken from the patriotic lyric”Awake” written by Sarojini Naidu. She is fondly remembered as the Nightingale of India. This is the last poem in the part of “The Broken wing: Songs of Love”.

Context and meaning: This poem is a fervent appeal for unity and action. That call is at once artistic and rhetoric. First, the nation is personified as our mother. All Indians become her children. The poet visualizes all children praying to he mother. They, entreat her to give them a chance to serve her and admire her glory. This kind of visualization is a powerful way to persuade everyone to tread that path.

Critical Comment: The poem expresses strong nationalistic feelings. It was written more than a century ago. Yet, it is very relevant today.

కవి పరిచయం : ఈ అందమైన వాక్యం సరోజినీ నాయుడు వ్రాసిన దేశభక్తి గీతం “మేల్కొను” నుండి తీసుకోబడింది. ఆమెను భారతదేశపు గానకోకిలగా ప్రేమగా గుర్తుంచుకుంటారు. “The Broken Wing: Songs of Love” లో ఇది చివరి కవిత.

సందర్భ౦ మరియు అర్థం : పద్య౦ ఐక్యత మరియు కార్యాచరణ కోసం తీవ్రమైన విజ్ఞప్తి. ఆ పిలుపు ఒకేసారి కళాత్మకంగా మరియు వాక్చాతుర్యంగా ఉంటుంది. మొదటగా, దేశం మన తల్లిగా వ్యక్తీకరించబడింది. భారతీయులందరూ ఆమె బిడ్డలయ్యారు. పిల్లలందరూ తల్లిని ప్రార్థిస్తున్నట్లు కవి దృశ్యమానం చేస్తుంది. ఆమెకు సేవ చేయడానికి మరియు ఆమె కీర్తిని మెచ్చుకునే అవకాశం ఇవ్వాలని వారు ఆమెను వేడుకుంటారు. ఈ రకమైన దృశ్యమానం ప్రతి ఒక్కరినీ ఆ మార్గంలో నడపడానికి ఒప్పించే శక్తివంతమైన మార్గం.

విమర్శ : బలమైన జాతీయ భావాలను తెలియజేస్తుంది ఈ పద్య౦. శతాబ్దపు కాలానికి ముందే ఇది వ్రాయబడింది. అయినప్పటికీ, ప్రస్తుతానికి చాలా సంబంధం కల్గి ఉంది. అది నేటికీ చాలా సందర్భోచితమైనది.

b) Awaken and sever the woes that enthral us (Revision Test – IV)

Introduction: This patriotic line is taken from the patriotic lyric, ” Awake” Written by Sarojini Naidu, the nightingale of India. This is the last poem of “The broken wings: Songs of Love”.

Context and meaning: It is a soul stirring plea for action and zenity. The woes of bondage are to be cut. Mother India should gain its glory gain and grow and glow. Hence, the children implore the mother to rouse them from their slumber and to cut off the ties of bondage in which they are bound at present their hands are to be purified by her so that they might be prompted to undertake more and more deed of victory and triumph.

Critical Comment: The poem expresses strong nationalistic feelings. It was written more than a century ago. Yet, it is very relevant today.

కవి పరిచయం : ఈ దేశభక్తిని తెలియజేయు వాక్యం దేశభక్తి గేయం. భారత గానకోకిల సరోజినీ నాయుడు వ్రాసిన “మేల్కొను” నుండి తీసుకోబడింది. “ది బ్రోకెన్ వింగ్: సాంగ్స్ ఆఫ్ లవ్” లో ఇది చివరి కావ్యం. ఈ పద్య౦.

సందర్భ౦ మరియు అర్థం : ఐక్యతను మరియు కార్యాచరణను ఇది మనస్సును కదిలించే కావ్యం. బానిస బాధలు తుంచివేయాలి. భరతమాత మరల తన కీర్తిని, ఎదుగుదలను మరియు కాంతి, వెలుగు పొందాలి. కాబట్టి భరతమాత బిడ్డలు, ప్రస్తుతం వారు బందీలై ఉన్న బానిస సంకెళ్ళను తుంచివేయుటకు ఆమెను నిద్రావస్థితి నుండి మేల్కొలపమని భరతమాతను ప్రార్థించుచున్నారు. విజయోన్ముకులు అవ్వటానికి, వారి చేతులను పవిత్రం చేయమని తల్లిని కోరుకుంటున్నారు.

విమర్శ: గట్టి జాతీయభావాలను ఈ పద్య౦ తెలియజేస్తుంది. ఒక శతాబ్దపు కాలానికి క్రితమే ఈ పద్య౦ వ్రాయబడింది. అయినప్పటికీ, ఈ రోజులకు బాగా సంబంధించింది.

TS Inter 2nd Year English Study Material Chapter 4 Awake

c) Ne’er shall we fail thee, forsake thee or falter,
whose hearts are thy home and thy shield and thine altar

Introduction: These patriotic lines are taken from the patriotic lyric, “Awake”, written by Sarojini Naidu the nightingale of India. This is the last poem in the part of “The Broken wing: songs of Love”

Context and meaning: Here, the poetess says that the children of the mother land love her very much. They have a great devotion for their mother land. They are the true children they have inherited her pride, moral and spiritual strength. They want to preserve these qualities. The y will never fail to protect her. They will never desert her. Their hearts are the home of the mother. They are her shield with which they would protect her. They are her altar at which they would worship her. It is with her help that they will be set again in the forefront of glory.

Critical Comment: The poem expresses strong nationalistic feelings It was written more than a century ago. Yet it is very relevant today.

కవి పరిచయం : ఈ దేశభక్తి పంక్తులు భారత గానకోకిల సరోజినీ నాయుడుచే రచించబడిన “మేల్కొను” అను దేశభక్తి గేయం నుండి తీసుకొనబడింది. “The Broken Wing: Songs of Love” లో ఇది చివరి వాక్యం.

సందర్భ౦ మరియు అర్థం : ఇక్కడ, మాతృభూమి బిడ్డలు ఆమెను అమితంగా ప్రేమిస్తున్నారంటుంది కవయిత్రి. వారికి మాతృభూమిపట్ల గొప్ప భక్తి ఉంది. వారి ఆమె (భరతమాత) నిజమైన బిడ్డలు. భరతమాత గర్వం, హుందా, నీతి మరియు ఆధ్యాత్మిక బలాన్ని వారసత్వంగా వారు పొందారు. ఈ లక్షణాలను నిలుపుకోవాలనుకుంటున్నారు వారు. ఎప్పటికీ ఆమెను వదులుకోరు వారు. వారి హృదయాల్లో భరతమాత విలువైనది. వారు ఆమె రక్షణ కవచాలు ఆమెను రక్షించుకుంటారు. ఆమెను పూజించే పూజాపీఠం వారు ఆమె (ఖర్చులు) సహాయంతో వారు (భారతీయులు) ఆమెను మురళీ కీర్తి ప్రతిష్టలు ముందు నిలుపుతారు.

విమర్శ : ధృఢమైన జాతీయ భావాలను ప్రదర్శిస్తుంది ఈ పద్య౦. ఒక శతాబ్దకాలం క్రితమే వ్రాయబడింది ఈ పద్య౦. అయినప్పటికి

d) …………………. Hearken,
O queen and O goddess, we hail thee! (Revision Test – IV)

Introduction: These are the concluding lines of the patriotic lyric, “Awake” written by sarojini Naidu, the Nightingale of India. This is the last poem of “The Broken wing! Songs of Love”.

Context and meaning: Here, people of all Indian religions assure her chat they would serve her with devotion to the best of their ability. They call upon their great mother, their queen and their goddess to listen to their prayer and awake from her present slumber. Their fearless united and devoted efforts shall certainly be sufficient to free her from her present bondage.

Critical Comment: This lyric is a soul stirring call for unity and action. It expresses strong nationalistic feelings.

కవి పరిచయం : భారత గాన కోకిల సరోజినీ నాయుడిచే రచించబడిన “మేల్కొను” అను దేశభక్తి గేయంలోని ముగింపు పంక్తులు ఇవి “ది బ్రోకెన్ వింగ్ : సాంగ్స్ ఆఫ్” లో ఇది చివరి కావ్యం.

సందర్భ౦ మరియు అర్థం : ఇక్కడ, అన్ని భారతీయ మతాల ప్రజలు ఆమెను తమ శక్తి మేరకు భక్తితో సేవ చేస్తామని ఆమెకు భరోసా ఇస్తారు. వారు తమ గొప్పతల్లిని, వారి రాణిని మరియు వారి దేవతను తమ ప్రార్థనను వినమని మరియు ఆమె ప్రస్తుత నిద్ర నుండి మేల్కొలపమని పిలుపునిచ్చారు వారు. వారి నిర్భయమైన ఐక్యత మరియు అంకితమైన ప్రయత్నాలు ఆమె ప్రస్తుత బానిసత్వం నుండి ఆమెను విడిపించడానికి ఖచ్చితంగా సరిపోతాయి అంటారు.

విమర్శ : ఈ గీతం ఐక్యత మరియు కార్యచర్య కోసం ఒక ఆత్మను (మనస్సు) కదిలించే పిలుపు. ఇది బలమైన జాతీయ భావాలను వ్యక్తపరుస్తుంది.

Paragraph Questions & Answers (Section A, Q.No.4, Marks: 4)
Answer the following Questions in about 100 words

a) Substantiate the critical comment that the poem”Awake” is a patriotic lyric. (Revision Test – IV)
Answer:
Yes, Sarojini Naidu’s song”Awake’ has been acclaimed as a patriotic lyric. A lyric is poem. A single speaker expresses emotions and thoughts in it. It is noteworthy for its musical quality and rhythm. And its theme is losty. According to it, the poem, “Awake”, has all these qualities in abundance. It is rich in musical element. If expresses strong nationalistic feelings. Thus, the content is patriotic and lience noble. With its preformed theme and artistic and lyrical form the poem is very much entitled to be applauded as patriotic lyric. It is a touching call to all Indians for unity and action. People of all religions pledge to come together to guard their mother, queen and goddess. Hence, this is a patriotic lyric.

అవును, సరోజినీ నాయుడు యొక్క పాట “మేల్కొను” పాట దేశభక్తి గీతంగా ప్రశంసించబడింది. గీతం అనేది ఒక పద్య౦. ఒకే వక్త దానిలో భావోద్వేగాలను మరియు ఆలోచనలను వ్యక్తపరుస్తారు. ఇందులో. ఇది దాని సంగీత నాణ్యత మరియు లయకు ముఖ్యమైంది. మరియు దీని ఇతివృత్తం ఉన్నతమైంది. దీనికి అణుగుణంగా, ఈ పద్య౦ ‘మేల్కొను” ఈ లక్షణాలన్నీ పుష్కలంగా ఉన్నాయి. ఇది సంగీత లక్షణంలో సమృద్ధిగా ఉంటుంది.

ఇది బలమైన జాతీయ భావాలను వ్యక్తపరుస్తుంది. అందువలన, ఇతివృత్తం దేశభక్తి మరియు సాహిత్య రూపంతో ఈ పద్య౦ దేశభక్తి గీతంగా ప్రశంసించబడటానికి చాలా అర్హతగల్గినది. ఐక్యత మరియు కార్యచరణ కోసం భారతీయులందరికీ ఇది హత్తుకునే పిలుపు. అన్ని మతాల ప్రజలు తమ తల్లి, రాణి మరియు దేవతను కోపాడుకోవడానికి కలిసి వస్తామని ప్రతిజ్ఞ చేస్తారు. అందుకే, ఇది దేశభక్తి గీతం.

TS Inter 2nd Year English Study Material Chapter 4 Awake

b) What do the children implore the mother in the poem, “Awake” ?
Answer:
The patriotic lyric, “Awake” is written by Sarojini Naidu, The nightingale of India. The poem is a servent appeal for unity and action. The nation is personified as our mother. All Indians becomes her children. The poet visualizes all children praying to the mother. They are invoking her to rise from her sound and long sleep and to bless them. So that they may succeed in their holy object of setting her free from the chain of slavery under the British rule. Thus, they implore her to awake from her sleep. They are ready to serve her. It is a soul storing plea for action and unity. The poem stands out for its strong nationalistic feelings.

దేశభక్తిగేయం “మేల్కొను” భారత గాన కోకిల, సరోజినీ నాయుడుచే రచింపబడింది. ఈ పద్య౦ ఐక్యత మరియు కార్యచరణ కోసం తీవ్రమైన విజ్ఞప్తి. దేశం మన తల్లిగా వర్తించబడింది. భారతీయులందరూ ఆమె బిడ్డలయ్యారు. పిల్లందరూ తల్లిని ప్రార్థిస్తున్నట్లు కవి దృశ్యమానం చేస్తారు. బ్రిటీష్ పాలనలో బానిసత్య సంకెళ్ళు నుండి ఆమెను విడిపించే వారి పవిత్రమైన లక్ష్యంలో వారు విజయం సాధించాలని ఆమె దీర్ఘ, ఘాడమైన నిద్ర నుండి లేచి వారిని ఆశీర్వదించమని ప్రార్థిస్తున్నారు. అందువలన, వారు ఆమెను నిద్ర నుండి మేల్కొనమని వేడుకుంటున్నారు. వారు ఆమెకు సేవ చేయడానికి సిద్ధంగా ఉన్నారు. ఇది కార్యాచరణ మరియు ఐక్యత కోసం ఆత్మను, మనస్సును కదిలించే అభ్యర్ధన. ఈ పద్య౦దాని బలమైన, గట్టి, జాతీయవాద భావాలకు చిహ్నంగా నిలుస్తుంది.

c) How do Indians plan to set their mother again in the forefront of glory? (Revision Test – IV)
Answer:
The patriotic lyric, “Awake” is written by Sarojini Naidu, The nightingale of India. According to the poetess, the children of the motherland are invoking her to rise from her sound and long sleep and bless them so that they succeed in their holy object of setting her free from the chain of slavery under, the British rule. The poetess says that her children love her very much. They have a great devotion for their motherland.

They are her true children. They have inherited her pride, moral and spiritual strength they want to preserve these qualities. They will never fail to protect her. They will never desert her. Their hearts are the home of the mother. They are moreover her shield and her alfar. It is with her help that they will be set again in the forefront of glory. All medians should forget their mutual differences for the attainment of freedom and rise against British tyranny.

దేశభక్తి గీతం ‘మేల్కొను’ భారత గాన కోకిల సరోజినీ నాయుడుచే రచింపబడింది. రచయిత ఉద్దేశం, భరతమాత బిడ్డలు, బ్రిటీషు పాలనలో బానిస సంకెళ్ళ నుండి భరతమాతను విడిపిచు వారి పవిత్రమైన లక్ష్యంలో వారు విజయం సాధించాలని ఆమె ఘాడమైన, సుదీర్ఘ నిద్ర నుండి లేచి వారిని ఆశీర్వదించమని ప్రార్థిస్తున్నారు. వారు ఆమెను అమితంగా ప్రేమిస్తున్నారని కవయిత్రి చెప్తుంది. వారు ఆమె నిజమైన బిడ్డలు. ఆమె హుందాను, నీతి మరియు ఆధ్యాత్మిక శక్తిని వారు హక్కుగా పొందినారు. ఈ లక్షణాలను వారు నిలుపుతారు.

ఆమెను రక్షించుటలో వారు ఎన్నడూ విఫలమవ్వరు. ఎన్నడూ ఆమెను వదులుకోరువారు. తల్లికి వారి హృదయాల్లో నివాసం ఏర్పరిచారు. వారు, ఆత్మీయులుగా, ఆమె రక్షణ కవచం మరియు పూజాపీఠం. ఆమె సహాయంతో వారు ఆమెను కీర్తి, ప్రతిష్టల ముందు తిరిగి ఉంచుతారు. బ్రిటీషు వారి క్రూరత్వం, నిరంకుశంకు వ్యతిరేకంగా మరియు స్వాతంత్య్రం సాధించుటకును భారతీయులందరూ వారే పరస్పర వ్యత్యాసాలు మరచి పోవాలి అంటుంది ఐకమత్యంతో పోరాడాలి.

d) What do the children of all creeds promise their mother separately and collectively?
Answer:
The patriotic lyric, “Awake” is written by Sarojini Naidu, the nightingale of India. The poetess says that the children of old creeds promise their mother that they would serve her with devotion to the best of their ability. The Hindus Assure her that they will worship her usual.

The parsees will dedicate the fire of hope burning in their hearts to her service. The muslims will defend her with the sword of their love. The christians convince her that they will show the same devotion and faith in her as they have in Jesus and Mary.

Having assured her separately of their devotion, love and faith followers of different religions call upon their great mother, their queen and their goddess to listen to their prayer and rise from her current sleep they promise her that their united and collective efforts will succeed to set her free from the chains of bondage.

దేశభక్తి గీతం ‘మేల్కొను’ భారత గాన కోకిలచే రచింపబడింది. వారి శక్తి కొలది అంకిత భావంతో భరతమాతకు సేవచేస్తామని, ఆమెకై పోరాడతామని, అన్ని మతాల వారు వారి తల్లి (భరతమాత)కి ప్రతిజ్ఞ చేసారని కవి చెప్తుంది. హిందువులు వారి సహజమైన పద్ధతిలోనే పూలతో ఆమెను పూజిస్తామని చెప్తున్నారు. ఆమె సేవకు పార్శీనులు వారి హృదయాల్లోని ఆశాజ్యోతిని ఆమెకు అంకితం చేస్తామన్నారు. వారి ప్రేమ భక్తి ఖడ్గంతో ఆమెను కాపాడుకుంటామని ముస్లీంలు మారుస్తున్నారు. క్రీస్తు మరియు మేరిమాత పలు చూపిన విశ్వాసం, భక్తినే ఆమెపట్ల చూపుతామంటున్నారు.

క్రిస్టియన్లు వారి భక్తి ప్రేమ మరియు వివ్వాసం గురించి ఆమెకు విడివిడిగా హామీ ఇవ్వడంతో, వివిధ మతాల అనుచరులు తమ గొప్ప తల్లిని వారి రాణిని మరియు దేవతను తమ ప్రార్థనను వినాలని మరియు ఆమె ప్రస్తుత నిద్ర నుండి లేవాలని పిలుపునిచ్చారు. ఆమెను భానిస బంధాల నుండి విముక్తి చేయడానికి తమ ఐక్య మరియు సమిష్టి ప్రయత్నాలు సఫలమవుతాయని వారు ఆమెకు (భరతమాతకు) హామీఇచ్చారు.

Awake Summary in English

About Author

TS Inter 2nd Year English Study Material Chapter 4 Awake 1

Sarojini Naidu (13 February 1879 2 March 1949)was an Indian political activist and poet. A proponent of civil rights, women’s emancipation, and anti-imperialistic ideas, she was an important person in India’s struggle for independence from colonial rule. Naidu’s work as a poet earned her the sobriquet ‘the Nightingale of India’, or ‘Bharat Kokila’ by Mahatma Gandhi because of colour, imagery and lyrical quality of her poetry.

Few of his notable works in English:

1905: The Golden Threshold, London: William Heineman[18]
1917: The Broken Wing: Songs of Love, Death and the Spring[19]
1961: The Feather of the Dawn, edited by Padmaja Naidu, Bombay: Asia Publishing House

The poem,”Awake” is written by Sarojini Naidu, the Nightingale of India. It is a patriotic Song. It is the last poem in the first section of “The Broken wing: Songs of Love”. It is dedicated to Mohammad Ali Jinnah, a trusted friend of Sarojini Naidu, It was revived at the Indian National Congress Meetings in Bombay, 1915. In this poem, the worshippers of all Indian religions are brought together to the alfar of mother India to hail her as a “Queen” and a “Goddess”.

It conveys the idea that mother India could be awakened from her slumber and emancipated from her bondage by the united efforts of her sons. To get Independence from the British rule Indians must forget their religions and communal differences and rise like one against the British tyranny. It is a call for unity and action.

In the first stanza, the poetess speaks on behalf of the children of India who implore her to awake from her sleep and who worship her and are ready to serve her. She must lead their prays the dark night of slavery is now about with the hope of freedom. She must awake and cut the chains of slavery which cause so much grief to her children. Their hands will be purified by her blessings. So, that their holy cause may triumph.

