TS Inter 2nd Year

Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A Measures of Dispersion Important Questions Long Answer Type to help strengthen their preparations for exams.

TS Inter Second Year Maths 2A Measures of Dispersion Important Questions Long Answer Type

Question 1.
Find the mean deviation about the mean for the following continuous distribution. [AP – Mar. 2015]

Solution:

Here, Σf_i = 100
Σf_ix_i = 7100
∴ Mean \(\frac{\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\mathrm{N}}=\frac{7100}{100}\) = 71
∴ Mean deviation about mean is M.D = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\mathrm{N}}=\frac{1040}{100}\) = 10.4.

Question 2.
Find the mean deviation about the mean for the following continuous distribution.

Solution:
We can form the following table from the given data :

Here, N = Σf_i = 100
Σf_ix_i = 12530
∴ Mean, \(\bar{x}=\frac{\sum_{i=1}^n f_i x_i}{N}=\frac{12530}{100}\) = 125.3
∴ Mean deviation about mean is M.D = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\mathrm{N}}\)
= \(\frac{1128.8}{100}\)
= 11.288 = 11.28.

Question 3.
Find the mean deviation from the mean of the following data, using the step deviation method. [AP – Mar. ’18; TS – Mar. 2016]

Solution:
We can form the following table from the given data :

Let the assumed mean be A = 35
Here, C = 10
N = Σf_i = 50
Mean, \(\bar{x}=A+\frac{\sum_{i=1}^n f_i d_i}{N} \times C\)
= \(A+\frac{\sum_{i=1}^7 f_i d_i}{N} \times C\)
= 35 + \(\frac{(-8)}{50}\) × 10
= 35 – 1.6 = 33.4
∴ Mean deviation about mean is M.D = \(\frac{\Sigma f_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\mathrm{N}}=\frac{659.2}{50}\) = 13.185.

Question 4.
Find the mean deviation about the mean for the given data using step deviation method. [AP – Mar. ’17, ’16, AP – May 2015] [TS – Mar. ’18, May ’16]

Solution:
We can form the following table from the given data:

Let the assumed mean be A = 25
Here, C = 10
N = Σf_i = 50
\(\bar{x}=A+\frac{\sum_{i=1}^n f_i d_i}{N} \times C\)
= \(\mathrm{A}+\frac{\sum_{\mathrm{i}=1}^5 \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\mathrm{N}} \times \mathrm{C}\)
= 25 + \(\frac{10}{50}\) × 10 = 27
∴ Mean deviation about mean is M.D = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\mathrm{N}}=\frac{472}{50}\) = 9.44.

Question 5.
Find the mean deviation about the median for the following continuous distribution.

Solution:

Median class = Class containing \(\frac{N}{2}^{\text {th }}\) item
= \(\frac{50}{2}\) = 25 item
= 20 – 30 class
Here, l = 20, C = 10, f = 14, m = 14, N = 50
∴ Median, M = l + \(\frac{\frac{N}{2}-m}{f}\) × C
= 20 + \(\frac{25-14}{14}\) × 10
= 20 + 7.86 = 27.86
∴ Mean deviation about median is M.D = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right|}{\mathrm{N}}\)
= \(\frac{517.16}{50}\) = 10.34

Question 6.
Find the mean deviation about median for the following data.

Solution:

Median class = Class containing \(\frac{\mathrm{N}^{\text {th }}}{2}\) item
= \(\frac{1000}{2}\) = 500th item = 35 – 40 class
Here, l = 35, C = 5, f = 160, m = 420, N = 1000
∴ Median, M = l + \(\frac{\frac{N}{2}-m}{f}\) × C
= 35 + \(\frac{500-420}{160}\) × 5
= 35 + 2.5 = 37.5
∴ Mean deviation about median is M.D = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right|}{\mathrm{N}}=\frac{8175}{1000}\) = 8.175.