They are the true children of their beloved Bharata Mata, and they have inherited her own pride and her own moral and spiritual courage. They will never fail to protect her. Thus will never desert her and they will ever sing the story of her greatness and glory. Their united efforts make her great and stories once again. Such as the dedication of her children and she must respond to their call.

In the next four lines, the followers of different Indian religions assure her that they would serve her with devotion to the best of their ability. The Hindus assure her that they will always worship her as they have been doing so far the parsees assure her that the fire of hope burning on their hearts will be dedicated to her service. The musalmans assure her that they would defend her with the sword of their love. The Christians assure her that they will show the same devotion and faith in their as they have in Jesus and Mary.

In the last two lines, having assured her separately of their devotion and service, the followers of all the different religions call upon their great mother, their green and their goddess to listen to their prayer and awake from her present slumber. Their fearless united and devoted efforts shall certainly be sufficient to free her from her present bondage. The lyrics a soul – stirring call for unity and action. It came from the heart of the poetess and so goes straight to the hearts of her readers.

Awake Summary in Telugu

Note: This summary is only meant for Lesson Reference, not for examination purpose

“మేల్కొను” అను కావ్యం భారత గానకోకిల సరోజినీ నాయుడుచే రచింపబడింది. ఇది దేశభక్తి గీతం “The Broken Wing: Songs of Love” యొక్క మొదటి విభాగంలో ఇది చివరి పద్య౦. ఇది సరోజినీ నాయుడు యొక్క నమ్మకమైన స్నేహితుడు, మహ్మద్ ఆలీ జిన్నా కు అంకితం చేయబడింది. ఇవి 1915లో బోంబేలో జరిగిన జాతీయ కాంగ్రెస్ సమావేశంలో పఠించబడింది. ఈ పద్య౦లో అన్ని భారతీయ మతాల ఆరాధనలు ఆమెను (భరతమాతను) రాణిగా మరియు దేవతగా కీర్తించడానికి భారతమాత యొక్క బలిపీఠం వద్దకు తీసుకురాబడ్డారు.

ఆమె కుమారుల పెక్కు ప్రయత్నాల ద్వారా భారతమాత తను నిద్ర నుండి మేల్కొనబడుతుందని మరియు ఆమె బానిసత్వం నుండి విముక్తి పొందగలదనే ఆలోచనను ఇది తెలియజేస్తుంది. బ్రిటీషు పాలన నుండి స్వాతంత్ర్యం పొందడానికి భారతీయులు తమ మతాలు మరియు మతభేదాలను మరచిపోయి బ్రిటీష్ దౌర్జన్యానికి వ్యతిరేఖంగా ఒక్కటిగా ఎదగాలి. ఐక్యత మరియు కార్యచరణకు ఇది పిలుపు.

మొదటి చరణంలో, కవయిత్రి తన నిద్ర నుండి మేల్కొనమని మరియు ఆమెను ఆరాధించి మరియు ఆమెకు సేవచేయడానికి సిద్ధంగా ఉన్న భారతమాత ముద్దుబిడ్డల తరుపున మాట్లాడుతుంది. ఆమె వారి ప్రార్థనను వినాలి. బానిసత్వం యొక్క చీకటి రాత్రి ఇప్పుడు స్వేచ్ఛ, స్వాతంత్య్రపు యొక్క ఆశతో ఉంది. ఆమె మేల్కొని తన పిల్లలకు చాలా దుఃఖం కలిగించే బానిసత్వపు గొలుసులను తెంచుకోవాలి. ఆమె ఆశీర్వాదం ద్వారా వారి చేతులు శుద్ధి చేయబడతాయి, తద్వారా వారి పవిత్రమైన కారణాన్ని విజయవంతం చేస్తారు.

ప్రియమైన భరతమాత యొక్క నిజమైన బిడ్డలు వారు మరియు ఆమె హుందా మరియు ఆమె నీతివంతమైన మరియు ఆధ్యాత్మిక ధైర్యాన్ని వారసత్వంగా వారు పునికి పుచ్చుకున్నారు. ఆమెను రక్షించుటలో వారు ఎన్నడూ విఫలమవ్వరు. వారు ఆమెను ఎన్నడూ త్యజించరు మరియు ఆమె కీర్తి మరియు గొప్పతనం గురించి ఎల్లప్పుడూ స్తుతిస్తారు. వారి యొక్క పెక్కు ప్రయత్నాలు తిరిగి ఆమెను కీర్తివంతం చేస్తాయి. అలాంటి అంకితభావానికి, ఆమె వారి పిలుపుకి ప్రతిస్పందించాలి.

తర్వాత నాల్గు శ్రేణుల్లో, వివిధ భారతీయ మతాల ఆరాధకులు వారి శక్తికొలది అంకితభావంతో ఆమెకు సేవ చేస్తామని నమ్మకంగా చెప్తున్నారు. ఇప్పటి వరకూ ఎలా పూజించారో అలానే ఆమెను ఎల్లప్పుడూ పూజిస్తారని నొక్కి చెప్తున్నారు హిందువులు. ఆమె సేవకు, పాటకు వారి హృదయాల్లోని ఆశాజ్యోతిని ఆమెకు అంకితం చేస్తామన్నారు. వారి ప్రేమ, భక్తి ఖడ్గంతో ఆమెను కాపాడుకుంటామని ముస్లీములు నొక్కి చెప్తున్నారు. క్రీస్తు మరియు మేరిమాత పట్ల చూపిన విశ్వాసం, భక్తినే ఆమె పట్ల చూపుతున్నామంటున్నారు క్రిష్టియన్లు.

చివరి రెండు శ్రేణుల్లో, వారి భక్తి, ప్రేమ, విశ్వాసం గురించి ఆమెకు విడివిడిగా హామీ ఇవ్వడంతో వివిధ మతాల ఆరాధనలు తమ గొప్ప తల్లివి, వారి రాణిని మరియు దేవతను తమ ప్రార్థనను వినాలని మరియు ఆమె ప్రస్తుత నిద్ర నుండి లేవాలని పిలుపునిచ్చారు. ప్రస్తుత బానిసత్వం నుండి ఆమెను విముక్తి చేయుటకు, వారి నిర్భయ ఐక్య మరియు అంకిత ప్రయత్నాలు సరిపోతాయి కాబట్టి, ఇవి కార్యచరణ మరియు ఐక్యత కోసం ఇది మనస్సును కరిగించే అభ్యర్ధన. ఇది కవయిత్రి మనస్సు నుండి వస్తుంది మరియు పాఠకుల హృదయాలకి నేరుగా వెళ్ళింది.

TS Inter 2nd Year English Study Material Chapter 4 Awake

Awake Summary in Hindi

Note: This summary is only meant for Lesson Reference, not for examination purpose

‘अवेक’ नामक कविता भारत को किला सरोजिनी नायडु से लिखी गई है । वह देशभक्ति गीत है । यह “द ब्रोकेन विंग’ : ‘सांग्स ऑफ़ लव’ के अंतिम खंट की आखनी कविता है । यह सरोजिनी नायडु के विश्वासनीय मित्र मुहम्मद अली जिन्ना को समर्पित है । 1915 में बंबई में भारतीय कांग्रेस की बैठक में गाया गया । इस कविता में भारत माता को रानी और देवता के रूप में परिवर्तितकर उसे अभिवादन करने के लिए भारतीय सभी धर्मों के उपासकों को एकता में लाया गया ।

यह इस विचार को व्यक्त करता है कि भारतमाता को नींद से जगाया जा सके और आपनी संतानों के सम्मिलित प्रयास से अपने दास्य बंधन से मुक्त किया जा सके । ब्रिटिशशासन से स्वतंत्रता प्राप्त करने के लिए । भारतीयों को अपने धर्मों, सांप्रदायिक संघर्ष करना चाहिए । यह एकता और क्रिया का आहवान है ।

परले छंद में, कवयित्री भारत के बच्चों की ओर से कहती हैं, जो भारत माता के उसकी नींद से जगाते हैं, जो उसकी पूजा करते हैं, और जो उसकी सेवा करने तैयार होते हैं । दासता का घोर अंधकार को दूर करने और स्वतंत्रता की आशा करने के लिए उसे बच्चों को बहुत परेषानी होती है । उसके आशईर्वात से उसके बच्चों के बहुत परेशानी होती है । उसके आशीर्वाद से उसके बच्चों के हाथ शुद्ध हो जाएँगे ताकि उनके पवित्र उद्देश्य की निजय हो सके ।

वे अपने प्यारी भारढकतमाता के सच्चे बच्चे हैं ओर उनमें उसका अहम अभिमान है और उसके नैतिक मूल्य हैं और आध्यात्मिक साहस है। उसकी रक्षा में वे कभी असफल नहीं होते । वे उसे कभी नहीं छोडेंगे। वे उसकी महानता और महिमा की कहानी गाएँके । उनके सम्मिलित प्रयास फिरसे उसकी महानता और चमक प्रदान करते हैं । उसकी संतानों का समर्पण एसा ही है । उसे उनके निमंत्रण की प्रतिक्रिया दिखानी चाहिए ।

यह गीत एकता और प्रतिक्रिया की आत्म प्रेरित पुकार है । यह कवियत्री के हृदय से निकला है और उसके पाठकों के हृदयों में सीधा घुसता हैं ।

Meanings and Explanations

waken (v) (వెయికన్)/ weɪ.kən/ : wake from sleep : నిద్ర నుండి మేలు కొలుపు, भारतीय

thy(adj) / (మీ యొక్క)/ ðaɪ/ : your : మీ యొక్క (old and poetic form of you), प्रार्थना करना, विनती करना

implore(v)/(ఇంప్లో (ర్))/ɪm’plɔ:r/ : request, plead: విన్నవించుకొను : तु, तुम

thee (pron)/ (దీ)/θri:/ : you : మిమ్ము (old from ‘you’) ; घुटने टेकना

kneel (v)/ (నీల్) /ni:l/ : rest on one’s bent knees : మోకాళ్ళను నేలకు ఆనించి వాని ఆధారంగా కూర్చొను, (భక్తిని సూచిస్తూ ), पूजा करना, प्रेम करना

adore (v) (అడో(ర్)) /ədɔ:r/ : worship; ఆరాధించు, పూజించు, शरमाना, उत्साह से लाल होना

TS Inter 2nd Year English Study Material Chapter 4 Awake

aflush(adj) /(అప్లష్)/ə’fl\(\Lambda \int\) : red with excitement ; ఆనందంతో ఎర్రబారిన : बंधन, गुलामी

morrow(n)/ (మోర ఉ)/ ‘mɒr.əʊ/ : tomorrow (poetic form) : काटना, भागों में बाँटना

thou (pron)/ (దౌ) //ðaʊ / : మీరు, दुःख, बाधाएँ

bondage (n)/ (బొండిజ్)/’bɒn.dɪdʒ/ : Slavery : బానిసత్యము; బంధము, गुलाम बनाना, अधीनस्थ बनाना, विनम्र बनाना

sever(n) / (సెవ(ర్)) /’sev.ər/ : cut, tear : కోసివేయు; త్రుంచు ; पवित्र बनाना

woes(n-pl): (వఉజ్)) : troubles /wəʊz/: కష్టములు, शानदार जीत

enthral(v)(also-enthrall)(ఇన్ త్రోల్ ) : /ɪn’θrɔ:l/ :make slaves ; బానిసలుగా చేయు ; bound ; బంధించు, निकट भविष्य में हम उसे हासिल करते हैं ।

hallow (v) : (హ్యాల ఉ) /’hæl.əʊ/ : sanctify; make pious : పవిత్రము చేయు, अधिकार के रूप में कुछ लेना था प्राप्त करना

triumphs(n-pl)/ (టై అమ్ ఫ్ జ్) /’traɪ.əmf/ : successes : విజయములు, कई अलग- अलग प्रकार के :

thine (pron)/(దైన్)/ ðaɪn/ : your’s (old form) మీవారము ; మీవి, गर्व – प्रसन्न या संतुष्ट होने की भावना); व्यकित-निष्ठा

inherit(v)/ (ఇన్ హెరిట్)/ɪn’her.ɪt/ : receive something as a right : వారసత్వముగా, హక్కుగా పొందు ; त्यागः छोड़ देना, छूटना

manifold(adj) : (మ్యానిఫ ఉల్డ్ )/(mæn.ɪ.fəʊld/ : of various types; हिलना, असफल होना

forsake(v)/(ఫ(ర్)సె ఇక్) /fɔ:seɪk/ : leave: వదిలివేయు, सुरक्षात्मक हिस्सा, बचाव की बात

falter (y)! (ఫ(ర్)సె ఇక్)/’fɒl.tər/ : fail : విఫలమగు, తప్పచేయు, स्तेमाल की जानेवाली एक सपाट संरचना

shield (n) (ఫోల్లట (ర్))/ʃi:ld/ : something that protects : ఆదారణంగా ఉండునది, महान राष्ट्रों की प्रथम पंक्ति में

altar(n) (షీల్ డ్ ) / ɔ:l.tə(r)/ : pieous place in temples : గర్భ గుడి పవిత్ర స్థలము, ताज पहनाय

dauntless (adj) / (డోంట్ లస్) /dɔ:nt.ləs/ : fearless : నిర్భయంగా, विश्वास, भरोसा

TS Inter 2nd Year English Study Material Chapter 4 Awake

avails (v) / (అవె ఇల్)/ ‘əveɪl/ : serve : assist : సేవించు, సహాయపడు, निर्भय भक्ति

hearken (v) / (హా (ర్)కన్) / ‘ha:kən/ : listen విను, सुनाना

hail (v) (హెయిల్) /heɪl/ : call, salute : పిలుచు, గౌరవించు, स्वागत, निमंत्रण

TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Telangana TSBIE TS Inter 1st Year Physics Study Material 12th Lesson Thermal Properties of Matter Textbook Questions and Answers.

TS Inter 1st Year Physics Study Material 12th Lesson Thermal Properties of Matter

Very Short Answer Type Questions

Question 1.
Distinguish between heat and temperature. [TS Mar. ’15]
Answer:
Differences between heat and temperature :

HeatTemperature
1. Heat is a form of energy.1. It represents relative degree of hotness (or) Coldness of a body.
2. Unit: joule (or) calorie2. Unit: °C or °F
3. It is cause3. It is effect.
4. Heat is measured by calorimeter4. It is measured by thermometer.
5. Quantity of heat supplied Q = m st5. Change in temperature of a body ∆ t = \(\frac{Q}{ms}\)

Question 2.
Explain triple point of water.
Answer:
The temperature of a substance remains constant during its change of state.

A graph plotted between temperature (T) and pressure (P) of a substance during change of state is called “phase diagram”.
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 1

In phase diagram of water, the P – T plane is divided into three regions.

The line ‘AO’ is called fusion curve. It gives equilibrium temperature and pressure between solid and liquid states.

The line ‘CO’ is called vaporisation curve. It gives equilibrium temperature and pressure between liquid and vapour states.

The line ‘BO’ is called sublimation curve. It gives the relation between of temperature and pressure between solid and vapour states.

Triple Point:
At point ‘O’ the curves AO, BO and CO will intersect.

It gives the temperature and pressure at which solid, liquid and vapour states of water are in equilibrium.

Coordinate of triple point of water a temperature = 273.16 K, pressure = 0.006 atmos (or) 611 pascals.

Question 3.
What are the lower and upper fixing points in Celsius and Fahrenheit scales? [TS Mar. ’16; AP Mar. ’19, ’18 ’16, May ’14]
Answer:
Centigrade (Celsius) scale of temperature:
In centigrade scale of temperature lower fixed point is freezing point of water at one atmosphere pressure, as 0°C. The upper limit is boiling point of water at 1 atm pressure, as 100°C.

Fahrenheit scale of temperature :
In Fahrenheit scale, the lower fixed point is freezing point of water at one atmosphere pressure, as 32°F. The upper fixed point is boiling point of water at 1 atm pressure, as 212°F.

TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 4.
Do the values of coefficients of expansion differ, when the temperatures are measured on Centigrade scale or on Fahrenheit scale?
Answer:
Yes. Coefficients of expansion α, αa and αv are not same in Celsius scale and in Fahrenheit scale. In Fahrenheit scale values of α, αa and αv are less than those in Celsius scale.

Since magnitude of 1°C > magnitude of 1°F this change takes place. The values of Celsius scale are 1.8 times more than the values in Fahrenheit scale.
αF = 5/9 αc

Question 5.
Can a substance contact on heating? Give an example. [AP Mar. ’18, ’16, May ’16; TS May, ’18, ’16]
Answer:
Yes. Some substances will contact on heating. Ex: Leather, rubber, cast Iron type metal.

Question 6.
Why gaps are left between rails on a railway track? [TS Mar. ’19; AP Mar. ’19, ’17, ’16, ’09; May ’16; June ’15]
Answer:
To allow the linear expansion rails.

In summer temperature of atmosphere increases so rails will expand. If no gap is given between rails then the rails will bend it leads to accidents. If gap is given the rails will expand into that gap and that track is safe.

Question 7.
Why do liquids have no linear and areal expansions? [TS Mar. ’19]
Answer:
Liquids have only volume expansion. No linear expansion or areal expansion. Because liquids does not have any independent shape they must be taken in a container. So we will consider only volume of liquid.

Question 8.
What is latent heat of fusion? [AP & TS May ’17]
Answer:
Latent heat of fusion (melting):
It is defined as the amount of heat energy absorbed or rejected by unit mass of substance while converting from solid to liquid or from liquid to solid.

Question 9.
What is latent heat of vapourisation? [AP Mar. ’13]
Answer:
Latent heat of vapourisation :
It is defined as the amount of heat energy absorbed or rejected by unit mass of substance while converting from liquid to vapour or from vapour to liquid state.

Question 10.
Why are utensils coated black? Why is the bottom of the utensils made of copper? [AP May ’18; TS Mar. ’18]
Answer:
Lower portion of the utensils is in contact with fire. Black bodies are good heat absorbers. So, a black bottom will absorb more heat.

Copper is a good conductor of heat. So, copper is used at the bottom of cooking utensils.

TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 11.
What is triple point of Water? Mention the values of temperature and pressure at triple point of water. [TS June ’15]
Answer:
Triple point:
The temperature and pressure where a substance can coexist in all its three states is called the “triple point”.

i.e., The substance will exist as a solid, as liquid and as vapour at that particular temperature and pressure.

For water the triple point is at a temperature of 273.16 K and at a pressure of 6.11 × 10-3 atmospheres or nearly 610 pascals.

Question 12.
State Boyle’s law and Charles law. [AP June 15; TS Mar. 15]
Answer:
Boyle’s Law :
At constant temperature, the volume (V) of a given mass of a gas is inversely proportional to its pressure (P).
∴ V ∝ \(\frac{1}{P}\) ⇒ PV = constant = K.

Charles Law:
At constant pressure, the volume (V) of a given mass of a gas is directly proportional to its absolute temperature (T).
∴ V ∝ T ⇒ \(\frac{V}{T}\) = K (constant)

Question 13.
State Wein’s displacement law. [AP Mar. ’17]
Answer:
Wein’s Displacement Law :
The wavelength (Ain) of maximum intensity of emission of black body radiation is inversely proportional to absolute temperature (T) of the black body.
i.e., λm ∝ \(\frac{1}{T}\) (or) λm = \(\frac{b}{P}\)
where ‘b’ is called ‘Wein’s constant”.

Question 14.
Ventilators are provided in rooms just below the roof. Why?
Answer:
Density of hot air is less. So in a room hot air goes to top layers i.e., nearer to the roof.

When ventilators are provided nearer to the roof hot air will escape easily from room. So we feel that the room is cool and circulation of air will become easy.

Question 15.
Does a body radiate heat at 0 K? Does it radiate heat at 0°C?
Answer:
According to Precost’s theory, every body above zero Kelvin will radiate heat energy to the surroundings.
So, i) A body at O’ Kelvin does not radiate heat energy.
ii) A body at 0°C i.e., at 273Kwill radiate heat energy.