Question 7.
Calculate the variance and standard deviaon,of the following continuous frequency distribution.
[AP – May 2016; TS – Mar. 2017; Mar. ‘14, Board Paper]

Solution:
We can form the following table from the given data:

∴ Mean, \(\bar{x}=\frac{1}{N} \sum_{i=1}^7 f_i x_i=\frac{3100}{50}\) = 62
Variance, σ² = \(\frac{1}{N} \sum_{i=1}^7 f_i\left(x_i-\bar{x}\right)^2\)
= \(\frac{1}{50}\) (10050) = 201
Standard deviation, σ = \(\sqrt{\frac{1}{N} \sum_{i=1}^7 f_i\left(x_i-\bar{x}\right)^2}\)
= \(\sqrt{\text { variance }}=\sqrt{201}\)= 14.17 (approximately).

Question 8.
The following table gives the daily wages of workers in a factory. Compute the standard deviation and the coefficient of variation of the wages of the workers.

Solution:
We shall solve this problem using the step deviation method, since the midpoints of the class intervals are numerically large.

From the table, N = \(\sum_{i=1}^9\) f_i = 72
Let the assumed mean, A = 300
Here, C = 50
From the table, \(\sum_{i=1}^9\) f_id_i = – 31
Mean, \(\bar{x}=A+\frac{\sum_{i=1}^9 f_i d_i}{N}\) × C
= 300 + \(\frac{-31}{72}\) × 50
= 300 – \(\frac{1550}{72}\)
= 278.5 (approximately)

Question 9.
The scores of two cricketers A and B in 10 innings are given below. Find who is a better run getter and who is a more consistent player.

Solution:
For cricketer A : Scores of A are 40, 25, 19. 80, 38, 8, 67, 121, 66, 76
∴ Mean, \(\overline{\mathrm{x}}_{\mathrm{A}}=\frac{\text { Sum of the scores }}{\text { No. of innings }}\)
= \(\frac{40+25+19+80+38+8+67+121+66+76}{10}\)
= \(\frac{540}{10}\) = 54
The deviations of the respective observations from the mean i.e., x_i – \(\overline{\mathrm{x}}\) are: – 14, – 29, – 35, 26, – 16, -46, 13, 67, 12, 22
Variance, σ² = \(\frac{1}{n} \sum_{i=1}^{10}\left(x_i-\bar{x}\right)^2\)
= \(\frac{196+841+1225+676+256+2116+169+4489+144+484}{10}\)
= \(\frac{10596}{10}\) = 1059.6
Standard deviation, σ_A = \(\sqrt{1059.6}\) = 32.55
C.V of A = \(\frac{\sigma_{\mathrm{A}}}{\overline{\mathrm{x}}_{\mathrm{A}}}\) × 100
= \(\frac{32.55}{54}\) × 100 = 60.28.

For cricketer B:
Scores of B are 28, 70, 31, 0, 14, 111, 66, 31, 25, 4
Mean \(\overline{\mathrm{x}}_{\mathrm{B}}=\frac{\text { Sum of the scores }}{\text { No. of innings }}\)
= \(\frac{28+70+31+0+14+111+66+31+25+4}{10}=\frac{380}{10}\)
= 38

The deviations of the respective observations from mean i.e., (x_i – \(\overline{\mathbf{x}}\)) are: – 10, 32, – 7, – 38, – 24, 73, 28, – 7, – 13, – 34
Variance, σ_B² = \(\frac{1}{n} \sum_{i=1}^{10}\left(x_i-\bar{x}\right)^2\)
= \(\frac{100+1024+49+1444+576+5329+784+49+169+1156}{10}\)
= \(\frac{10680}{10}\) = 1068
Standard deviation. σ_B = \(\sqrt{1068}\) = 32.68
C.V of B = \(\frac{\sigma_B}{\bar{x}_B}\) × 100
= \(\frac{32.68}{38}\) × 100 = 86
Since, \(\overline{\mathbf{x}}_{\mathrm{A}}\) is greater than \(\overline{\mathbf{x}}_{\mathrm{B}}\) then cricketer A is better run getter (scorer).
Since, C.V of A is less than C.V of B then cricketer ¡s also a more consistent player.

Question 10.
From the prices of shares X and Y given below, for 10 days of trading, find out which share is more stable?

Solution:
Share X:
Prices of share X are 35, 54, 52, 53, 56, 58, 52, 50, 51, 49
Mean, \(\overline{\mathrm{x}}_{\mathrm{X}}=\frac{1}{\mathrm{~m}} \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}\)
= \(\frac{510}{10}\) = 51
The deviations of the respective observations from mean i.e., (x_i – \(\overline{\mathrm{x}}\))² are: – 16, 3, 1, 5, 7, 1, – 10, – 2
Variance, σ_X² = \(\frac{1}{n} \sum_{i=1}^{10}\left(x_i-\bar{x}\right)^2\)
= \(\frac{256+9+1+25+49+1+1+0+4}{10}\)
= \(\frac{350}{10}\) = 35
Standard deviation, σ_x = √35 = 5.9 (approximately)
C.V of X = \(\frac{\sigma_X}{\bar{x}_X}\) × 100
= \(\frac{5.9}{51}\) × 100 = 11.5 (approximately)

Share Y:
Prices of share Y are 108, 107, 105. 105, 106, 107, 104, 103, 104, 101
Mean, \(\overline{\mathrm{x}}_{\mathrm{Y}}=\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}\)
= \(\frac{1050}{10}\) = 105
The deviations of the respective observations from mean i.e.. (x_i – \(\overline{\mathrm{x}}\)) are 3, 2, 0, 0, 1, 2, – 1, – 2, – 1, – 4
Variance, σ_Y² = \(\frac{1}{n} \sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2\)
= \(\frac{9+4+0+0+1+4+1+4+1+16}{10}\)
= \(\frac{40}{10}\) = 4
Standard deviation, σ_Y = √4 = 2
C.V. of Y = \(\frac{\sigma_Y}{\bar{x}_Y}\) x 100
= \(\frac{2 \times 100}{105}\) x 100 = 1.904
Since, C.V of X is greater than C.V of Y then the share of Y is more consistent (stable).

Question 11.
Find the mean and variance using the step deviation method of the following tabular data, giving the age distribution of 542 members.

Solution:
We form the following table From the given data:

Let the assumed mean, A = 55
Here, C = 10

Some More Maths 2A Measures of Dispersion Important Questions

Question 12.
Students of two sections A and B of a class show the following performance in a test (conducted 100 marks).

Which section of students has greater variability In performance?
Solution:
Given that,

The variance of distribution of marks of section A is σ₁² = 64
∴ The standard deviation of distribution of marks of section A is σ₁ = 8
The variance of distribution of marks of section B is σ₂² = 81
∴ The standard deviation of distribution of marks of section B is σ₂ = 9
The average marks in the test of section A is \(\bar{x}_1\) = 45
The average marks in the test of section B is \(\bar{x}_2\) = 45
∵ The average marks in the test of both sections of students is the same i.e., 45.
The standard deviation of the distribution of marks in section B is greater than the standard deviation of (he distribution of marks in section A.
Hence, the section B has greater variability in the performance.

Question 13.
Find the variance and standard deviation of the following frequency distribution.

Solution:
From the given data, we can form the following table:

Mean of the given data is (\(\overline{\mathrm{x}}\)) = \(\frac{1}{N} \sum_{i=1}^7 f_i x_i\)
= \(\frac{760}{40}\) = 19
Variance, σ² = \(\frac{1}{N} \sum_{i=1}^7 f_i\left(x_i-\bar{x}\right)^2\)
= \(\frac{1}{40}\) (1736) = 43.4
Standard deviation, σ = \(\sqrt{\text { variance }}\)
= \(\sqrt{43.4}\) = 6.58 (approximately).

Question 14.
Calculate the variance and standard deviation for a discrete frequency distribution.