Question 16.
State the different modes of transmission of heat. Which of these modes require medium? [TS May ’18]
Answer:
Transmission of heat is of three types.
1) Conduction 2) Convection 3) Radiation

For propagation of heat energy medium is required in case of conduction and convection.

TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 17.
Define coefficient of thermal conductivity and temperature gradient.
Answer:
“The coefficient of thermal conductivity”
(K) it is the quantity of heat flowing normally per second through unit area of the substance per unit temperature gradient.
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 2

‘Temperature gradient” is defined as the change in temperature along the conductor per unit length.

Temperature gradient
Change in temperature
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 3

Question 18.
Define emissive power and emissivity.
Answer:
Emissive power:
The emissive power of a body is defined as the energy radiated by the body per second per unit area at a given temperature and wavelength.

Emissivity:
Emissivity is defined as the ratio of the emissive power of a body to that of a black body at the same temperature.

Question 19.
Is there any substance available in nature that contracts on heating? If so, give an example. [TS May ’16]
Answer:
Yes. Some substances will contact on heating.
Ex: Leather, rubber, cast Iron type metal.

Question 20.
What is greenhouse effect? Explain global warming. [AP Mar. ’15, ’13; TS Mar. & May ’16]
Answer:
Green house effect: Earth will absorb heat radiation and reradiate heat energy of longer wavelength. This longer wave length heat radiation is reflected back to earth due to green house gases such as Carbon dioxide [CO2], Methane (CH4) Chloroflurocarbons, Ozone (O3), etc. As a result temperature of earth’s atmosphere is gradually increasing. This is known as “green house effect.”

Global warming:
Earth receives heat energy during day time from sun. It reradiates heat energy in the form of longer electromagnetic waves.

But due to presence of green house gases the longer electromagnetic waves were reflected back to earth. As a result temperature of earth’s atmosphere is gradually increasing.

This process will increase with the increased content of green house gases in atmosphere. As a result temperature of earth’s atmosphere increases gradually.

Question 21.
Define absorptive power of a body. What is the absorptive power of a perfect black body?
Answer:
Absorptive power of a body is defined as the ratio of energy absorbed by the body within the wave length range of A and A + dA to the total energy flux following on the body.
Absorptive power,
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 4

Question 22.
State Newton’s law of cooling. [AP Mar. ’18, ’16, May ’18, ’17, June ’15; TS Mar. ’18, TS May ’16]
Answer:
Newton’s Law of cooling states that the rate of loss of heat of a hot body is directly pro-portional to the difference in temperature between the body and its surroundings pro-vided the difference in temperatures is small and the nature of the radiating surface remains same.
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 5
Where k is the proportionality constant

Question 23.
State the conditions under which Newton’s law of cooling is applicable. [AP May ’16; TS June ’15]
Answer:
Newton’s law of cooling is applicable

  1. loss of heat is negligible by conduction and only when it is due to convection.
  2. loss of heat occurs in a streamlined flow of air i.e., forced convection.
  3. temperature of the body is uniformly distributed over it.
  4. temperature difference between the body and surroundings is moderate i.e., upto 30 K.

TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 24.
The roofs of buildings are often painted white during summer. Why? [TS Mar. ’17, ’15; AP May ’16]
Answer:
When roofs of buildings are coated white we will feel relatively cold during summer.

Absorptive power of white surface is less. So roofs coated white will absorb less heat energy. So less quantity of heat is transmitted into the house. So we feel less hot or cold inside the house.

Question 25.
What is thermal expansion? [TS Mar. ’16]
Answer:
The increase in interatomic distance due to thermal energy is called “thermal expansion”.

As a result the length solids or volume of liquids or pressure of gases will increase.

Question 26.
Why is it easier to perform the skating on the snow? [TS Mar. ’16]
Answer:
Due to increase of pressure melting point decreases, So it is easier to perform the skating on the snow.

Short Answer Questions

Question 1.
Explain Celsius and Fahrenheit scales of temperature. Obtain the relation between Celsius and Fahrenheit scales of temperature.
Answer:
Celsius (Centigrade) scale of temperature :
In centigrade scale of temperature lower fixed point is freezing point of water at one atmosphere pressure, as 0°C. The upper limit is boiling point of water at 1 atm pressure, as 100°C.

The interval between lower limit and upper limit [100 – 0 = 100] is divided into 100 equal parts and each part is called 1°C.

Fahrenheit scale of temperature :
In Fahrenheit scale, the lower fixed point is freezing point of water at one atmosphere pressure, as 32°F. The upper fixed point is boiling point of water at 1 atm pressure, as 212°F.

The interval between upper fixed point and lower fixed point (212 -32 = 180) is divided into 180 equal parts and each part is called 1°F.

Relation between Fahrenheit and Celsius scale of temperatures:
In both scales, lower limit and upper limit are same. The only change is in numerical values of lower and upper limits.

In Fahrenheit lower limit = 32, upper limit = 212, difference of limits = 180

In Celsius scale lower limit = 0, upper limit = 100, difference of limits = 100
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 6
C = Temperature in Celsius scale.
F = Temperature in Fahrenheit scale.

Question 2.
Two identical rectangular strips one of copper and the other of steel, are riveted together to form a compound bar. What will happen on heating?
Answer:
When two dissimilar metals say copper and steel are riveted together that arrangement is called “bimetallic strip”.

When a bimetallic strip is heated copper strip will expand more than steel due to more expansion coefficient.
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 7

Since they are riveted they must expand as a common piece. As a result bimetallic strip will bend in the form of an arc. For the metallic strip with high a its length is more so it is on the outer side. For the strip with less a its length is less. It will be at the inner side of the arc.

TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 3.
Pendulum clocks generally go fast in winter and slow in summer. Why? [TS Mar. ’19, ’17]
Answer:
In summer due to increase in temperature of atmosphere length of pendulum will increase.

Time period of pendulum T = 2π\(\sqrt{\frac{l}{g}}\)

T ∝ √l. So it will make less number of oscillations per day. So clock will run slowly in summer.

In winter temperature of atmosphere decreases. So length of pendulum decreases.

Hence time period of oscillation will also decrease. As a result, pendulum will make more oscillations per day so clocks will run fast in winter.

Question 4.
In what way is the anomalous behaviour of water advantageous to aquatic animals? [AP Mar. ’18, 17, 14; May 18. 17, 14; TS May ’18]
Answer:
In cold countries and at polar region temperature falls below 0°C at winter. So surface of water will be frozen. Due to anomalous expansion of water even though the surface of lakes, and sea are frozen water will exist at bottom layers at 4°C.

Different layers in between ice and bottom will have different temperatures like 1°C, 2°C or 3°C. In these layers, aquatic animals are able to survive even in winter.

Anomalous expansion of water helps for the survival of aquatic life at polar region and in cold countries.

Question 5.
Explain conduction, convection and radiation with examples. [TS Mar. ’18, ’16, ’15, June ’15; AP Mar. ’19, ’15, May, ’16]
Answer:
Conduction :
It is a mode of transfer of heat from one part of the body to another, from particle to particle in the direction of fall of temperature without any actual movement of the heated particles.
Ex: When one end of a metal rod is heated, its other end becomes hot. Here, the heat goes from hot end of the metal rod towards cold end, by conduction.

Convection :
It is a mode of transfer of heat from one part of the medium to another part by the actual movement of the heated particles of the medium.
Ex : Seabreeze, Tradewind, etc.

Radiation :
It is a mode of transfer of heat from the source to the receiver without any actual movement of source or receiver and also without heating the intervening medium.
Ex : Heat from sun comes to us through radiation. On standing near fire, we feel hot as heat comes to us through radiation.

Long Answer Questions

Question 1.
Explain thermal conductivity and coefficient of thermal conductivity.
A copper bar of thermal conductivity 401 W (mK) has one end at 104°C and the other end at 24°C. The length of the bar is 0.10 m and the cross-sectional area is 1.0 × 10-6 m-2. What is the rate of heat conduction along the bar?
Answer:
The ability to conduct heat in solids is called ‘Thermal conductivity.”

Consider a bar with rectangular cross-section as shown in the figure. The faces ABCD and EFGH are maintained at θ1 and θ2 respectively (θ1 > θ2). Heat passes from one end to the other.
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 8

The amount of heat conducted (Q) depends on,

  1. Amount of heat conducted Q is proportional to area of cross-section A perpendicular to flow.
    ∴ Q ∝ A ………. (1)
  2. is proportional to temperature difference between the two ends.
    ∴ Q ∝ (θ2 – θ1) …………. (2)
  3. is proportional to the time of flow.
    Q ∝ t ………. (3) and
  4. is inversely proportional to the length of the conductor.
    Q ∝ \(\frac{1}{l}\) …………. (4)

TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 9
where k = constant called coefficient of thermal conductivity.

Coefficient of thermal conductivity (k) :
It is defined as the amount of heat conducted normally per sec per unit area of cross-section per unit temperature gradient.
S.I. Unit w m k-1
Dimensional formula = [ M¹L¹T-3θ-1]

Problem:
Thermal conductivity of copper,
Kc = 401 W/m-k
Temperature at one end, θ2 = 104°C
Temperature of 2nd end, θ1 = 24°C
Length of copper bar, l = 0.1 m; Area,
A = 1.0 × 10-6 m-2
Rate of conduction,
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 10

TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 2.
State the explain Newton’s law of cooling. State the conditions under which Newton’s law of cooling is applicable.
A body cools down from 60°C to 50°C in 5 minutes and to 40°C in another 8 minutes. Find the temperature of the surroundings. [TS May ’17, ’16; AP May ’13]
Answer:
Newtons’ Law of cooling :
The rate of loss of heat of the body is directly proportional to the difference of temperature of the body and the surroundings.

Explanation :
The law holds good only for small difference of temperature. Also, the loss of heat by radiation depends upon the nature of the surface of the body and the area of the exposed surface. We can write
– \(\frac{dQ}{dt}\) = k (T2 – T1) (sign indicates loss) …….. (1)

where k is a positive constant depending upon the area and nature of the surface of the body. Suppose a body of mass ‘m’ and specific heat capacity ‘s’ is at temperature T2. Let T1 be the temperature of the surroundings. If the temperature falls by a small amount dT2 in time dt, then the amount of heat lost is
dQ = ms dT2
∴ Rate of loss of hfeat is given by
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 11
where K = k/ms

Conditions (under which Newton’s law of cooling is applicable):
Newton’s law of cooling is applicable

  1. loss of heat is negligible by conduction and only when it is due to convection.
  2. loss of heat occurs in a streamlined flow of air i.e., forced convection.
  3. temperature of the body is uniformly distributed over it.
  4. temperature difference between the body and surroundings is moderate i.e., upto 30 K.

PROBLEM :
Let ‘θo‘ be the temperature of the surroundings.

In first case :
Initial temperature, θ1 = 60°C
Final temperature, θ2 = 50°C
Time of cooling, t = 5 minutes = 5 × 60 = 300s
From Newton’s law of cooling we can write,

In secoend case :
Initial temperature, θ1 = 60°C
Final temperature, θ2 = 40°C
Time of cooling, t = 13 minutes = 13 × 60 = 780s
Again from Newton’s law of cooling we can write,
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 13
On solving equations (1) and (2) we get, θo = 33.33°C

Problems

Question 1.
What is the temperature for which the readings on Kelvin Fahrenheit scales are same?
Solution:
On the Kelvin and Fahrenheit scales
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 14
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 15

Question 2.
Find the increase in temperature of aluminium rod if it’s length is to be increased by 1%. (α for aluminium = 25 × 10-6/°C) [AP Mar. ’15; June ’15]
Solution:
Coefficient of linear expansion of aluminium, α = 25 × 10-6/°C
We know that percentage increase in length
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 16

Question 3.
How much steam at 100°C is to be passed into water of mass 100g at 20°C to raise its temperature by 5°C? (Latent heat of steam is 540 cal / g and specific heat of water is 1 cal / g°C)
Solution:
Latent heat of steam, Ls = 540 cal/g
Specific heat of water, Lw = 1 cal / g°C
Mass of water, mw = 100g

According to method of mixture or from the principle of calorimetry we can write, Heat lost by steam = heat gained by water
i.e., msLs + msSw(100 – t) = mwSw (t – 20)
∴ ms × 540 + ms × 1(100 – 25)
⇒ 100 × 1 × (25 – 20)
⇒ 615ms = 500(or)ms = \(\frac{500}{615}\) = 0.813 g

TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 4.
2 kg of air is heated at constant volume. The temperature of air is increased from 293 K to 313 K. If the specific heat of air at constant volume is 0.718 kJ/kg K, find the amount of heat absorbed in kJ and kcal. (J = 4.2 kJ/kcal.)
Solution:
Mass of air, m = 2 kg
Change in temperature, ∆T = 313 – 293 = 20K.
Specific heat at constant volume, Cv = 0.718 k.J/kg – K.
Heat mechanical equivalent, J = 4.2 kJ/k.cal.
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 17
∴ Heat energy absorbed,
Q = 2 × 0.718 × 10³ × 20. = 28.72 kJ
= 6.838 k calories.

Question 5.
A clock, with a brass pendulum, keeps correct time at 20°C, but loses 8.212 s per day, when the temperature rises to 30°C. Calculate the coefficient of linear expansion of brass.
Solution:
Temperature of correct time, t1 = 20°C
Loss or gain of time in seconds per day = 8.212 sec.
Final temperature, t2 = 30°C
∴ ∆t = 30 – 20 = 10
α of pendulum material = ?

In pendulum loss or gain of time in seconds per day = 43,200. α ∆t
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 18

Question 6.
A body cools from 60°C to 40°C in 7 minutes. What will be its temperature after next 7 minutes if the temperature of its surroundings is 10°C? [AP May ’13]
Solution:
In first case :
Initial temperature, θ1 = 60°C
Final temperature, θ2 = 40°C
Time of cooling, t1 = 7 minutes
= 7 × 60 = 420s
Temperature of surroundings, θ0 =10°C
From. Newton’s law of cooling, we can write,
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 19

In second case:
Initial temperature, θ1 = 40°C
Time of cooling, t2 = 7 minutes = 420s
Again, from Newton’s law of cooling we can write,
TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 20
on solving equations (1) & (2) we get, 0 = 28°C

TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 7.
If the maximum intensity of radiation for a black body is found at 2.65 µ m, what is the temperature of the radiating body? (Wein’s constant = 2.9 × 10-3 mK)
Solution:
Wavelength corresponding to maximum intensity, λmax = 2.65 µm = 2.65 × 10-6 m.
Wein’s constant, b = 2.90 × 10-3 mK.
From Wein s Law, T = \(\frac{\mathrm{b}}{\lambda_{\mathrm{m}}}=\frac{2.90 \times 10^{-3}}{2.65 \times 10^{-6}}\)
= 1094 K.

TS Inter 2nd Year English Study Material Chapter 3 Hiroshima Child

Telangana TSBIE TS Inter 2nd Year English Study Material 3rd Lesson Hiroshima Child Textbook Questions and Answers.

TS Inter 2nd Year English Study Material 3rd Lesson Hiroshima Child

Annotations (Section A, Q.No. 2, Marks: 4)
Annotate the following in about 100 words each.

a) I knock and yet remain unseen. (Revision Test – III)

Introduction: This line is taken from the poem, “Hiroshima Child”, written by Nazim Hikmet. He is a turkish poet, play wright and novelist. He is recognised as one of the greatest poets of the twentieth century. Most of his writings are about war. Content and Meaning: Here, the speaker is a seven-year old. Hiroshima girl. She died when an atom bomb was dropped on Hiroshima during the world war II. The soul of the girl knocks on every door to warn them about the adverse effects of war. But nobody pays attention to her as she invisible.

Critical comment: The speaker begs people to fight for peace and to let children grow happily. The war against war touches our hearts.

కవి పరిచయం : ఈ వ్యాసం నజీమ్ హిక్మెట్ వ్రాసిన “హిరోషిమా చైల్డ్” అను పద్య౦ నుండి తీసుకొనబడింది. అతను ఒక టర్కిష్ కవి, నాటక కర్త మరియు నవలా రచయిత. ఇతడు 20వ శతాబ్దపు గొప్ప కవులలో ఒకడుగా గుర్తింపు పొందాడు. ఇతని రచనలలో చాలా వరకు యుద్ధం గురించినవే.

సందర్భ౦ మరియు అర్థం : ఇక్కడ వ్యక్తి 7 సం॥ల హిరోషిమా చిన్నారి. రెండవ ప్రపంచయుద్ధ సమయంలో హిరోషిమాపై అణుబాంబు వేసినప్పుడు ఈ చిన్నారి చనిపోయింది. ఈ చిన్నారి ఆత్మ ప్రతి తలుపును తడుతూ యుద్ధం యొక్క దుష్ప్రభావాలను గురించి వారిని హెచ్చరిస్తుంది. కానీ, ఎవ్వరూ ఆమెను పట్టించుకోరు. ఆమె కనిపించదు కాబట్టి.

విమర్శ : పిల్లలు సంతోషంగా ఎదగడానికి మరియు శాంతి కోసం పోరాడాలని స్పీకర్ ప్రజలను వేడుకుంటున్నారు. యుద్ధానికి వ్యతిరేఖంగా యుద్ధం మన హృదయాలను తాకుతుంది.

b) I am seven now as I was then
when children die they do not grow.

Introduction: These heart touching lines are taken from the poem.”Hiroshima child”, written by Nazim Hikmet. He is a turkish poet, playwright and novelist. He is recognised as one of the greatest poets of the twentieth century. Most of his writing are about war.

Content & Meaning: Here, the little child knocks on every door. No one hears or sees as the child died at seven in the Hiroshima bomb blast. Since then, the child has felt neither growth nor hunger, nor any wants. The child continues to be in the same state. The vehement plea is for peace, the poet uses a dead child to press upon our guilty conscience.

Critical comment: Through the soul of the little child, the poet begs people to fight for peace and to let children grow happily.

కవి పరిచయం : ఈ హృదయాన్ని కదిలించే పంక్తులు నజీమ్ హిక్మేట్చే రచించబడిన “హిరోషిమా చైల్డ్” అను పద్య౦ నుండి తీసుకొనబడింది.

సందర్భ౦  మరియు అర్థం : ఇక్కడ చిన్నారి ప్రతి తలుపును తడుతుంది. ఏడు సం||ల వయస్సులోనే హిరోషిమా అణుబాంబు దాడిలో చనిపోవటంతో, ఎవ్వరూ ఆ చిన్నారిని చూడలేదు, వినలేరు. అప్పటి నుండి, ఆ చిన్నారి కి ఎదుగుదలలేదు. ఆకలి లేదూ ఏమీలేదు అలానే ఉండిపోయింది. శాంతి కోసం ఒక గట్టి ప్రయత్నం ఇది. కవి ఎదుగుదలలేదు. ఆకలి లేదూ ఏమీలేదు అలానే ఉండిపోయింది. శాంతి కోసం ఒక గట్టి ప్రయత్నం ఇది. కవి చనిపోయిన చిన్నారిని మన మనస్సాక్షిని నొక్కడానికి ఉపయోగించాడు.

విమర్శ : చిన్నారి ఆత్మ ద్వారా శాంతి కోసం పోరాడాలని మరియు చిన్నారులను సుఖంగా ఎదగనివ్వమని ప్రజలను కవి వేడుకుంటున్నారు.

TS Inter 2nd Year English Study Material Chapter 3 Hiroshima Child

c) I ask for nothing for myself
For I am dead, for I am dead

Introduction: These heart touching lines are taken from the poem.”Hiroshima child”, written by Nazim Hikmet. He is a turkish poet, playwright and novelist. He is recognised as one of the greatest poets of the twentieth century.

Content & Meaning: The little visits every home, seeks neither food nor things. The child no longer needs food or water to survive because we have taken away her life in which these things were actually prevalent. We have taken away everything from this child. She was innocent all along. The poet repeats the lines ” For I am dead for I am dead” throughout the poem to remind us that we have killed this innocent child with our unnecessary violence. He expects our fight for peace.

Critical comment: The poet uses the technique of repetition to convey his theme. The begs us to fight for peace.