Solution:
From the given data, we can form the following table:

Mean of the given data is (\(\overline{\mathrm{x}}\)) = \(\frac{1}{N} \sum_{i=1}^n f_i x_i\)
= \(\frac{1}{N} \sum_{i=1}^7 f_i x_i=\frac{420}{30}\) = 14
Variance, σ² = \(\frac{1}{N} \sum_{i=1}^7 f_i\left(x_i-\bar{x}\right)^2\)
= \(\frac{1}{30}\) (1374) = 45.8
Standard deviation, σ =\(\sqrt{\text { variance }}\)
= \(\sqrt{45.8}\) = 6.76 (approximately).

Question 15.
Find the mean deviation about the median for the following data.

Solution:
We can form the following table from the given data:

Median class = Class containing \(\frac{N^{t h}}{2}\) item
= \(\frac{100}{2}\) = 50th item = 40 – 50 class
Here, l = 40, f = 28, m = 32, C = 10, N = 100
∴ Median, M = l + \(\frac{\frac{\dot{N}}{2}-\mathrm{m}}{\mathrm{f}}\) × C
= 40 + \(\frac{50-32}{28}\) × 10
= 40 + \(\frac{180}{28}\)
= 40 + 6.42 = 46.42
∴ Mean deviation about median is M.D = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right|}{\mathrm{N}}=\frac{1428.4}{100}\) = 14.284

Question 16.
Lives of two models of refrigerators A and B obtained ¡n a survey are given below.

From the above data suggest which model to purchase.
Solution:
Model – A:
To find the mean and variance of lives of model A refrigerators, we shall construct the following table:

From the table, N = \(\sum_{\mathbf{i}=1}^5\) f_i = 46, \(\sum_{\mathbf{i}=1}^5\) f_ix_i = 212
Mean, \(\bar{x}_A=\frac{1}{N} \sum_{i=1}^5 f_i x_i\)
= \(\frac{212}{46}\) = 4.6
From the table, \(\sum_{i=1}^5 f_i\left(x_i-\bar{x}\right)^2\) = 244.96
Variance, σ_A² = \(\frac{1}{N} \sum_{i=1}^5 f_i\left(x_i-\bar{x}\right)^2\)
= \(\frac{244.96}{46}\) = 5.32
Standard deviation, σ_A = \(\sqrt{5.32}\)
= 2.3 (approximately)
Coefficient of variation of model A = \(\frac{\sigma_{\mathrm{A}}}{\mathrm{x}_{\mathrm{A}}}\) × 100
= \(\frac{2.3}{4.6}\) × 100
= 50 (approximately).

Model B:
To find the mean and variance of lives of model B refrigerators, we shall construct the following table:

From the table, N = \(\sum_{i=1}^5\) f_i = 49
\(\sum_{i=1}^5\) f_ix_i = 297
Mean, \(\bar{x}_B=\frac{1}{N} \sum_{i=1}^5 f_i x_i\)
= \(\frac{297}{49}\) = 6.06
From the table, \(\) = 224.8164
Variance, σ_B² = \(\frac{1}{N} \sum_{i=1}^5 f_i\left(x_i-\bar{x}\right)^2\)
= \(\frac{224.8164}{49}\) = 4.58 (approximately)
Standard deviation, σ_B = \(\sqrt{4.58}\)
= 2.1 (approximately)
Coefficient of variation of model B = \(\frac{\sigma_B}{x_B}\) × 100
= \(\frac{2.1}{6.06}\) × 100 = 34.65 (approximately)
Since C.V. of model B is less than C.V. of model A then the model B is more consistent than the model A with regard to life in years.
Hence, we suggest model B for purchase.

Telangana TSBIE TS Inter 2nd Year English Study Material Grammar Writing Descriptions Exercise Questions and Answers.

TS Inter 2nd Year English Grammar Writing Descriptions

Q.No.16 (4 Marks)

Students learning a language go ahead step by step-sounds/letters → words → sentences → composition. “Writing Descriptions” is a kind of composition. Here, we try to learn how to describe a person, a place, a thing or a process. One should first be clear about the points (content) that are to be included.