కవి పరిచయం : హృదయాన్ని తాకేటటువంటి ఈ పంక్తులు నజీమ్ హిక్ మెట్చే రచించబడిన “హిరోషిమా చైల్డ్” అను కావ్యం నుండి గ్రహించబడినవి. అతను టర్కిష్ కవి నాటక రచయిత మరియు నవలా రచయిత. అతను 20వ శతాబ్దపు గొప్ప కవులలో ఒకడుగా గుర్తింపబడ్డాయి.

సందర్భ౦ మరియు అర్థం : చిన్నారి ప్రతి ఇంటిని సందర్శిస్తుంది. ఆహారం కానీ, వస్తువులు కానీ వెతకటం లేదు. జీవించడానికి ఆహారం లేదా నీరు ఆ చిన్నారికి అవసరం లేదు. ఎందుకంటే ఆ చిన్నారి జీవితాన్ని మనం తుడిచివేసాము. ఆ చిన్నారి సమస్తం మనం తీసుకున్నాము.

ఆ చిన్నారి ఏమీ తెలియని అమాయకురాలు. “నేను చనిపోయాను, నేను చనిపోయాను” అను చిన్నారి పంక్తులను కవి పదేపదే పునరావృత్తం చేస్తున్నాడు. ఎందుకంటే మనం అనవసర హింసతో అమాయకురాలి ప్రాణం తీసాము అన్న విషయాన్ని గుర్తు చేయటానికి ఈ నైపుణ్యాన్ని కవి ఉపయోగించాడు. శాంతి కోసం మన పోరాటాన్ని కవి ఆశిస్తున్నాడు.

విమర్శ : తన ఇతివృత్తాన్ని తెలియజేయుటకు, కవి పునరావృత్తం అను నైపుణ్యాన్ని ఉపయోగిస్తున్నాడు. శాంతికోసం పోరాడమని కవి మనల్ని వేడుకుంటున్నాడు.

d) All that i need is that for peace (Revision Test – III)
you fight today you fight today

Introduction: These heart touching lines are taken from the poem. “Hiroshima child”, written by Nazim Hikmet. He is a turkish poet, playwright and novelist. He is recognised as one of the greatest poets of the twentieth century.

Content & Meaning: The child is the speaker in the poem. She lost her life at seven in the Hiroshima bomb blast. Since then, she has felt neither growth nor hunger. She continues to be in the same state, she visits every home, seek, neither food nor things. She warns us, about the evils of war. As she doesn’t want anything, she begs us to fight for peace. Her desire is to promote peace. She urges us to let every child grow, play and laugh. The only thing she want is peace: Fight for peace.

Critical comment: The poet uses the repetition to demonstrate that we must fight for peace and nothing else, or else innocent people and children would die for our unnecessary injustices.

కవి పరిచయం : ఈ పంక్తులు నజీమ్ హిక్మేట్చే రచించబడిన “హిరోషిమా చైల్డ్” అను పద్య౦లోని చివరి చరణంలోనివి. అతను ఒక టర్కిష్ కవి, నాటక రచయిత మరియు నవలా రచయిత. అతను 20వ శతాబ్దపు గొప్ప కవులలో ఒకడుగా గుర్తింపబడ్డాయి.

సందర్భ౦ మరియు అర్థం : చిన్నారి ఈ పద్య౦లో వక్త. 7 సం||ల చిరు ప్రాయంలోనే హిరోషిమా బాంబుపేలుడులో తన జీవితాన్ని కోల్పోయింది. అప్పటి నుండి ఎలాంట ఎదుగుదల లేదు, ఆకలి లేదు. అలానే ఉండిపోయింది. ప్రతి ఇంటానా సంచరిస్తుంది ఏమీ కోరటం లేదు. ఆహారం లేదా ఏ వస్తువులు అడగటం లేదు. యుద్ధ దుష్ప్రభావాలను గురించి మనల్ని హెచ్చరిస్తుంది. ఆమె ఏమి కోరుకోవటంలేదు.

మనల్ని శాంతి కోసం పోరాడమంటుంది. శాంతి స్థాపన ఆమె కోరిక. ప్రతి చిన్నారిని ప్రశాంతంగా ఎదగనివ్వమని, ఆడుకోనివ్వమని మరియు సంతోషంగా ఉండనివ్వమని మనల్ని అర్జిస్తుంది. కేవలం మన నుండి శాంతి కోసం పోరాడమని ఆ చిన్నారి కోరుకుంటుంది.

విమర్శ : మనందరం శాంతి కోసం పోరాడాలి లేదంటే అమాయక ప్రజలు, చిన్నారులు మన అనవసర అన్యాయాలకు ప్రాణాలు కోల్పోతారు అన్న విషయాన్ని ప్రదర్శించటానికి కవి పునరావృత్తం (మరల, మరల) ను ఉపయోగించాడు.

Paragraph Questions & Answers (Section A, Q.No.4, Marks: 4)
Answer the following Questions in about 100 words

a) What is the theme of the poem,”Hiroshima Child”?
Answer:
Nazim Hikmet is recognised as one of the greatest poets of the 20th century. Most of his writings are about war. His present poem, “Hiroshima Child”, is about a seven year old girl who was killed in the bombing of Hiroshima in the world war II. The little girl is the speaker in the poem. The soul of the girl knocks on every door to warn them about adverse effects of war. She begs them to fight for peace.

But, nobody pays attention to her as she is invisible. The theme of the poem is to promote peace. The poet uses a character, a standpoint and language to drive home the theme. This technique serves the desired purpose. The reader is made to plunge into thought first and action next. Thus, the vehement plea is for peace. Fight for Peace.

నజీమ్ హిక్మెట్ 20వ శతాబ్దపు గొప్పకవులలో ఒకరుగా గుర్తింపు పొందాడు. ఇతని రచనలలో చాలా వరకు యుద్ధం గురించినవే. ఇతని ప్రస్తుత పద్య౦ “హిరోషిమా చైల్డ్” రెండవ ప్రపంచయుద్ధంలో హిరోషిమాపై అణుబాంబు దాడిలో చనిపోయిన ఏడు సంవత్సరాల చిన్నారి గురించిన కావ్యం. ఆ చిన్నారే వక్త ఈ పద్య౦లో, యుద్ధ దుష్ప్రభావాల గురించి ప్రజలను హెచ్చరించడానికి, ఆ చిన్నారి ఆత్మ ప్రతి తలుపు తడుతుంది. వారిని శాంతి కోసం పోరాడమని ఆ చిన్నారి వేడుకుంటుంది.

కానీ, ఆమెకానరాదు కాబట్టి, ఎవ్వరీమెను పట్టించుకోరు. శాంతి స్థాపనే ఈ పద్య౦ యొక్క ఇతి వృత్తం. దీనికోసం కవి ఒక పాత్రను మరియు చక్కటి సరళమైన భాషను ఉపయోగించాడు. తాను అనుకున్న ఉద్దేశ్యం నెరవేరుస్తుంది ఈ నైపుణ్యం. పాఠకుడిని మొదట ఆలోచింపజేసి, తర్వాత కార్యానుకుడిని చేస్తుంది. అలా, శాంతి కోసం ఒక గట్టి ప్రయత్నం ఇది. శాంతి కోసం పోరాటం.

TS Inter 2nd Year English Study Material Chapter 3 Hiroshima Child

b) Why does the poet appeal for peace through a dead child?
Answer:
Nazim Hikmet is recognised as one of the greatest poets of the 20th century. Most of his writings are about war. His present poem, “Hiroshima Child” is about a seven year old girl who died in the Hiroshima bomb blast during the World War II. The little girl is the speaker in the poem. The soul of the girl knocks on every door to warn them about to let children grow happily.

The poet to fight for peace and to let children grow happily. The poet appeals for peace through the mouth of the little girl, a dead child. It is because the poet wants to press upon our guilty conscience. The child no longer needs food or water to survive because we have taken away her life with our unnecessary injustices. We have taken away everything from her. Thus, the poet uses a dead child to touch our hearts to promote peace.

20వ శతాబ్దపు గొప్ప కవులలో ఒకరుగా నజీమ్ హిక్మెట్ గుర్తిపు పొందాడు. ఇతని రచనలలో అధికం యుద్ధం గురించినవే. ఇతని ప్రస్తుత పద్య౦ “హిరోషిమా చైల్డ్”, రెండవ ప్రపంచయుద్ధ సమయంలో హిరోషిమాపై అణుబాంబు దాడిలో ప్రాణాలు కోల్పోయిన ఏడు సం॥ చిన్నారి గురించి. ఆ చిన్నారే ఈ పద్య౦లో వక్త. యుద్ధ దుష్ప్రభావాల గురించి ప్రజలను హెచ్చరించడానికి ఆ చిన్నారి ప్రతి తలుపు తడుతుంది. శాంతి కోసం పోరాడమని మరియు చిన్నారులను సంతోషంగా, ఆనందంగా ఎదగనివ్వమని జనాన్ని వేడుకుంటుంది. చనిపోయిన, చిన్నారి నోటి నుండి శాంతి కోసం పోరాడమని జనాన్ని కవి అర్జిస్తున్నాడు.

ఎందుకంటే, మన మనస్సాక్షికి కవి నొక్కి చెప్పాలనుకుంటున్నాడు. మనం మన అనవసరమైన అన్యాయాలతో ఆ చిన్నారి ప్రాణాలను తీసుకున్నాము కాబట్టి ఆ బిడ్డ జీవించడానికి ఆహారం లేదా నీరు అవసరంలేదు. మనము ఆమె నుండి సమప్తం తీసుకున్నాము. అలా శాంతిని పెంపొందించడానికి, మన హృదయాలను తాకడానికి చనిపోయిన చిన్నారిని కవి
ఉపయోగించాడు.

c) Describe the feelings of the child when she knew that she was dead at the age of seven. (Revision Test – III)
Answer:
Nazim Hikmet is recognised as one of the greatest poets of the 20th century. Most of his writings are about war. His present poem, “Hiroshima Child” is about a seven year old girl who died in the Hiroshima bomb blast during the World War II. The little girl is the speaker in the poem. The girl knocks on every door.

She says that she doesn’t want anything. It is because she has already died. She no longer needs food or water to survive. She continues to be in the same state. She visits every home and begs them to fight for peace and to let children grow happily. Her feelings touch our hearts and awake us plunge into thought first and action next through her mouth, the poet emphasizes the need for peace.

20వ శతాబ్దపు గొప్ప కవులలో ఒకరుగా నజీమ్ హిక్మెట్ గుర్తిపు పొందాడు. ఇతని రచనలలో అధికం యుద్ధం గురించినవే. ఇతని ప్రస్తుత పద్య౦ “హిరోషిమా చైల్డ్”, రెండవ ప్రపంచయుద్ధ సమయంలో హిరోషిమా పట్టణంపై అణుబాంబు దాడిలో ప్రాణాలు కోల్పోయిన ఏడు సం॥ చిన్నారి గురించినది ఈ పద్య౦. చిన్నారి ఈ పద్య౦లో వక్త. ఈ చిన్నారి ప్రతి తలుపు తడుతుంది. తనకు ఏమి వద్దని చెప్తుంది. ఎందుకంటే ఆమె చనిపోయింది అప్పటికీ.

ఆమె జీవించడానికి ఆహారం లేదా నీరు అవసరంలేదు. ఆ పరిస్థితిలోనే ఉంటుంది. ప్రతి గడప సందర్శిస్తుంది. అందరినీ శాంతి కోసం పోరాడమని, చిన్నారులను ఆనందంగా ఎదగనివ్వమని వేడుకుంటుంది. ఆమె అభిప్రాయాలు మన హృదయాన్ని తాకుతున్నాయి. మరియు మనల్ని మొదటిగా ఆలోచింపచేస్తాయి. తర్వాత కార్యోన్ముకులను చేస్తున్నాయి.

d) “I ask for nothing for myself” Why do you think the child asked nothing for herself?
Answer:
Nazim Hikmet is recognised as one of the greatest poets of the 20th century. Most of his writings are about war. His present poem, “Hiroshima Child” is about a seven year old girl who died in the Hiroshima bomb blast during the World War II. The little girl is the speaker in the poem. The soul of the girl knocks on every door to warn them about the evils of war. She begs people to fight for peace. She needs neither food nor water to survive. It is because we have taken away everything from this child.

Therefore, she says that she needs, nothing for herself the only thing she wants is our fight for peace which makes the future generation happy to live, Her plea is for peace. Fight for peace. She urges us to fight for peace and nothing else or else innocent people and children will die for our injustices.

20వ శతాబ్దపు గొప్ప కవులలో ఒకరుగా నజీమ్ హిక్మెట్ ఒకరుగా పేరు పొందాడు. ఇతని రచనలలో అధికం యుద్ధం గురించినవే. ప్రస్తుత కావ్యం “హిరోషిమా బాలిక”, రెండవ ప్రపంచయుద్ధ కాలంలో హిరోషిమా బాంబు దాడిలో మరణించిన ఏడు సం॥ చిన్నారి గురించి. ఆ చిన్నారి ఈ పద్య౦లో వక్త. ఈ చిన్నారి ప్రతి తలుపు తడుతూ యుద్ధం దుష్ప్రభావాల గురించి హెచ్చరిస్తుంది. శాంతి కోసం పోరాడమని జనాన్ని వేడుకుంటుంది.

తాను చనిపోయినందున, తనకేమీ అవసరం లేదు అంటుంది. జీవించడానికి అవసరమైన ఆహారం లేదా నీళ్ళ సహితం అవసరం లేదు ఆ చిన్నారికి ఎందుకంటే మనమే ఆ చిన్నారి జీవితాన్ని చిదిమివేశాము. కావున, తనకేమి అవసరం లేదు అంటుంది. తాను కోరుకున్నది కేవలం శాంతి కోసం మన పోరాటం. అదే మన భవిష్యత్తు.

One the Grasshopper and Cricket Summary in English

About Author

TS Inter 2nd Year English Study Material Chapter 3 Hiroshima Child 1
Nazım Hikmet Ran (15 January 1902 – 3 June 1963) commonly known as Nazim Hikmet was a Turkish-Polish poet, playwright, novelist, screenwriter, director and memoirist. He was acclaimed for the “lyrical flow of his statements”. Described as a “romantic communist” and “romantic revolutionary”, he was repeatedly arrested for his political beliefs and spent much of his adult life in prison or in exile. His poetry has been translated into more than fifty languages.

1961: Legend of Love (by Arif Malikov)
1935: Letters to Taranta-Babu (Poem)
1966-67, Human Landscapes from My Country (Poem)
1965: The Epic of the War of Independence(Poem)

The heart touching poem, ‘Hiroshima Child’ is written by Nazim Hikmet. He is Turkish poet, playwright and novelist. His present poem is about a seven-year-old girl who died in the Hiroshima bomb attack during the world war II. It deals with the adverse effects of war loss of life, innocence and destruction. It is a call for peace.

This is a short poem of only five stanzas. The little girl, who is no longer alive, is the speaker in the poem. The poet describes the experiences of the little girl during the war. He wants to remind us of the innocent lives that were killed in the bombing of Hiroshima in the world War II. Therefore, he uses a character to drive home theme. The poem begins with the little girl knocking on every door. The soul of the girl knocks on every door to warn them about the evils of war. She requests them not fight as she was victim of it.

She begs them to fight for peace. But, No one hears or sees as she is invisible. It is because she died at seven in the bomb blast since then the child has felt neither growth nor hunger, nor any wants. She continues to be in the same state. The poet respects the lines “For I am dead for I am dead” throughout the poem to remind us that we have killed this innocent child with our unnecessary violence.

In the third stanza, the poet depicts the child’s hair, eyes and bones to drive into our heads that we hard stained a poor child who was just like us. Here, the child visits every home and seeks, neither food nor things. The poet respects that the wild no longer needs any material things because she is dead. The vehement plea is for peace.

In the final stanza, the girl begs people to fight for peace and to let children grow, play and laugh happily. The poet has delivered a very single but serious message in a clear and short manner. Therefore, we must fight for peace and nothing else, or else innocent people and children will die for our wicked actions or unnecessary injustices. The war against war touches our hearts.

TS Inter 2nd Year English Study Material Chapter 3 Hiroshima Child

One the Grasshopper and Cricket Summary in Telugu

Note: This summary is only meant for Lesson Reference, not for examination purpose

మనస్సును రంజింపజేసే కావ్యం “హిరోషిమా చైల్డ్” నజీమ్ హిక్మేట్చే రచింపబడింది. అతను ఒక టర్కిష్ కవి, నాటకరచయిత మరియు నవలారచయిత. రెండవ ప్రపంచ యుద్ధ సమయంలో హిరోషిమా పట్టణం పై అణుబాంబు దాడి జరిగినపుడు చనిపోయిన ఏడు సం॥ల చిన్నారి గురించి ఇతని ప్రస్తుత పద్య౦. యుద్ధ దుష్ప్రభావాలు ముఖ్యంగా జీవితాలు కోల్పోవడం, చిన్నార్లు చనిపోవటం మరియు వినాశనం జరగటం మొ॥వి ఈ పద్య౦ ఇతివృత్తం. ఇది శాంతి కోసం పిలుపు.

ఇది కేవలం ఐదు stanzas గల చిన్న పద్య౦. చనిపోయిన చిన్నారి ఈ పద్య౦లో వ్యక్తి. యుద్ధ సమయంలో ఆ చిన్నారి అనుభవాలు, వేదనలు గురించి కవి వివరిస్తున్నాడు. రెండవ ప్రపంచయుద్ధ సమయంలో హిరోషిమాకై అణుబాంబు దాడిలో చంపబడిన అమాయక జీవితాలు గురించి మనకు తెలియజేయాలనుకుంటున్నాడు. కావున, ఒక చిన్నారి పాత్ర ద్వారా తన సందేశాన్ని అందిస్తున్నాడు. చిన్నారి ఆత్మ యొక్క దుష్ప్రభావాలను గురించి మనల్ని హెచ్చరించడానికి ప్రతి తలుపు తడుతుంది.

అలా గడపగడపకు తిరుగుతూ యుద్ధం చేయవద్దని ప్రాదేయపడుతుంది. తాను దాని బాధితురాలుగా, శాంతి కోసం పోరాడమని వేడుకొంటుంది. కానీ, ఆమెను ఎవ్వరూ పట్టించుకోరు. ఆమె కానరాదు కాబట్టి. ఎందుకంటే ఆ చిన్నారి 7 సం॥ల వయస్సులోనే బాంబుదాడిలో మరణించింది కాబట్టి అప్పటి నుండి, ఆ చిన్నారికి ఎదుగుదల లేదు, ఆకలి లేదు, ఏమిలేదు. అలానే ఉండిపోయింది. మన అనవసర హింసతో ఆ అమాయక చిన్నారిని చంపివేశామని గుర్తుచేయటానికి పద్య౦ ఆసాంతం కవి “For Iam dead for Iam dead” అను పంక్తులను ఉపయోగిస్తున్నాడు మళ్ళీ మళ్ళీ.

మూడవ stanza లో మనలాంటి చిన్నారిని మనం చంపేశాం అని ఆ చిన్నారి జుట్టు, కళ్ళు మరియు ఎముకలు మన తలల్లోకి ఎక్కిరించేలా కవి వర్ణించాడు. ఇక్కడ ఆ చిన్నారి ప్రతి ఇంటిని సందర్శిస్తుంది. ఆహారం కానీ వస్తువులు కానీ ఏదీ వెతకదు. ఆ చిన్నారి చనిపోయినందున ఆమెకు ఇకపై ఎటువంటి పదార్థాలు, వస్తువులు అవసరం లేదని కవి పునరావృత్తం చేస్తాడు. శాంతి కోసం గట్టి విన్నపం ఇది.

ఆఖరి చరణంలో ఆ చిన్నారి శాంతి కోసం పోరాడాలని మరియు పిల్లలు పెరగడానికి, ఆడుకోవడానికి మరియు సంతోషంగా నవ్వడానికి ప్రజలను వేడుకుంటుంది. కవి చాలా సరళమైన కానీ గంభీరమైన సందేశాన్ని స్పష్టంగా మరియు తేలికగా అందించారు. కాబట్టి, మన ప్రపంచ శాంతికోసం పోరాడాలి. లేదంటే అమాయక ప్రజలు మరియు పిల్లలు మన దుష్ట చర్యలను లేదా అనవసర అన్యాయాలను చనిపోతారు. యుద్ధానికి వ్యతిరేఖంగా యుద్ధం, మన హృదయాలను తాకుతుంది.

One the Grasshopper and Cricket Summary in Hindi

Note: This summary is only meant for Lesson Reference, not for examination purpose

दिल को छूनेवाली ‘हीरोशिमा चाइल्ड’ कविता नज़ीम हिकमेट से लिखी गई है । वे एक तुर्की कवि, नाटककार और उपन्यासकार हैं । यह वर्तमान कविता एक सात वर्षीय लड़की के बारे में है, जो द्वितीय विश्व युद्ध के दौरान हीरोशिमा बम हमले में मर गई थी । यह युद्ध के प्रतिकूल प्रभावों, जीवन की हानि और विनाश से संबंधित है । यह शांति का आहवान है ।

यह केवल पाँच छदों की एक छोटी कविता के रूप में है। छोटी लड़की, जो अब जीवित नहीं है, कविता में वक्ता है । कवि युद्ध में गोता लगानेवाले छोटी लड़की के अनुभवों का वर्णन करते हैं । वे हमें उन निर्दोष लोगों की याद दिलाना चाहते हैं, जो द्वितीय विश्व युद्ध में हिरोशिमा की बमबारी में मारे गए थे । वे होम थीम को चलाने के लिए एक चरित्र का उपयोग करते है । कविता का आरंभ छोटी लड़की के हर दरवाजे पर दस्तक देने से होता है। लड़की की आत्मा लड़ाई की बुराई के बारे में चेतावनी देनी के लिए हर दरवाजे पर दस्तक देती है, वह उनसे मतलड़ने का अनुरोध करती है क्यों कि वह उसका शिकार थी ।

वह उनसे शांति के लिए लड़ने की भीख माँगती है । लेकिन कोई नहीं सुनता था नहीं देखता है क्यों कि वह अदृश्य है । ऐसा इसलिए है कि वह सात साल की उम में बम विस्फोट में मरगई । तब से लडकी को न तो कोई विकास हुआ और न ही भूख, न ही कोई इच्छाएँ । वह उसी स्थिति में बनी हुई है । कवि पंक्तियों को दोहराते हैं, ‘फ़ार आई एम डेड, फ़ार आई एम डेड’, पूरी कविता में हमें याद दिलाने के लिए कि हमने इस मासूम बच्चे को अपनी अनावश्यक हिंसा से मार डाला है ।

तीसरे छंद में कवि ने बच्ची के बालों, आँयों और हड्डियों को हमारे म्स्तष्कों में घुसाने के लिए दर्शाया है कि हम ने एक बेचारी बच्ची को मार डाला है जो हमारी जैसी ही थी । इधर बच्ची हर घर जाती है और न तो खाना माँगती है और न चीजें । कवि दोहराते हैं कि बच्ची को किसी भौतिक वस्तु की आवश्यकार नही हैबथाकि वह मन चुकी है । उसका प्रगाढ अनुनय विनय शांति का है ।

अंतिम धंद में, लड़की लोगों से माँगती है कि वे शांति के लिए लड़ें और बच्चों को बढ़ने, खेलने और खुशी से हँसने दें। कवि ने स्पष्ट और संक्षिप्त तरीके से एक बहुत ही सरल लेकिन गंभीर संदेश दिया है । अतः शांति के लिए अवश्य लड़ें और कुछ नहीं वरना हमारे दुष्ट कार्यों और अनावश्यक प्रवृत्ति से मासूम लोग और बच्चे मरेंगे । युद्ध के खिलाफ युद्ध हमारे दिलों को छूता है ।

Meanings and Explanations

Hiroshima (prop.n)/ (హిరోషిమా)/ : a city in Japan; became known well to the world as powerful (atom and hydrogen) bombs were dropped on it an 06.08.1945 during the World War II; జపాన్ నగరం, 1945 ఆగస్ట్ 6న అణుబాంబుల తాకిడితో (రెండవ ప్రపంచ యుద్ధంలో) ప్రపంచానికి ఈ పట్టణం బాగా తెలిసింది. ఆ అంశమే ప్రస్తుత పద్యానికి మూలాధారము.
जापान में एक शहर : दुनिया के लिए अच्छी तरह जाना जाता है क्योंकि द्विती ( ) विश्वयुद्ध के दौरा न. 06-08-1945 को शक्तिशाली (परमाणु और, हाइड्रोजन) बम गिराए गए थे :

tread (n)/(ట్రడ్)/tred : a step ; అడుగు, నడక ; एक कदम : एक सैर

TS Inter 2nd Year English Study Material Chapter 3 Hiroshima Child

knock(v) /(నోక్) / nɒk: తలుపు పై కొట్టు తట్టు లోపలికి వెళ్ళుటకు అనుమతికై;

scorched (స్కో(ర్)చ్ ట్)/skɔ:tʃt : burned ; కాల్చబడును, जला हुआ

swirling (v+ing )(adj) / (స్వ(ర్)లింగ్) / swз:lŋ () : twisting ; సుడులు తిరుగుతున్న, घूमना

scattered (స్క్యా ట(ర్)డ్) : dispersed; spread; వెదజల్లబడెను, बिखरा हुआ, फैलाव

TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Telangana TSBIE TS Inter 1st Year Physics Study Material 11th Lesson Mechanical Properties of Fluids Textbook Questions and Answers.

TS Inter 1st Year Physics Study Material 11th Lesson Mechanical Properties of Fluids

Very Short Answer Type Questions

Question 1.
Define average pressure. Mention it’s unit and dimensional formula. Is it a scalar or a vector? [AP Mar. ’17]
Answer:
Average pressure (Pav) :
The normal force acting per unit is called average pressure.
⇒ Pav = \(\frac{F}{A}\)
Units : In SI = Nm-1
The dimensional formula : ML-1T-2
It is a scalar quantity.

Question 2.
Define Viscosity. What are its units and dimensions? [AP May ’16, ’13; TS May ’18, June ’15)
Answer:
Viscosity :
The property of a fluid which opposes the relative motion between the layers is called viscosity.

Units in SI:
Coefficient of viscosity Nm-2 s (or) Pa – s. (or) Poiseuille
Units in C.G.S : Coefficient of viscosity = poise.
The dimensional formula = ML-1 T-1.

Question 3.
What is the principle behind the carburetor of an automobile? [TS Mar. ’18, ’17; AP Mar. ’19, ’15; June ’15]
Answer:
Carburetor of an automobile is based on the principle of “Bernoulli’s theorem”.

Question 4.
What is magnus effect? [AP May ’18, ’17, Mar. ’15; TS Mar. ’19, ’16]
Answer:
Magnus effect :
When a spinning ball is thrown it deviates from its usual path in flight. This effect is called Magnus effect.

Question 5.
Why are drops and bubbles spherical? [AP Mar. ’18, ’17, ’16, ’14, May ’18, ’17, ’16, ’14, ’13; TS May ’18, ’17, ’16]
Answer:
Due to property of surface tension, the surface of liquid behaves like a stretched membrane and has a tendency to acquire minimum surface area. The sphere has minimum surface area when compared to other shapes of same volume.

Therefore, drops and bubbles acquire spherical shape in order to have the minimum surface area.

TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 6.
Give the expression for the excess pressure in a liquid drop. [TS Mar. ’17]
Answer:
Excess of pressure in a liquid drop is,
p = \(\frac{2s}{r}\) where ‘s’ = surface tension and ‘r’ = the radius of the liquid drop.

Question 7.
Give the expression for the «<cess pressure in an air bubble inside the iquid. [AP Mar. ’19]
Answer:
Excess of pressure in an air buble inside the liquid is, P = \(\frac{2S}{R}\)
where S = Surface Tension, R = Radius of air bubble of liquid.

Question 8.
Give the expression for the excess pressure in a soap bubble in air. [TS Mar. ’16]
Answer:
Excess of pressure in a soap bubble is, p = \(\frac{4s}{r}\) where
‘s’ = surface tension and
‘r’ = radius of the drop.

Question 9.
What are water proofing agents and water wetting agents? What do they do?
Answer:
Water proofing agents :
The substances which are used to increase the angle of contact are called “water proofing agents”.

Wetting agents:
The substances which are used to decrease the angle of contact are called “wetting agents”.
Ex: Soaps, detergents and dying substances.

Question 10.
Why water droplets wet the glass surface and does not wet lotus leaf? [TS Mar. ’15]
Answer:
Angle of contact between water drop and glass is less than 90° so water drop will wet glass surface.
Angle of contact between water and lotus leaf is greater than 90°. So water drops cannot wet lotus leaf.

Question 11.
What is angle of contact? [AP May ’14, Mar. 16]
Answer:
Angle of contact: It is the angle between the walls of the container and the tangent drawn over the surface of the liquid. This angle must be. measured in the interior side of the liquid.

TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 12.
Mention any two examples (or) applications that Obey Bernoullis theorem and justify them. [AP Mar. ‘ 18; TS Mar. 15]
Answer:
Applications of Bernoulli’s theorem :

  1. Dynamic lift on the wings of an aeroplane
  2. Swinging of a spinning cricket ball is a consequence of Bernoulli’s theorem.
  3. During cyclones, the roof of thatched houses will fly away. This is a consequence of Bernoulli’s theorem.

Question 13.
When water flows through a pipe, which of the layers moves fastest and slowest? [TS June ’15]
Answer:
When water is flowing through a pipe water layers in contact with bottom layers of pipe will have lowest velocity and water layer just below the top of inner layer of pipe will have highest velocity.

Question 14.
“Terminal velocity is more if surface area of the body is more.” Give reasons in support of your answer.
Answer:
Yes, Terminal velocity of a body is more when surface area of a body is more.

According to Stokes formula, terminal velocity of a smooth spherical body is,
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 1

The surface area of a spherical body A = 4πr². So when surface increases, ‘r²’ value increases. Hence from Stoke’s formula, Teminal velocity increases.

Short Answer Questions

Question 1.
What is atmospheric pressure and how is it determined using Barometer?
Answer:
The atmospheric pressure at any point is equal to the weight of a vertical column of air of unit cross-sectional area extending from that point to the top of tfie earth’s atmosphere.
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 2

Determination of atmospheric pressure using Barometer :
A long tube closed at one end and filled with mercury is inverted into a trough of mercury. This device is known as mercury barometer. The space above the mercury column in the tube contains only mercury vapour whose pressure p is so small that it may be neglected. Otherwise there is a perfect vacuum, which is called Torricellian vacuum.

The pressure inside the column at point A must equal the pressure at point B, which is at the same level.

‘P’ at A = Pressure at B = atmospheric pressure = Pa
Pa = ρgh …………. (1)
where ρ = density of mercury
h = height of mercury column

In this experiment, it is found that the mercury column in the barometer has a height of about 76 cm at sea level equivalent to one atmosphere (1 atm).

At sea level, atmospheric pressure is the pressure exerted by 0.76 m of mercury column, i.e., h = 0.76 m
ρ = 13.6 × 10³ kg m-3 And g = 9.8 ms-2.
∴ Atmospheric pressure, Pa = hρg
= 0.76 × (13.6 × 10³) × 9.8
= 1.013 × 105 Nn-2 (or) Pa

A common way of stating pressure is in terms of cm or mm of mercury (Hg). A pressure equivalent to 1 mm is called a torr.
1 torr = 133 Pa.

TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 2.
State Dalton’s law of partial pressures. [AP Mar. ’14]
Answer:
Dalton’s law of partial pressures :
For a mixture of non interacting ideal gases at same temperature and volume total pressure in the vessel is the sum of partial pressures of individual gases.
i.e. P = P1 + P2 + ……….. total pressure
P1, P2, ……… etc. are individual pressures of each gas.

Question 3.
What is gauge pressure and how is a manometer used for measuring pressure differences?
Answer:
Gauge Pressure :
The pressure p, at depth below the surface of a liquid open to the atmosphere is greater than atmospheric pressure by an amount ρgh. The excess pressure (P – Pa), at depth h is called as “gauge pressure at that point.”

Measurement of pressure difference using a Manometer :
An open tube manometer is a useful instrument for measuring pressure differences. It consists of a U-tube containing a suitable liquid i.e., a low density liquid (such as oil) for measuring small pressure differences and high density liquid (such as mercury) for large pressure differences. One end of the tube is open to the atmosphere and other end is connected to the system whose pressure to be measured.
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 3

The pressure P at A is equal to pressure at point B. If the pressure in the vessel is more than the earth’s atmospheric pressure, then the level of liquid in arm-I of U-tube will go down upto point A and the level of liquid in arm-II of U-tube will rise up to point C. Then the pressure of air in vessel is equal to pressure at point A. Let ‘h’ be the difference of liquid levels in the two arms of U-tube. Let ρ be the density of liquid in U-tube and Pa be the atmospheric pressure.

Since, the pressure is same at all points, at the same level, so pressure at point A,
PA = pressure at point B
= pressure at C + pressure due to column of liquid of height ‘h’.
So, PA = PC + hρg or PA – PC = hρg ………… (1)
Here, PC = Pa = atmospheric pressure.
If PA = P, then from eq ………. (1)
P – Pa = hρg
Here, P – Pa = Pg = gauge pressure = hρg.

Question 4.
State Pascal’s law and verify it with the help of an experiment.
Answer:
Pascal’s law:
It states that “the pressure in a fluid at rest is the same at all points if they are at the same height”.

Proof of Pascal’s law:
Consider an element in the interior of a fluid at rest as shown in the figure. The element ABC – DEF is in the form of a right-angled prism.

In principle, this prismatic element is very small so that every part of it can be considered at the same depth from the liquid surface and therefore, the effect of gravity is the same at all these points.

The forces on this element are those exerted by the rest of the fluid and they must be normal to the surfaces of the element.
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 4

Thus, the fluid exerts pressures Pa, Pb and Pc on this element of area corresponding to the normal forces Fa, Fb and Fc as shown in fig. on the faces BEFC, ADFC and ADEB denoted by Aa, Ab and Ac respectively. Then,
Fb sin θ2 = Fc and Fb cos θ2 = Fa (by equilibrium)
Ab sin θ2 = Ac and Ab cos θ2 = Aa (by geometry)
Thus, \(\frac{F_b}{A_b}=\frac{F_c}{A_c}=\frac{F_a}{A_a}\) ⇒ Pb = Pc = Pa
Hence, pressure exerted is same in all directions in a fluid at rest.

This proves the Pascal’s law.

Question 5.
Explain hydraulic lift and hydraulic brakes.
Answer:
Hydraulic lift and Hydraulic brakes are based on the Pascal’s law. The principle states that “whenever external pressure is applied on any part of a fluid contained in a vessel, it is transmitted undiminished and equally in all directions

Hydralic lift:
In a hydraulic lift, two pistons are separated by the space filled with a liquid as shown in fig.

A piston of small cross-section A1 is used to exert a force F, directly on the liquid.

The pressure, P = \(\frac{F_1}{A_1}\) is transmitted throughout the liquid to the larger cylinder attached with a larger piston of area A2, which results in an upward force of P × A2.
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 5

Therefore, the piston is capable of supporting large force.
F2 = PA2 = \(\frac{F_1A_2}{A_1}\)

By changing the force at A1, the platform can be moved up or down. Thus, the applied force has been increased by a factor of \(\frac{A_2}{A_1}\) and this factor is the mechanical advantage of the device.

Hydraulic brakes :
Hydraulic brakes in automobiles also work on Pascal’s principle. When we apply a little force on the pedal with our foot, the master piston moves inside the master cylinder, and the pressure caused is transmitted through the brake oil to act on a piston of larger area. A large force acts on the piston and is pushed down expanding the brake shoes against brake lining. In this way, a small force on the pedal produces a large retarding force on the wheel.

An important advantage of the system is that the pressure set up by pressing pedal is transmitted equally to all cylinders attached to the four wheels so that the braking effort is equal on all wheels.

TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 6.
What is hydrostatic paradox?
Answer:
Hydrostatic paradox :
This is useful to prove that the liquid pressure is the same at all points at the same horizontal level.
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 6

Consider three vessels A, B and C of different shapes as shown in the figure. They are connected at the bottom by a horizontal pipe. On filling with water, the level in the three vessels is the same though they hold different amounts of water. This is so, because water at the bottom has the same pressure below each section of the vessel.

Thus, it proves that the height of the fluid column is independent of the cross sectional or base area and the shape of the container

Question 7.
Explain how pressure varies with depth.
Answer:
Variation of pressure with depth:
Consider a fluid at rest in a container. Let point 1 is at a height h’ above a point 2 as shown in the figure. Consider a cylindrical element of fluid having area of base ‘A’ and height ‘h’. As the fluid is at rest, the resultant horizontal forces should be zero and the resultant vertical forces should balance the weight of the element. The forces acting in the vertical direction are due to the fluid pressure at the top (P1A) acting downward, at the bottom (P2A) acting upward.
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 7

If ‘mg’ is weight of the fluid in the cylinder, we have
(P2 – P1) A = mg …………. (1)
Now, if ρ is the mass density of the fluid, then mass of fluid, m = ρv
⇒ m = ρhA …………. (2)
∴ From (1) and (2)
P2 – P1 = ρgh …………. (3)

Pressure difference depends on the vertical distance ‘h’ between the points (1 and 2), mass density of the fluid p and acceleration due to gravity ‘g’.

If the point 1 under discussion is shifted to the top of the fluid, which is open to the atmosphere, P1 may be replaced by atmospheric pressure (Pa) and we replace P2 by P2 then eq.(2) becomes
P – Pa = ρgh ⇒ P = Pa + ρgh

Thus, the pressure P, at depth below the surface of a liquid open to the atmosphere is greater than atmospheric pressure by an amount ρgh.

Question 8.
What is Torricelli’s law? Explain how the speed of efflux is determined with an experiment.
Answer:
Torricelli’s law :
Torricelli discovered that the speed of efflux from an open tank is given by a formula identical to that of a freely falling body. The word efflux means ‘fluid outflow’.
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 8

Determination of speed of Efflux:
Consider a tank containing a liquid of density ‘ρ’ with a small hole at a height ‘y1‘, from the bottom as shown in the figure.

The air above the liquid, whose surface is at height ‘y2‘, is at pressure, P.

From the equation of continuity, we have
V1 A1 = V2A2 ⇒ V2 = \(\frac{A_1}{A_2}\)V1 …….. (1)

It the cross sectional area of the tank, A2 is much larger than that of the hole (i.e., A2 >> A1), then we may consider the fluid to be approximately at rest at the top. i.e., V2 = o.

Now, applying the Bernoulli equation at points (1) and (2) and noting that, at the hole P1 = Pa, the atmospheric pressure, we get.
Pa + \(\frac{1}{2}\)ρv²1 + ρgy1 = P + ρgy2
Taking y2 – y1 = h, we have
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 9

When P > > Pa and 2gh may be ignored, the speed of efflux is determined by the container pressure. Such a situation occurs in rocket propulsion. On the other hand if the tank is open to the atmosphere, then P = Pa and from eq (2), we get
V1 = \(\sqrt{2gh}\) ………….. (3)

This is the speed of a freely falling body, at any point of height ‘h’ during its fall. This equation is known as “Torricelli’s law”.

Question 9.
What is Venturimeter? Explain how it is used.
Answer:
Venturi-meter :
The venturi meter is a device to measure the flow speed of incompressible fluid.

It consists of a tube with a broad diameter and a small constriction at the middle as shown in the figure.
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 10

The manometer contains a liquid of density ρm. The speed ν1 of the liquid flowing through the tube at the broad neck area A is to be measured. From equation of continuity, the speed at the constriction, ν2 = \(\frac{A}{a}\) ν1
According to Bernoulli’s equation,
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 11

This pressure difference causes the fluid in the U-tube connected at the narrow neck to rise in comparison to the other arm.