Then, one should try t. present those points in correct language and in a systematic order. Keen obse ation of suitable models first and regular practice in writing descriptions nex nelp one master this art. Constant assessment of one’s own performance and necesaryremedial measures go a long way in enhancing this competency.

I. DESCRIBING A PERSON:

While describing a person we include details like the person’s appearance, qualities, nature, typical traits, etc.

Now, see a model.

The person I always remember most is the principal of my college. He is not very tall, but he is well-built in his physique. He has grey hair and wears spectacles but only while reading. While he speaks to us, at the assembly, he has the habit of pushing his spectacles to the top of his head. He rarely smiles and always looks stern. However, he is really very kind and always treats the students in a friendly manner. When a student is in trouble, he always tries to help him. All the students and teachers respect our principal.

Examine another model.
This disheartening introspection was interrupted by the waiting-room door opening. An old man stood on the threshold, looking at me silently. He wore a heavy black jacket buttoned high in the chest, narrow trousers, and a two-inch collar. In his hand, he held a pair of gold-rimmed pince-nez, which were attached to his right lapel by a thick black silk ribbon.

He was so thin, so old, so pale, and so slow he could have taken his place in the near-by post-mortem room without attracting attention. He clipped his glasses on to his nose with a slow, shaky movement and inspected me more carefully. I leapt to my feet and faced him.
(From the short story ‘An Interview’)

Exercise

Question 1.
Describe your mother using the hints given below.

tall and thin – black hair – important person in life – wake up at 5 am – prepares us for college – takes care of father, grandparents too – works without complaining – emotionally strong – want to be like her – symbol of love and sacrifice
Answer:
MY MOTHER

Of all persons on this earth, I love my mother most. She is my first and best teacher. I love to be like her and if I can, I will be the blessed one. With her tall and slim appearance, she looks impressive at once. Her thick dark hair is an added attraction. She wakes up at five in the morning and works hard till late in the evening with a never fading smile. She gets ready for us everything we need, takes care of father’s and grandparents’ needs. She is very strong both physically and emotionally. If someone is looking for a true symbol of love, sacrifice and contentment, my mother stands as that symbol!

Question 2.
Describe your best friend Deepthi using the hints given below.

fair and tall – 165 ems – aged 25 – slim – thick hair – large, expressive eyes – wears western clothes – sari on special occasions – practises yoga – a good athlete – hardworking – honest – helpful – my best friend
Answer:
DEEPTHI- MY REST FRIEND

Deepthi – the name rings music in my ears! She is my most beloved friend. She fits perfectly into the definition of a true friend – as offered by Khalil Gibran. She is. twenty-five. She looks fair. And she is tall indeed and very slim. With her thick black hair and ever-present smile, she impresses everyone. She practises Yoga. She is fond of athletics. She prefers modem western clothes. Yet, she puts on traditional Indian costumes on special occasions. She is honest to the core and loves hard but smark work. I am proud to be her friend!

II. DESCRIBING A PLACE:

The description of a place intludes in it details like its location, special features, Significance, people, crops, landmarks and so on.

Examine the model given under carefully.
Yadagirigutta is a small town in the Yadadri Bhuvanagiri District of Telangana State. It is 60 KMs from Hyderabad city. The famous Hindu Temple, Sri Lakshmi Narasimha Swamy Temple is situated on a hillock in the town. It is a unique pleasant hillock that enjoys moderate climate in allseasons. The place is visited by an average of five thousand to eight thousand pilgrims every day. After the formation of Telangana State, the Hon’ble Chief Minister Sri Kalvakuntla Chandra shekhar Rao has been personally supervising the development of the temple to make it one of the famous tourist attractions not only in India but also all over the world.