The difference in height h measures the pressure difference.
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 12

Uses: Venturi meter is used for measuring the speed of incompressible liquid and rate of flow of liquid through pipes.

TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 10.
What is Reynold’s number? What is its significance?
Answer:
Reynold’s number Re :
Reynold’s number is a pure number which determines the nature of flow of liquid through a pipe.
Re = \(\frac{\rho v d}{\eta}\)
where
η = coefficient of viscosity of the liquid.
ρ = density of liquid
ν = critical velocity of the liquid flowing through the pipe.
d = diameter of the pipe.

Significance :
Re is dimensionless number and therefore, it remains same in any system of units.

The critical Reynold’s number for the onset of turbulence is in the range 1000 to 10000, depending on the geometry of the flow. For most cases

Re < 1000 signifies laminar flow.
1000 < Re < 2000 is unsteady flow.
Re < 2000 implies turbulent flow.

Reynold’s number describes the ratio of the inertial force per unit area to the viscous force per unit area for a flowing fluid.

Question 11.
Explain dynamic lift with examples.
Answer:
Dynamic lift on a spinning ball :
Consider the motion of a spinning ball. Its motion consists of two parts 1) Translatory motion 2) Self rotation called spinning.

1) Translatory motion:
Due to translatory motion it passes through the medium air with a velocity say (V). Due to translatory motion the number of streamlines on the top of the ball and at the bottom of the ball are equal. So there is no resultant force on the ball due to translatory motion through the fluid. Hence dynamic lift is zero.
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 13

2) For spinning motion :
Let the ball rotates about its axis with a velocity say ∆V in clockwise direction since surface of ball is not perfectly smooth it will drag air molecules with it. So at the top layers the velocity of air is V + ∆V due to air drag. At bottom layers the velocity of air is V – ∆V.

As velocity is more at top layers pressure is less and velocity is less at bottom layers, so pressure is high at bottom layers. This is due to Bernoulli’s theorem.

Due to the pressure difference at bottom layers and top layers some upward thrust will act on the ball. So some dynamic lift will act on a spinning ball. As a result the path of a spinning ball is curved.

Question 12.
Explain Surface Tension and Surface energy. [AP Mar. ’13]
Answer:
Surface tension the force per unit length on an imaginary line drawn on the surface of the liquid and acting perpendicular to it.
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 14

Surface energy:
The work done to increase the surface area of a liquid is called surface energy.
Surface energy = Surface tension × Increase in surface area.

Question 13.
Explain how surface tension can be measured experimentally.
Answer:
To find surface tension of a liquid in laboratory torsional balance is used. It is as shown in figure. It consists of a movable metallic rod fixed on a stretched wire. The position of rod can be adjusted. A glass plate is attached at one end of the rod and weight’s pan is connected at the other end of the rod.
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 15

Procedure :
A cleaned glass plate is taken. Its length ‘l’ and thickness ‘t’ is measured. Since thickness ‘l’ is very small when compared with length t, thickness ‘t’ is ignored.

Glass plate is fixed to metallic rod. Necessary weights are placed in the pan and glass plate is made horizontal to the table. Weights in pan W0 is measured. Pure liquid whose surface tension is to be determined is taken in a glass beaker. The liquid is poured until it just touches the glass plate. Now plate is pulled down with some force due to surface tension of liquid. Weights in the pan are gradually increased until the glass plate is just escaped from forces of surface tension. Weights W1 are noted. The experiment is repeated for three to four times and average weight W1 is noted.
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 16

Long Answer Questions

Question 1.
State Bernoulli’s principle. From conservation of energy in a fluid flow through a tube, arrive at Bernoulli’s equation. Give an application of Bernoulli’s theorem.
Answer:
Bernoulli’s theorem :
Bernoulli’s theorem states that “when a non viscous liquid flows between two points then the sum of pressure energy, kinetic energy and potential energy per unit mass is always constant at any point in the path of that liquid”.
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 17

Bernoulli’s theorem is applicable to non viscous, incompressible and irrotational liquids in streamline flow only.

Proof :
Let us consider that a liquid of density ‘ρ’ is flowing through a pipe of different area of cross sections A1 and A2 as shown.
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 18

Let the liquid enters at A1 with a velocity V1 and with a pressure P1, density of liquid at A1 is say ρ. Let the liquid leaves the pipe through A2 with a velocity V2 and pressure P2. Density of liquid at A2 is say ρ.

Since liquid is incompressible, p is con-stant.

At region 1 the liquid will move a distance of V1 ∆t where ∆t is very small time interval. Similarly at region 2 the liquid will move through a distance V2 ∆t.
Work done on fluid at region 1 = W1 = P1 A1
V1 ∆t = P1∆V

Work done on fluid at region 2 = W2 = P2 A2 V2 ∆t = P2 ∆V

Since same volume of liquid pass through the pipe, ∆ is constant.

∴ Work done by fluid = W1 – W2 = (P1 – P2) ∆V → 1

∵ Liquid is uncompressible ‘ρ’ is constant.

So mass of liquid entering the pipe and leaving the pipe ∆m is given by
∆m = ρA1V1∆t = ρ∆V
Change in potential energy of liquid
∆U = ρg∆v(h2 – h1) → 2
Change in kinetic energy of liquid
∆K = \(\frac{1}{2}\)ρ∆v(V²1 – V²2)
From work energy theorem work done = change in energy
∴ ∆W = ∆U + ∆K
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 19

i.e., sum of pressure energy, potential energy and kinetic energy of the fluid is always constant.

Limitations :
Bernoulli ‘s theorem is applicable to non-viscous and uncompressible liquids only.

Applications of Bernoulli’s theorem :

  1. Dynamic lift on the wings of an aeroplane is due to Bernoulli’s theorem.
  2. Swinging of a spinning cricket ball is a consequence of Bernoulli’s theorem.
  3. During cyclones, the roof of thatched houses will fly away. This is a consequence of Bernoulli’s theorem.

TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 2.
Define coefficient of viscosity. Explain Stoke’s law and explain the conditions under which a rain drop attains terminal velocity, υt. Give the expression for υt.
Answer:
Coefficient of viscosity (η) :
The viscous force acting tangentially on unit area of the liquid when there is a unit velocity gradient in the direction perpendicular to the flow is defined as “Coefficient of viscosity.”

Coefficient of viscossity η = \(\frac{-F}{A}\frac{dx}{dv}\)
Unit Nm-2 – s (or) pascal – second.

According to Stoke’s law, the viscous force acting on a freely falling, smooth spherical body of radius ‘a’ is proportional to the coefficient of viscosity η, radius ‘a’ and velocity ‘υ’ of the body.
∴ F ∝ ηav or F = 6 π η av, where 6π is the proportionality constant.
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 20

A rain drop of radius ‘ω’, density ρ falling under gravity through air of density a experiences a force of buoyancy equal to the weight of displaced air which is (\(\frac{1}{2}\)πa³) σg.

The weight of the rain drop acting downwards = (\(\frac{1}{2}\)πa³) ρg.

∴ Resultant force acting downwards = \(\frac{1}{2}\)πa³ρg – \(\frac{1}{2}\)πa³σg

When this force is equal to the viscous drag acting upwards, then the rain drop acquires a constant velocity called terminal velocity, vt.

At terminal velocity viscous drag = 6πηav.
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 21

Definition:
Terminal velocity of a body falling through a liquid is defined as that constant velocity which the body acquires when it falls in a fluid.

Problems

Question 1.
Find the excess pressure inside a soap bubble of radius 5 mm. (Surface tension is 0.04 N/m). [TS May ’16]
Solution:
Radius r = 5 mm = 5 × 10-3 m.
Surface tension ST = 0.04 N/m
= 4 × 10-2 N/m.
Excess pressure inside soap bubble
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 22

Question 2.
Calculate the work done in blowing a soap bubble of diameter 0.6 cm. against the surface tension force. (Surface tension of soap solution = 2.5 × 10-2 Nm-1)
Solution:
Work done = Surface tension (S) × increase of area (2 × 4πr²)
∴ W = S ( 4πr² ) × 2
(since the bubble has two surfaces)
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 23

Question 3.
How high does methyl alcohol rise in a glass tube of diameter 0.06 cm? (Surface tension of methyl alcohol = 0.023 Nm-1 and density = 0.8 gmcm-3. Assume that the angle of contact is zero)
Solution:
Surface tension of methyl alcohol (S) = 0.023 N/m.
Density, ρ = 0.8 gr/cm³ = 800 kg/m³
Diameter of tube, D = 0.06 cm
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 24

TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 4.
What should be the radius of a capillary tube if water has to rise to a height of 6 cm in it ? (Surface tension of water – 7.2 × 10-2 Nm-1)
Solution:
Surface Tension of water,
S = 7.2 × 10-2 N/m
Height of water, h = 6 cm = \(\frac{6}{100}\) m
Radius of capillary tube r = ?
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 25

Question 5.
Find the depression of the meniscus in the capillary tube of diameter 0.4 mm dipped in a beaker containing mercury. (Density of mercury = 13.6 × 10³ Kg m-3 and surface tension of mercury = 0.49 Nm-1 and angle of contact = 135°).
Solution:
Diameter of tube = 0.4 mm ;
∴ Radius, r = 0.2 mm = \(\frac{0.2}{10^3}\) m
Density of mercury = 13.6 × 10³ kg/m³
Angle of contact, 0 = 135°
Surface tension of mercury, S = 0.49 N/m.
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 26

Question 6.
If the diameter of a soap bubble is 10 mm and its surface tension is 0.04 Nm-1, find the excess pressure inside the bubble. [TS Mar. ’18, My ’16; AP June ’15; Mar. ’14]
Answer:
Diameter of soap bubble =10 mm
Radius, r = 5 mm = 5 × 10-3m
Surface tension, S = 0.04 Nm-1
Excess pressure inside the soap bubble, P = \(\frac{4S}{r}\)
TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 27

Question 7.
If work done by an agent to form a bubble of radius R is W, then how much energy is required to increase its radius to 2R?
Solution:
Energy required to form soap bubble of radius, R = W .
∴ w = 8πR²
Work done to blow a bubble of Radius 2R = 8π(2R)² = 4w
∴ Work done to increase the radius from R to 2R = 4w – w = 3w

TS Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 8.
If two soap bubbles of radii R1 and R2 (in vacuum) coalasce under isothermal conditions, what is the radius of the new bubble. Take T as the surface tension of soap solution.
Solution:
When joined in isothermal condition change in temperature of system is zero. So change in internal energy of the system is zero.
Let
Surface energy of 1st bubble U1 = 8πR2S
Surface energy of 2nd bubble U2 = 8πR²1S
Surface energy of new bubble U = 8πR²2S
But 8πR²S= 8πR²1S + 8πR²2S
⇒ R² = R²1 +R²2
Radius of new bubble R² = \(\sqrt{\mathrm{R}_1^2+\mathrm{R}_2^2}\)

TS Inter 2nd Year English Study Material Chapter 2 One the Grasshopper and Cricket

Telangana TSBIE TS Inter 2nd Year English Study Material 2nd Lesson One the Grasshopper and Cricket Textbook Questions and Answers.

TS Inter 2nd Year English Study Material 2nd Lesson One the Grasshopper and Cricket

Annotations (Section A, Q.No. 2, Marks: 4)
Annotate the following in about 100 words each.

a) The poetry of earth is never dead.

Reference: This beautiful line is taken from the sonnet, “On the Grasshopper and Cricket” written by John Keats, an English Romantic poet. He devoted his life to the perfection of poetry.

Context and Meaning: In the poem, John Keats depicts the beauty of Nature. He expresses his feelings regarding Natures song. He refers to it as “The poetry of Earth” which becomes the main them of the poem. During the summer heat birds stop singing. Then Nature’s poetry offers us comfort and joy. The grasshopper’s songs represent Nature’s poetry. Nature is brimming with elements that help living things flourish. As a result even in intense heat, natural elements such as the “Cooling Tree” and “Pleasant weed” can be discovered. The grasshopper songs tirelessly, bringing relief to all those who have grown restless due to the hot sun.

Critical Comment: The poet says that the poetry of earth is never dead. It always keeps singing irrespective of seasons.

కవి పరిచయం : ఈ అందమైన కావ్యం ఆంగ్ల కాల్పనిక కవి జాన్ కీట్స్ చే రచించబడిన “On the Grasshopper and Cricket” అను సోనెట్ నుండి తీసుకొనబడింది. అతను తన జీవితాన్ని కవిత్వం యొక్క పరిపూర్ణతకు అంకితం చేశాడు.

సందర్భ౦ మరియు అర్థం : ఈ పద్య౦లో ప్రకృతి అందం గురించి కీట్స్ వివరిస్తాడు. ప్రకృతి పాటకు సంబంధించి తన భావాలను వ్యక్తపరుస్తున్నాడు. అతను దానిని భూమి యొక్క కావ్యంగా పేర్కొన్నాడు. ఇది పద్యం యొక్క ప్రధాన ఇతివృత్తంగా మారుతుంది. వేసవి వేడి సమయంలో పక్షులు పాడటం మానేస్తాయి. అప్పుడు, ప్రకృతి మనకు సౌఖ్యం మరియు ఆనందాన్ని అందిస్తుంది. మిడత యొక్క పాటలు ప్రకృతి కావ్యాన్ని (ప్రతిబింబిస్తుంది) సూచిస్తాయి. ప్రకృతి జీవుల వృద్ధికి సహాయపడే అంశాలతో నిండి ఉంది. ఫలితంగా తీవ్రమైన వేడిలో కూడా శీతలీకరణ చెట్టు మరియు ఆహ్లాదకరమైన కలుపు మొక్క వంటి సహాజ మూలకాలు కనుగొనబడతాయి. మిడత అలసిపోకుండా పాడుతూ వేడి ఎండల కారణంగా చంచలమైన వారందరికీ ఉపశమనం కలిగిస్తుంది.

విమర్శ : ప్రకృతి/భూమి కావ్యం ఎప్పటికీ కనుమరుగు అవ్వదని కవి చెప్తున్నాడు. ఋతువులతో సంబంధం లేకుండా ప్రకృతి ఎల్లవేళల పాడుతూనే ఉంటుంది.

b) He rests at ease beneath some plesant weed. (Revision Test – I)

Introduction: The above line is taken from the sonnet “On the Grasshopper and Cricket” written by John Keats. He denoted his life to the perfection of poverty.

Context and Meaning: Here the poet expresses his feelings, regarding natures song. The Grasshopper and the Cricket are used as symbols. Seasons may come and go. But Nature never fails to inspire us with its songs. When birds, stop singing in extreme heat, during the summer. The earth is filled with songs of a grasshopper. We can hear the voice of the grasshopper who runs from hedge to hedge. He keeps singing tiredlessly and when he gets tired with fun, he goes under some pleasant weed to take rest.

Critical Comment: The poet sends the message that nature is beautiful all the line, irrespective of the season.

కవి పరిచయం : ఈ వాక్యం జాన్ కీట్స్ చే రచించబడిన “On the Grasshopper and Cricket” అను సోనెట్ (14 పంక్తులు గల ప్రత్యేక పద్య౦ నుండి తీసుకొనబడింది. ఇతను తన జీవితాన్ని కవిత్వం యొక్క పరిపూర్ణతకు అంకితం చేశాడు.

సందర్భ౦ మరియు అర్థం : ఇక్కడ కవి ప్రకృతి పాటకు (కావ్యం)కు సంబంధించి తన భావాలను వ్యక్తపరుస్తున్నాడు. మిడతను మరియు కీచురాయిని చిహ్నాలుగా వాడుతున్నాడు. ఋతువులు రావచ్చు మరియు పోవచ్చు. కానీ, తన పాటలతో మనల్ని ప్రేరేపించటంలో ప్రకృతి ఎన్నడూ విఫలము చెందదు. వేసవి విపరీతవేడిలో పక్షులు అన్నీ పాడటం ఆపినపుడు ప్రకృతి/భూమి మిడత యొక్క పాటలతో నింపబడుతుంది. ఒక పొద నుంచి మరొక పొదకు పరుగెడుతూ, పాడుతున్న మిడత యొక్క స్వరాన్ని మనం వినగలము. ఇది అలసట చెందకుండా, పాడుతూ ఉంటుంది మరియు ఆనందంతో అలసట చెందినప్పుడు సేదతీరుటకు ఆహ్లాదకరమైన కలుపు మొక్క క్రిందకు జారుతుంటుంది.

విమర్శ : ప్రకృతి అన్ని కాలాల్లో అందంగా, రమణీయంగా ఉంటుందని కవి సందేశాన్ని పంపుతున్నాడు ఈ పద్యం ద్వారా.

TS Inter 2nd Year English Study Material Chapter 2 One the Grasshopper and Cricket

c) On a lone winter evening, when the frost Has wrought a silence, from the stone there shrills the Cricket’s song, in warmth increasing ever.

Introduction: These lines are taken from the second part(sestet) of the sonnet ” On the grasshopper and the Cricket” written by John Keats, a romantic poet. He devoted his life to the perfection of poetry.

Context and Meaning: The poet expresses his feelings regarding Nature’s song. The Grasshopper and the cricket are used as symbols. Seasons may come and go. But Nature never fails to inspire us with its songs. During winter birds stop singing. Here is a deathly silence. Frost spreads its blanket over Nature Regardless, a shrill second comes from beneath stones and it is the cricket singing the cricket’s song restores warmth. We can hear the song of the cricket, which breaks this silence. Thus, the cricket takes up the responsibilities of singing the glory of Nature in winter.

Critical Comment: John keats sends the message that nature is beautiful all the time, irrespective of the season.

కవి పరిచయం : ఈ పంక్తులు రొమాంటిక్ కవి, జాన్ కీట్స్ చే రచించబడిన “On the Grasshopper and Cricket” అను సోనెట్ యొక్క రెండవభాగం, (చివరి ఆరు పంక్తులు) నుండి తీసుకొనబడినవి. కవిత్వం యొక్క పరిపూర్ణతకు ఇతను తన జీవితాన్ని అంకితం చేశాడు.

సందర్భ౦ మరియు అర్థం : ప్రకృతి పాటకు సంబంధించి కవి తన భావాలను వ్యక్తపరుస్తున్నాడు. మిడతను మరియు కీచురాయిలను చిహ్నాలుగా ఉపయోగించాడు. కాలాలు (ఋతువులు) వస్తూ, పోతూ ఉంటాయి. కానీ, ప్రకృతి తన పాటలతో మనల్ని ప్రేరేపించటంలో ఎన్నడూ విఫలమవ్వదు. శీతాకాలంలో, పక్షులు పాడటం ఆపివేస్తాయి. పూర్తి నిశ్శబ్దత ఉంటుంది. దట్టమైన మంచు తన దుప్పటితో ప్రకృతిని కప్పివేస్తుంది.

అయినప్పటికీ, సంబంధం లేకుండా, వాళ్ళ క్రింద నుండి ఒక గంభీరమైన శబ్దం, స్వరం వస్తుంది. కీచురాయి గానం, ఈ నిశ్శబ్ధాన్ని విచ్ఛిన్నం చేసే కీచురాయి పాటను మనం వినవచ్చు. ఈ విధంగా, కీచురాయి చలికాలంలో ప్రకృతి వైభవాన్ని పాడే/ఆలపించే బాధ్యతను తీసుకుంటుంది.

విమర్శ : ప్రకృతి అన్ని కాలాల్లో అందంగా, రమణీయంగా ఉంటుందన్న సందేశాన్ని జాన్కీట్స్ ఈ పద్య౦ ద్వారా తెలియజేస్తున్నాడు.

d) And seems to one in drowsiness halflost; (Revision Test – I)
The Grasshopper’s among some grassy hills

Introduction: These are the conducting lines of the poem “On the Grasshopper and Cricket” written by John Keats, a Romantic poet. He devoted his life to the perfection of poetry.

Context and Meaning: John Keats celebrates the music of the Earth. He finds beauty in hot summer as well as in the cold winter. Here, the grasshopper is symbol of hot summer and cricket is symbol of cold winter. During the winter season in the frosty evening, the birds stop singing songs. At that time the cricket begins to sing. He spreads the warmth of joy everywhere. The people who are half sleep feel that it is the grasshopper song which is coming from the grassy hills. Thus, he depicts the beauty of Nature.

Critical Comment: The poet sends the message that nature is beautiful all the time, irrespective of the season. In a similar way, we should be joyful in our life and be happy in all situations.

కవి పరిచయం : ఇది రొమాంటిక్ కవి జాన్ కీట్స్ చే రచించబడిన “On the Grasshopper and Cricket” అను పద్య౦ యొక్క ముగింపు పంక్తులు. ఇతను కవిత్వం యొక్క పరిపూర్ణతకు తన జీవితాన్ని అంకితం చేశాడు.

సందర్భ౦ మరియు అర్థం : జాన్ కీట్స్ భూమి/ప్రకృతి యొక్క సంగీతాన్ని ప్రశంసిస్తున్నాడు. అతి వేసవిలోను మరియు అతి శీతాకాలంలోను ప్రకృతి సౌందర్యాన్ని, రమణీయతను ఇతను గుర్తిస్తున్నాడు. ఇక్కడ మిడత తీవ్ర వేసవి కాలానికి చిహ్నం మరియు కీచురాయి చల్లని శీతాకాలానికి చిహ్నం. చలికాలంలో అతి శీతలమైన సాయంత్రం, పక్షులు పాడటం మానేస్తాయి. ఆ సమయంలో, కీచురాయి పాడటం ప్రారంభమౌతుంది. అతను ప్రతిచోటా ఆనందం యొక్క వెచ్చదనాన్ని సుఖాన్ని వ్యాప్తి చేస్తాడు. సగం నిద్రలో ఉన్నవారు పచ్చటి కొండల నుండి వచ్చే మిడతపాట అని భావిస్తారు. ఆ విధంగా, ప్రకృతి అందాన్ని సౌందర్యాన్ని అతను వర్ణిస్తున్నాడు.

విమర్శ : కాలంతో సంబంధం లేకుండా, ప్రకృతి ఎల్లవేళలా అందంగా ఉంటుందని కవి సందేశం పంపాడు. అన్ని సందర్భాలలో, మనం జీవితంలో ఆనందంగా ఉండాలి.

Paragraph Questions & Answers (Section A, Q.No.4, Marks: 4)
Answer the following Questions in about 100 words

a) What is the theme of the poem on the Grasshopper and Cricket? (Revision Test – I)
Answer:
The poem “On the Grasshoppers and Cricket’ is written by John Keats, an English Romantic poet. He has devoted his life to the perfection of poetry. In this poem, John Keats depicts the beauty of Nature. He says that the poetry of earth as symbols to praise Nature is never ending beauty.

Seasons may come and go but Nature never fails to inspire us with its songs. When birds stop singing in extreme heat, the earth is filled with the songs of a grasshopper. He sings endlessly, but when tired rests under some pleasant weed. During winter birds stop singing. There is a deathly silence. Frost spreads its blanket over Nature. Regardless, a shrill second comes from beneath stones and it is the cricket singing. Its song restores warmth. Thus, the small creatures prove to the world that the poetry of earth never ceases.

“On the Grasshopper and Cricket” అను పద్యం ఆంగ్ల రొమాంటిక్ కవి జాన్ కీట్స్ చే రచింపబడింది. కవిత్వం యొక్క పరిపూర్ణతకు ఇతను తన జీవితాన్ని అంకితం చేశాడు. ఈ పద్యంలో, ప్రకృతి యొక్క సౌందర్యాన్ని కీట్స్ వివరిస్తున్నాడు. ప్రకృతి/భూమి యొక్క కవిత్వం/సంగీతం, పాట ఎప్పటికీ నిలిచిపోదు అని అంటున్నాడు ఇతడు. శాశ్వతమైన ప్రకృతి సౌందర్యాన్ని స్తుతించుటకు, ప్రశంసించుటకు ఇతను మిడతను మరియు కీచురాయిలను చిహ్నాలుగా ఉపయోగించాడు. ఋతువులు, కాలాలు రావాలి, పోవాలి కానీ ప్రకృతి తన పాటలతో, సంగీతంతో మనల్ని ప్రేరేపించడంలో ఎన్నడూ విఫలం చెందదు. తీవ్రమైన వేసవి వేడిలో పక్షులు పాడటం ఆపినపుడు,  భూమి/ప్రకృతి మిడత పాటలతో నింపబడుతుంది. ఇది నిరంతరాయంగా పాడుతూ ఉంటుంది.

అయితే అలసట చెందినపుడు, ఆహ్లాదకరమైన కలుపు మొక్క కింద విశ్రాంతి తీసుకుంటుంది. శీతాకాలంలో పక్షులు పాడటం ఆపుతాయి. అక్కడ పూర్తి, ఘోరమైన నిశ్శబ్దం ఉంటుంది. మంచు తన దుప్పటిని ప్రకృతి మీద కప్పుతుంది. అయినప్పటికీ, రాళ్ళ క్రింద నుండి ఒక గంభీరమైన శబ్దం, స్వరం వస్తుంది. అది కీచురాయి గానం. దీనిపాట సౌఖ్యాన్ని పునరుద్దిస్తుంది. అలా, ప్రకృతి కవిత్వం ఎప్పటికీ కనుమరుగవ్వదని, నిలిచిపోదని ఈ సూక్ష్మజీవులు ప్రపంచానికి ఋజువు చేస్తున్నాయి.

TS Inter 2nd Year English Study Material Chapter 2 One the Grasshopper and Cricket

b) According to Keats, when does one hear a cricket’s song?
Answer:
The poem ” On the Grasshopper and Cricket” is written by John Keats. He is an English Romantic poet. He has developed his life to the perfections of poetry. According to him, the poetry of earth never ceases. He uses the Grasshopper and Cricket as symbols to praise Nature’s never ending beauty. Seasons may come and go. Nature never fails to inspire us with its songs. During cold winter, the birds stop singing.

There is deathly silence. Frost spreads fits blanket over Nature. Then a shrill sand comes from beneath stones and it is the Cricket singing. He breaks this silences. So, we can hear the song of the Cricket in winter. Its song restores warmth.

“On the Grasshopper and Cricket” అను పద్య౦ జాన్ కీట్స్ చే రచింపబడింది. ఇతను ఒక రొమాంటిక్ కవి. ఇతను తన జీవితాన్ని కవిత్వ పరిపూర్ణతకు అంకితం చేశాడు. ఈ పద్య౦లో ప్రకృతి యొక్క సౌందర్యాన్ని వర్ణిస్తున్నారు. ఇత) ఉద్దేశ్యంలో ప్రకృతి కావ్యం శాశ్వతం. ఎన్నడూ కనుమరుగు అవ్వదు. వాటన్నిటికీ కనుమరుగు అవ్వని ప్రికృ ) సౌం గర్యాన్ని ప్రశంసించుటకు, ఇతను మిడతను మరియు కీచురాయిలను చిహ్నాలుగా ఉపయోగించాడు. కీచురాయి అతి శీతాకాలానికి చిహ్నం ఇక్కడ. (కాలాలు) ఋతువులు రావచ్చు పోవచ్చు.

కానీ ప్రకృతి తన సంగీతంతో మనల్ని ప్రేరేపించుటలో ఎన్నడూ విఫలం చెందదు. అతి శీతాకాలంలో, పక్షులు పాడటం నిలిపివేస్తాయి. అక్కడ పూర్తిగా నిశ్శబ్దం. మంచు తన దుప్పటిని ప్రకృతి మీద పరచుతుంది. అప్పుడు ఒక గంభీరమైన స్వరం రాళ్ళ క్రింద నుండి వస్తుంది. అది కీచురాయి గానం. ఈ నిశ్శబ్దాన్ని ఆ స్వరం విచ్ఛిన్నం చేస్తుంది. శీతాకాల సమయంలో మన కీచురాయి పాటను వినగలం. దీని పాట సుఖం, వెచ్చదనాన్ని ఇస్తుంది. పునరుద్దరిస్తుంది.

c) When does a grasshopper sing?
Answer:
The poem” On the Grasshopper and Cricket” is written by John Keats. He is an English Romantic poet. He has developed his life to the perfections of poetry. In this poem, he depicts the beauty of nature. He says that the poetry of earth never ceases. He uses the Grasshopper and Cricket as symbols to praise Nature’s never ending beauty. Seasons may come and go. But Nature never fails to inspire us with its songs.

During hot summer, all the singing birds stop singing and take rest under the shady branches of trees. But the song of the nature goes on. We can hear the voice of the grasshopper, who runs from hedge to hedge. He sings endlessly, but when tired rests under some pleasant weed. Thus, the grasshopper sings during summer.

“On the Grasshopper and Cricket” అను కావ్యం జాన్ కీట్స్ చే రచింపబడింది. ఇతను ఒక ఆంగ్ల రొమాంటిక్ కవి. కవిత్వం యొక్క పరిపూర్ణతకు తన జీవితాన్ని అంకితం చేశాడు. ఈ పద్య౦లో ప్రకృతి సౌందర్యాన్ని ఇతను వర్ణిస్తున్నాడు. ప్రకృతి కావ్యం (భూమి) శాశ్వతం అది ఎప్పటికీ కనుమరుగు అవ్వదు అని చెప్తున్నాడు. ఎప్పటికీ కనుమరుగు అవ్వని, శాశ్వతమైన ప్రకృతి సౌందర్యాన్ని ప్రశంసించుటకు, కవి మిడతను మరియు కీచురాయిలను చిహ్నాలుగా ఉపయోగించాడు. ఋతువులు (కాలాలు) రావచ్చు, పోవచ్చు కానీ ప్రకృతి తన పాటలతో మనల్ని ప్రేరేపించుటలో ఎన్నడూ విఫలం చెందదు. తీవ్ర వేసవి సమయంలో పాడే పక్షులన్నీ పాడటం ఆపి చెట్ల కొమ్మల నీడలో విశ్రాంతి తీసుకుంటాయి.

కానీ, ప్రకృతి మాత్రం తన పోరును కొనసాగిస్తుంది. పొదల నుండి పొదలకు పరుగెడుతూ పాడే మిడతను మనం వినగలం. ఇది నిరంతరం పాడుతుంది. అయితే అలసిపోయినప్పుడు ఒక సుందరమైన కలుపు మొక్క క్రింద విశ్రాంతి తీసుకుంటుంది. అలా, మిడత వేసవికాలంలో పాడుతూ ఉంటుంది.

d) Discuss the common features between the grasshopper and cricket.
Answer:
The poem ” On the Grasshopper and Cricket” is written by John Keats. He is an English Romantic poet. He has developed his life to the perfections of poetry. In this poem, he depicts the beauty of nature. He says that the poetry of earth never ceases. He uses the Grasshopper and the Cricket as symbol to praise nature’s never ending beauty. Seasons may come and go. But, nature never fails to inspire us with its songs.

Therefore both the grasshopper r and the cricket are the representative voices of nature’s music or poetry. Both offer a soothing effect to the extremities of climate. The grasshoppers song balances the extreme heat during the summer by providing music that is comforting and pleasing the cricket does the same during winter.

“On the Grasshopper and Cricket” అను కావ్యం జాన్ కీట్స్ చే రచింపబడింది. ఇతను ఒక ఆంగ్ల రొమాంటిక్ కవి. కవిత్వం యొక్క పరిపూర్ణతకు తన జీవితాన్ని అంకితం చేశాడు. ఈ పద్య౦లో, అతను ప్రకృతి సౌందర్యాన్ని వర్ణిస్తాడు. ప్రకృతి కావ్యం పాట ఎల్లవేళలా ఉంటుంది. అది ఎన్నటికీ ఆగిపోదని చెప్తున్నాడు. ఎల్లవేళలా ఉం ప్రకృతి సౌందర్యాన్ని ప్రశంసించుటకు, మిడతను మరియు కీచురాయిని చిహ్నాలుగా కవి ఉపయోగించాడు. కాలాలు రావచ్చు, పోవచ్చు.

కానీ, ప్రకృతి తన మధురమైన పాటలతో మనల్ని ప్రేదెంచుటలో ఎన్నడూ విఫలమవ్వదు. కాబట్టి మిడత మరియు కీచురాయి రెండూ ప్రకృతి సంగీతం లేదా కవిత్వానికి ప్రాతినిధ్య స్వరాలు. రెండూ వాతావరణం యొక్క అంత్య భాగాలకు ఓదార్పు ప్రభావాన్ని అందిస్తాయి. మిడత పాట ఓ కారని మరియు ఆహ్లాదకరమైన సంగీతాన్ని అందించడం ద్వారా వేసవిలో తీవ్రమైన వేడిని సమతుల్యం చేస్తుంది. చలికాలంలో కీచురాయి కూడా అదే పని చేస్తుంది.

One the Grasshopper and Cricket Summary in English

About Author

TS Inter 2nd Year English Study Material Chapter 2 One the Grasshopper and Cricket 1

John Keats (31 October 1795 – 23 February 1821) was an English poet of the second generation of Romantic poets, with Lord Byron and Percy Bysshe Shelley, although his poems had been in publication for less than four years when he died of tuberculosis at the age of 25. They were indifferently received in his lifetime, but his fame grew rapidly after his death. The Encyclopaedia Britannica of 1888 called one ode “one of the final masterpieces”. Today his poems and letters remain among the most popular and analysed in English literature – in particular “Ode to a Nightingale”, “Ode on a Grecian Urn”, “Sleep and Poetry” and the sonnet “On First Looking into. Chapman’s Homer”.

The poem ‘On the Grasshopper and Cricket’ is written by John Keats in December 1816. It is a fine piece of Petrarchan Sonnet. It is inspired by the beauty of nature. It is about Nature and says that the poetry of earth never ceases the poet celebrates the music of the earth in this poem. He finds beauty in hot summer as well as in the cold winter. He Symbolizes the grasshopper as hot summer and the cricket as a very cold winter. He says that the music of nature or earth is always alive whether it is hot summer or cold winter.

Seasons may come and go. But nature never fails too inspire us with its songs. During the hot summer all the singing birds are tired and take rest under the shady branches of trees. They stop singing. But the song of nature goes on. We can still hear the voice of the grasshopper who runs from hedge to hedge. He sings endlessly and timelessly, and when tired, rests under some pleasant weed thus, the grasshopper carries on the duty of singing the everlasting song of nature. During summer, he is a fun loving and cheerful creature.

The poet further says that during the cold winter also the birds are silent. There is cutter silence on the frosty winter days. But the earth has its own way of expressing pleasure and joys. A cricket sings through the stones and breaks this silence. He sings from the stones but not from the trees. His songs appears to be increasingly the warmth every moment. People can hear it sitting in their houses. However, the poet says that to a person who is half sleep, it may appear to be a grasshoppers song coming from the grassy hills.

Thus, we can say that the grasshopper and the cricket perform a big responsibility. They carry on with nature’s continuous and everlasting music irrespective of the extreme climate. The poet has thus personified them. They are a symbol of constant joyous mood of nature. Hence the poem teaches us that we shall be joyful and pleasant no matter what the situations are in our life. With this attitude, we can easily over come all the obstacles in life.

TS Inter 2nd Year English Study Material Chapter 2 One the Grasshopper and Cricket

One the Grasshopper and Cricket Summary in Telugu

Note: This summary is only meant for Lesson Reference, not for examination purpose

“On the Grasshopper and Cricket” అను పద్య౦ జాన్ కీట్స్ చే 1816లో వ్రాయబడింది. ఇది పెట్రార్కన్ సొనెట్ యొక్క చక్కటి భాగం. ఇది ప్రకృతి అందంచే ప్రేరణపొందింది. ఇది ప్రకృతికి సంబంధించిన కావ్యం మరియు భూమి యొక్క కవిత్వం ఎప్పటికీ నిలిచిఉంటుంది ఆగిపోదని చెప్తుంది. కవి ఈ పద్య౦లో భూమి యొక్క సంగీతాన్ని గురించి గొప్పగా చెప్తున్నాడు. అతను తీవ్ర వేసవిలో మరియు చల్లని శీతాకాలంలో అందాన్ని చూస్తున్నాడు. అతను మిడతను వేడి వేసవిగా మరియు క్రికెట్ కీచురాయిని చాలా చల్లని శీతాకాలంగా సూచిస్తాడు. అది వేడి వేసవి లేదా చల్లని, తీవ్రమైన శీతాకాలమైనా, ప్రకృతి లేదా భూమి యొక్క సంగీతం ఎల్లప్పుడూ సజీవంగా ఉంటుందని చెప్తున్నాడు కవి.

ఋతువులు రావచ్చు మరియు పోవచ్చు. కానీ ప్రకృతిని పాటలతో మనల్ని ప్రేరేపించడంతో అది ఎప్పుడూ విఫలం కాదు. తీవ్ర వేసవి సమయంలో, పాడే పక్షులనీ అలిసిపోయి చెట్ల కొమ్మలో నీడ క్రింద విశ్రాంతి తీసుకుంటాయి. అవి పాడటం ఆపి వేస్తాయి. కానీ ప్రకృతి పాట మాత్రం కొనసాగుతుంది. కంచె నుండి కంచె వరకు పరిగెత్తుతూ పాడే మిడతగొంతు మనము ఇప్పటికీ వినగలము. ఇది అలసిపోకుండా మరియు నిరంతరం పాడుతూ మరియు అలసిపోయినప్పుడు ఒక సులభమైన కలుపు మెలికల క్రింద విశ్రాంతి తీసుకుంటాడు. ఈ విధంగా మిడత ప్రకృతి యొక్క శాశ్వతమైన పాటను పాడే బాధను నిర్వహిస్తుంది వేసవిలో, ఇది సరదాను ప్రేమించే మరియు ఉల్లాసంగా ఉండేవి.

చలికాలంలో కూడా పక్షులు మౌనంగా ఉంటాయి. కవి ఇంకా చెప్తాడు. అతిశీతలమైన శీతాకాలపు రోజులలో పూర్తిగా నిశ్శబ్దం. కానీ, భూమికి ఆనందం మరియు ఆక్రందాలను వ్యక్తీకరించడానికి దానికి స్వంతమార్గం ఉంది అంటున్నాడు. ఒక కీచురాయి రాళ్ళలో పాడుతూ ఈ నిశ్శబ్దాన్ని ఛేదిస్తుంది. ఇది చెట్లు నుండి కాదు, రాళ్ళ నుండి పాడుతోంది. అతని పాట ప్రతిక్షణం వెచ్చదనాన్ని పెంచుతుంది. జనం తమ ఇళ్ళలో కూర్చుని వినగలరు దీన్ని. ఏది ఏమైనప్పటికీ, సగం నిద్రలో ఉన్న వ్యక్తికి ఇది (గడ్డి కొండలు) పచ్చటి కొండల నుండి వచ్చే Grasshopper పాటగా కనిపిస్తుందని కవి చెప్తాడు.

అలా, మిడత మరియు కీచురాయి పెద్ద బాధ్యతను నిర్వర్తిస్తాయని మనం చెప్పగలం. వారు తీవ్రమైన వాతావరణాలతో సంబంధం లేకుండా ప్రకృతి యొక్క నిరంతర మరియు శాశ్వతమైన సంగీతాన్ని కొనసాగిస్తారు. కవి ఈ విధంగా వాటిని వ్యక్తీకరించాడు. అవి ప్రకృతి యొక్క స్థిరమైన ఆనందకరమైన స్థితికి చిహ్నం ప్రతీక. కాబట్టి జీవితంలో ఎలాంటి పరిస్థితులు ఎదురైనా మనం ఆనందంగా, ఆహ్లాదకరంగా ఉండమని ఈ పద్య౦ బోధిస్తుంది. ఈ వైఖరితో మనం జీవితంలోని అన్ని అడ్డంకులను అయినా సులభంగా అధిగమించవచ్చు. ఇది ఈ కావ్యం సందేశం.

One the Grasshopper and Cricket Summary in Hindi

Note: This summary is only meant for Lesson Reference, not for examination purpose

‘ऑन द ग्रासहोपर एंड क्रिकेट” नामक कविता दिसंबर 1816 में जॉन कीट्स द्वारा लिखा गया है । यह पेट्रार्चन सॉनेट की बेहतरीन रचना है । यह प्रकृति की सुंदरता से प्रेरित है। यह प्रकृति के बारे में है । कवि कहते हैं कि पृथ्वी की कविता कभी समाप्त नहीं होती । इस कविता में कवि पृथ्वी के संगीत का जश्न मनाते हैं। वे तपती गर्मी में तथा कड़ाके की ठंड में सुंदरता पाते हैं। वे टिड्डे को भीषण गर्मी और झींगुर को बहुत ठंडी सर्दी के रूप में दशति है। उनका कहना है कि प्रकृति या पृथ्वी का संगीत हमेशा जीवित रहता है, चाहे वह गर्मी हो या सर्दी ।

मैसम आ सकते हैं और जा सकते हैं। लेकिन प्रकृति अपने गीतों से हमें प्रेरित करन में कभी असफल नहीं होती है । तपती गरमी में सभी गानेवाले पक्षी थक जाते हैं और कृक्षों की शाखाओं के नीचे छाया में आराम लेते हैं । ने गाना बंद कर देते हैं । लेकिन प्रकृति का गाना जारी होता है । टिड्डे की आवाज हमें अभी भी सुनाई देती है, जो बाड़े से बाड़े तक दौड़ता है। वह लगातार और अथक जाता है और जब वह थक जाता है तब सुखद अपतृण के नीचे आराम लेता है । इस प्रकार टिड्डा प्रकृति का चिरस्थायी गीत गाने का कर्तव्य निभाता है । गर्मी के दौरान वह कौतुक – प्रिय और प्रमुदित प्राणी है ।

कवि आगे कहते हैं कि बहुत सर्दी के दौरान पक्षी खामोश होते हैं। वहत ठंडी सर्दियों के दिनों मे पूरी खानेशी होती है। लेकिन धरती को अपनी प्रसन्नता और हर्ष व्यक्त करने का तरीका है । झीर पत्थरों से गाता है और इस चुप्पी को तोड़ता है । वह पत्थरों से गाता है । न कि पेड़ों से । उसका गाना हर पल सरगर्मी और जोश बढ़ाता है । लोग अपने घर में बैठकर इसे सुन सकते हैं। हालाँकि, कवि कहते हैं कि जो व्यक्ति आधी नींद में है, उसे लगता है कि टिड्डे का गाना घास की पहाड़ियाँ से आता है ।

इस प्रकार, हम कह सकते हैं कि टिड्डा और झींगुर बड़ी जिम्मेदारी निभाते हैं । तीव्र जलवायु की परवाह किए बिना वे प्रकृति का निरंतर और चिरस्थायी संगीत जारी रखते हैं। इस प्रकार कवि ने उन्हें साकार किया । वे निरंतर खुशियों के प्रतीक हैं । इस दृष्ट से यह कविता हमें सिखाती है कि हमारी जिंदगी में चाहे जो भी परिस्थितियाँ हों, हमें हर्षित और सुखद होना चाहिए । इस दृष्टि कोण से हम जिंदगी में सभी बाधाएँ आसानी से पार कर सकते है |

Meanings and Explanations

grasshopper(n)/ (గ్రస్ హూప(ర్)/’gra:shɒp.ə(r)/ : A plant eating insect, మిడుత, पैधा खानेवाला एक कीड़ा

cricket(n)/ (క్రికెట్) / krɪk.ɪt : a small jumping insect that makes a loud sound, కీచురాయి, ज्यादा आवाज करने वाला छलाग मालनेवाला एक छोटा कीडा

faint(adj)/ (ఫెఇంట్)/ feɪnt : Feeling weak and tired, అలసిపోయిన, कमजोर और थका हुआ महसूस करना

hide (v)/ (హైడ్)/ haɪd : be behind; వెనుక, చాటున, నీడన ఉండు, छिपाना, छिपना

hedge (హెజ్)/hedʒ : a thick bush ; పొదల పరంపర

mown(v-pp)(ofmow)(మఉన్) / məʊn : Trionmed with a sharp blade or machine cut; కత్తిరించబడిన

mead(n) : Meadows; పచ్చిక బయలు , घास के मैदान

TS Inter 2nd Year English Study Material Chapter 2 One the Grasshopper and Cricket

new mown mead : Freshly cut grassland; కొత్తగా మట్టసమంగా కత్తిరించబడిన పచ్చిక బయలు, अभी – अभी काटा गया घास का मैदान

he has never done : ఆయన పని కొనసాగుతూనే ఉంటుంది అంతము, ముగింపు ఉండదు

delights(n-pl)(డిలైటెస్)/dɪlaɪt : Joys; సరదాలు, వినోదాలు, खुशियाँ

weed(n)/ (వీడ్)/ wi:d : wild plant; కలుపు మొక్క , जंगली पौधा

ceasing(v+ing)/ (సీసింగ్) /si:sın : stop; ముగిస్తున్న, रोकना

frost(n)/frost(ఫ్రోస్ట్)/ frɒst : a cover of ice particles; మంచు పొరలు, बर्फ

wrought(v-pp)(of work)/(రోట్) / rɔ:t : worked ; done ; చేసెను, पूरा करना

shrills(v-pre.ten)/(షి ల్ జ్) / ʃrɪlz : Comes through loud and clear, స్పష్టంగా పెద్దగా వస్తున్న

drowsiness(n) / (డ్రౌజినెస్)/’draʊ.zi.nəs: sleepiness; నిద్ర మత్తు, आधी नींद

TS Inter 2nd Year English Grammar Dialogue Writing

Telangana TSBIE TS Inter 2nd Year English Study Material Grammar Dialogue Writing Exercise Questions and Answers.

TS Inter 2nd Year English Grammar Dialogue Writing

Q.No.20 (4 Marks)

A conversation is an informal spoken exchange of information, feelings, thoughts and ideas. It usually takes place among persons who know each other. Conversations use speech as the medium of language and are, therefore, spontaneous and unplanned.

A dialogue is a written piece of a conversation. It has, therefore, features of both speech and writing in it. Dialogue writing is a skill that helps us in developing both our speech and writing.

In a dialogue, speakers and listeners keep changing their roles. Dialogues can be very short or long. A dialogue need not always contain grammatically complete sentences. In informal style, we hardly ever use complete sentences.

For example,

A: Posted my letter?
B: Not yet.
It would be very artificial to have the same dialogue in complete sentences as:
A: Have you posted my letter ?
B: I have not posted it yet.

The levels of formality or informality in style depend on the topic, the purpose of communication and the relationship between the participants.

In dialogues, appropriateness is as -much necessary as correctness. A greeting, a question, a compliment are to be responded appropriately. Contracted forms like ‘i’m, it’S’, you’re’ etc are preferred in informal style.

Apart from responding appropriately to a compliment, it is also necessary to add some relevant information so that the dialogue is carried forward.

Look at the two examples of the same dialogue:

1. A: Your shirt is very nice.
B: Thanks for the compliment.

2. A: Your shirt is very nice.
B: Oh. I’m happy you like it. I bought it in Ahmedabad and it cost me only Rs. 190/-

A good dialogue has an element of surprise or shock. A dialogue must not be totally predictable.
Some useful tips to keep in mind:

  • The number of exchanges is to be a minimum of ten.
  • Greetings (opening and closing) are to be appropriate to the relationship between the speakers (For example : ‘Good morning’ to a teacher: ‘Hello’ to some elderly person and ‘Hal’ to a friend and no greeting at all to a member of the family.)
  • Dialogues to be brief; need not even be complete sentences.
    For example: The response to “Where are you going?” can be “To college”. You need not write “I am going to college.”
  • Depending on the relationship between the speakers, formal or informal language is to be used.
  • Use of linkers like ‘however’, ‘gap fillers like ‘mm’, polite expressions like ‘you’re right, but’ add to the good qualities of the given dialogue.
  • Keeping in view the given, topic for dialogue and the relationship between the speakers is most important.

Now observe the following examples carefully.

Dialogue between two friends who met at a shop:

Padmini : Hi Tripura!
Tripura : Hi Appi! How are you? Long time, no see.
Padmini : Well, rather busywith exams. And what’s the news at your end ?
Tripura : Nothing much. The same boring routine.
Padmini : Ok, then. I’ll get going. Bye.
Tripura : See you soon. Bye.

TS Inter 2nd Year English Grammar Dialogue Writing

Exercise

Observe the way the friends greet each other, how they say goodbye and the kind of language they use. Is the language the same as that which one uses in written English ?
Try to-spot any two ways in which.lt is different from written English. Now look at the following dialogue.
Answer:
Greetings : Hi
Taking Leave : Bye
No – It is not the same as that one used in written English. Two differences:

  1. contractions: I’llin the place of I will.
  2. incomplete sentences : I have been rather ‘busy with examinations. In its place -7 rather busy with exams.

Dialogue between a manager and a clerk:

Clerk : Good Morning, sir.
Manager : Good Morning. I see that you are late to work again.
Clerk Sir : I am sorry but I had some urgent work at home.
Manager : I’m getting tired of your excuses.
Clerk Sir : please excuse me. I will be on time from tomorrow.
Manager : This is the last warning. You can go to your desk now.
Clerk : Thank you, sir.

Excercise

i) How do the clerk and the manager begin and end the dialogue? What kind of language is used? How different is the language from written English ? Discuss.
Answer:
They begin with formal greetings (Good Morning) and end with formal leave taking (Thank you,’ sir). It is not much different from written English.

ii) If you compare the two pieces of dialogue given above, what do you notice ? Discuss whether” there are any similarities and differences and, if so, why?
Answer:
First dialogue uses informal style while the second one uses formal style.

Let’s now consider different types of situations.

1. Greeting
Informal: Hi, Neeta! Great to see you !/ Hi, Neeta! What a surprise! Hi, Neeta! How’re you?
Formal: Good morning; Ms Neeta! It is indeed a pleasure to see you.

2. Leave taking / farewell
Informal: Bye, See you soon! I Great seeing you. Bye! I Guess that’s all for now. Bye! – Formal: I must leave now. Good night, sir!/ It was a pleasure to have met you. Good day!

3. Asking a question
Informal: Hey! Quick question. Where’s Timbuktu ? I Hi, Renu! Know where Timbuktu is ?
Formal: Renuka, can you please tell me where Timbuktu is? I Renuka, I’m sorry to | bother you but could you tell me where Timbuktu is?

4. Giving an answer (to the above question) „
Informal: How the heck would I know? I Don’t know, don’t care/ No idea / Mali, West Africa.
Formal: I’m sorry, Meena, but I don’t know the answer. I Timbuktu is a city in Mali, West Africa.

5. Making a request
Informal: Hey, give me your book for a day? I Mohan, lend me your book for a day? (tone is one of rising inflexions)
Formal: Mohan, please lend me your book for a day./I would be obliged if you could ; lend me your book for a day. I Mohan, would you be kind enough to lend me your book for a day?

6. Making a suggestion
Informal: Coming for a walk ? I Let’s go for a walk.
Formal: I think it would be nice if we could go for a walk./ We can consider going for a walk.

7. Giving advice
Informal: Write neatly, you nitwit!/ Improve your writing, Uma.
Formal: Uma, I advise you to improve your writing’/ It would be nice if you could improve your writing, Uma.

TS Inter 2nd Year English Grammar Dialogue Writing

8. Offering an apology
Informal: Sorry it broke. I Awfully sorry I broke the plate.
Formal: I apologise for breaking the plate. II am extremely sorry for having broken the plate.

9. Expressing gratitude’
Informal: Thanks Partha, for the pen.

Formal: Thank you so much for the pen./ It was indeed nice of you, Parthasarathy, to have given me the pen.
Given below are some dialogues depicting what people might say in situations they encounter in their day-to-day life. Study them and observe the use of formal and informal language, abbreviations, tone and flow of ideas.

Observe the following Model Dialogues:

Dialogue between two newcomers to the college

Suresh : Hello, I think you are a newcomer to this college.
Praveen : Yes, you are right. I am Praveen.
Suresh : Oh, I am Suresh. I am also a newcomer.
Praveen : Nice to meet you. How are you?
Suresh : I am fine and you ?
Praveen : I’m fine; may I know why you have chosen this college?
Suresh : This college is supposed to be the best college in our town.
Praveen : You are right, Suresh. The academic system of this college is quite up to date, I hope Can we go to the class now?
Suresh : Sure, let’s go there.
Praveen : Here we are.

Dialogue between a teacher and a student

Student : Sir, good morning.
Teacher : Good morning, why were you absent yester -day?
Student : Sir, my brother was suffering from a fever. I went to see him.
Teacher : How is he now? What did the doctor say?
Student : He has to take rest.
Teacher : When your brother comes back, I want to talk to him.
Student : Sure, sir. Thank you sir.

Dialogue between father and son

Father : You have appeared for the Class XII exams
What are your plans for the future?
Son : I haven’t decided so far, dad.
Father : Don’t you feel ashamed of yourself. You should have decided your goal in life.
Son : I would like to join UG course in any ope of the reputed colleges in India.
Father : What’s your long-term goal?
Son : I want to become an actor
Father : Are you sure you want to take it as your profession?
Son : Yes, dad
Father : Ok, be serious and decide a suitable career for yourself.

Dialogue between a bookseller and customer

Bookseller : Good morning. How can I help you?
Customer : Good morning, I want to buy a few books. .
Bookseller : There are different kinds of books here. What kind of books do you want?
Customer : I want William Shakespeare’s Hamlet and R.K. Narayan’ The Guide.
Bookseller : We have R.K. Narayan’s The Guide. Sorry to say that Shakespeare’s books are out of stock,
Customer : When do you get the books?
Bookseller : Sir, next Monday.
Customer : 0k, thank you.
Bookseller : Do you want any other books?
Customer : No, thanks. How much do I have to pay for this?
Bookseller : 150 rupees, sir. We have given you 10% discount.
Customer : Thanks. Here is your payment.
Bookseller : Thanks for coming.

Dialogue between a passenger and a booking clerk .

Clerk : Good morning! How can I help you?
Passenger : I want to reserve four berths to Chennai on KCG Express.
Clerk : Have you filled the reservation form?
Passenger : Sir, here it is!
Clerk : Since it is festival season, there ¡s no seat available on 14 and 15 January.
Passenger : Oh, is that so Is there any seat available on Chennal Express?
Clerk : Yes. There are four seats available.
Passenger : Thank you sir. Please get me those four tickets.
Clerk : Welcome.

Complete the following dialogue between a girl and her mother.

Mother : I want to teach you how to cook ladies’ fingers curry today.
Daughter : No, mom. I have some work today, I am planning to go out.
Mother : a) __________
Daughter : Yesm mom. I was busy last week.
Mother : b) _______
Daughter : Is it compulsory to learn cooking?
Mother : c) _______
Daughter : Ok, mom. I go out and come back within an hour. Then I would learn.
Mother : d) ________
Daughter : Thank you mom.
Answer:
a) You said the same last week also.
b) But, when will you learn cooking at least a few items?
c) Absolutely. We need food and we cannot always depend on others.
d) That’s nice! That’s the spirit!

Exercises

Question 1.
Write a dialogue between two friends on the choice of career. (Revision Test – V)
Answer:
BETWEEN TWO FRIENDS:
Pavani : Hi, Sudha! What do you want to do after Intermediate?
Sudha : Hi, Vani! We are in MPC group. And our automatic choice is Engineering. The only thing to decide is which branch and which college. Do you have any other ideas?
Pavani : Yes. I’m not interested in Engineering.
Sudha : Very surprising I What else will you do ?.
Pavani : I’ll pursue B.Sc., course.
Sudha : Are you mad? Joining B.Sc., ?
Pavani : Why are you so excited? I just love Physics. I want to join B.Sc., now and then M.Sc., Physics. Later I want togo for research in Physics, particularly in Nanotechnology.
Sudha : Really, stunning! I don’t know anything about other options. All these days I’ve been under the impression that M.P.C in Intermediate means Engineering afterwards.
Pavani : There are lot many other options Sudha! You can consult any lecturer, or go through ‘education pages’ of newspapers or even browse the internet. Plenty of courses are available. We have to select the one that we are really interested in.
Sudha : Thank you Vani. You’ve really opened my eyes. That too, at the right time. I’ll follow your guidelines. Bye!

TS Inter 2nd Year English Grammar Dialogue Writing

Question 2.
Prepare a dialogue between yourself and your father discussing the secrets of success in business.
Answer:
Father : Where are you going Ajith?
Sai : Not particular, dad. Just like that.
Father : Hey, Ajith! It’s time you started picking up a lesson or two in running businesses.
Sai : I’ll certainly dad. Ever ready. Will you tell me one today ?
Father : Wonderful! Remember that two traits are very important to succeed in business.
Sai : Mm. Yes! What are they ?
Father : Honesty and wisdom.
Sai : Great. Then wisdom jneans!
Father : Never make such promises !
Sai : Dad! Stunning!

Question 3.
Imagine you got the first rank in the intermediate first year. Your close friend came to congratulate you. How do you share your experience with him? Write a dialogue between you and your friend. (Model Question Paper)
Answer:
Varun : Hi, Avinash. Welcome
Avinash : Hi, Akshay. Heartiest congratulations to you on securing the first rank.
Varun : Thank you!
Avinash : Welcome. You did work very hard and deserve such honours.
Varun : Not exactly hard work Avinash. Just smart work.
Avinash : What do you mean ?
Varun : Yes. I used to attend classes with prior preparation on the topic being discussed. That ensured my concentration. And I made notes of the topics discussed at the end of the day. That made revision easy. That never took much time. I used to pursue my hobbies. I found time for leisure activities, social relations
Avinash : Hey, that sounds great. It’s like a cakewalk. Then I’ll follow your example.
Varun : That’d be fine. You are always welcome. Bye for now.
Avinash : Thanks a bunch. Bye!

TS Inter 2nd Year English Grammar Dialogue Writing

Question 4.
You are working in ABCD Company. Ask your employer to enhance your salary from the next month. Write a dialogue between you and your employer.
Answer:
You : Good morning, sir.
Manager : Good morning, yes?
You : Sir, a humble submission. It’s three years since I had my last hike in my salary.
Manager : Yes, you’re right. Our company had tough times.
You : Yes, sir. But the cost of living has been going up. It’s really difficult to manage without a hike you from next month onwards.
Manager : Yes. We do really understand your difficulties. We are pooling our resources to hike salaries from the next month.
You : Thank you so much sir. Bye
Manager : You’re welcome. Bye!

Question 5.
Build a dialogue between a salesman and a’ customer who has sold a defective apparel. (Model Question Paper)
Answer:
Salesman : Good evening sir! How may I help you?
Customer : Good evening! This is the shirt I bought from you two days ago. And here is the bill.
Salesman : No need to show the bill sir. Any problem?
Customer : Yes, no buttons. And the sleeve’s are too loosely stitched.
Salesman : I’m very sorry sir. I’ll at once replace it with a new one.
Customer : But why this defective shirt? Lot of inconvenience and embarrassment.
Salesmam : I regret it very much sir. Normally our quality check is very stringent sir. I don’t know how this defective piece has reached here. Sorry once again sir. Here is the new shirt.
Customer : Thank you. But make sure such things don’t happen.
Salesman : Sure sir. Thank you.

Question 6.
As a student, you would like to open an account in the nearby State Bank of india. Write a conversation between you and an SBI bank manager about opening a savings bank account
Answer:
You : Good morning sir.
Manager : Good morning! How may I help you?
You : I would like to open a saving bank account here, sir.
Manager : Have you got your Aadhaar and PAN card?
You : Yes sir. I have brought attested copies and originals too.
Manager : Then, fill in this form correctly and completely. Enclose copies of those two documents. Deposit an amount of Rs two thousand in the cash counter. Hand over the receipt and filled in form to our Ass. Manager.
You : Thank you sir.
Manger : Welcome.
You : Sir, can I have the passbook and ATM card today?
Manager : No. You can get your passbook in a day or two. But, for ATM card you will have to apply later.
You : Right sir. Bye!
Manager : Bye.