Exercise

Question 1.
Describe the famous tourist place, Ooty using the hints given below.

town – hill station – Tamil Nadu – 265 KMs from Bangalore – located in the Nilgiris – 7500 ft above sea level – cool and a pleasant summer destination – lakhs of tourists – – coffee plantations – top boarding schools – Queen of the Hill Stations
Answer:
OOTY THE QUEEN OF HILL. STATIONS

Ooty or Udhagamandalam is rightly and deservedly regarded as the Queen of Hill Sta-tions. Situated in the Western. Ghats at an elevation of7500 feet in Nilgiris district of Tamil Nadu, Ooty is one of the most preferred tourist destinations in India. Nature’s, bounty, pleasant climate, hill slopes, waterfalls, lakes, churches, temples, Botanical and Rose Gardens, hill railways are some of the prominent Sights that mesmerise the visitors. Travel up the Ghat road or rail track itself is an experience to cherish for life time. It is, therefore, no wonder that the English people loved to spend here most of their time. (Before India became independent.)

Question 2.
Describe your own village or town based on your knowledge and understanding in not more than 100 words.
Answer:
MY VILLAGE

My village is away from the bustle of modern urban societies. With green-covered hill ridges on two sides and a stream on another side, our village is endowed with rich natural beauty. Climate round the year is pleasant. Villagers grow a variety of crops twice a year. All homes have at least one milching animal each. People lead peaceful, comfortable and contented lives. Maintaining cleanliness and greenery is a collective responsibility. Basic amenities like roads, drains are maintained well. A school and a hospital meet villagers’ educational and health needs. Our village is a model one.

III. DESCRIBING A THING :

In the descriptions of things, the particulars involved are the appearance, make up, uses and any other relevant details of the thing described. Observe the sample description that follows.

The National Flag of India is a horizontal tricolourof deep saffron at the top, white in the middle and dark green at the bottom in equal proportion. The ratio of width of the flag to its length is two is to three (2:3). In the centre of the white band is a navy-blue colour wheel with twenty-four spokes known as the Ashoka Chakra. This is taken from I he famous Ashoka Sthupa (pillar) of Saranath. Each spoke symbolises one principle of life and also the twenty-four hours In the day, which is why it is called the wheel of time.

Exercise

Question 1.
Describe about the Charminar using the hints given below.

monument location 1591 AD four minarets height materials used commein orate the end of deadly plague prime tourist attraction
Answer:
THE CHARMINAR

The Charminar has for long been the symbol of Hyderabad and the emblem of Telangana. Founded in 1591 by Muhammad Quli Qutb Shah, this 48.7 meter high monument has been the centre of attraction of tourists from across the globe. With Indo-Islamic architectual styles and rich ornate designs, its appearance is a feast to eyes. Situated on the eastern bank of the Musi River in the old city area, this tourist centre also serves as a religious centre on all important festive occasions. A very busy market place surrounds the landmark where a wide range of articles are sold at a’ reasonable price. The Charminar (four minarets – pillars) is a place to visit.

Question 2.
Describe about the Statue of Equality using the hints given below.

the statue -location – imposing size – posture – materials used – wide eyes – overflowing with mercy huge ears eager to listen – broad shoulders ready to take up respon sibilities – folded hands – appealing for Equality – for service to society – Supreme Being [Model Question Paper]
Answer:
THE STATUE OF EQUALITY

The latest feather in Telangana’s cap is the 216 – feet, golden-hued Panch aloha (five metals) Staute of Equality Situated on a sprawling 50-acre site at Muchintal, near Hyderabad, the statue stands for the image and preachings of the eleventh century Vaishnavite Shri Bhagavad Ramanuja charya. Bhagavad Ramanujacharya put his heart and soul to practise and propagate equality among all.

The statue is not just a monument; it is a symbol; it is a movement. The imposing statue in sitting posture has wide eyes that overflow with mercy. Huge ears stand for eagerness to listen and learn. Broad shoulders express readiness to take up responsibilities. Folded hands appeal for equality. A visit to the place is bound to bring in transformation!

Question 3.
Describe about the Telangana Martyrs Memorial using the hints given below.

monument – location – designer – materials used – height – completion date – dedicated to – symbol for
Answer:
TELANGANA MARlYRS MEMORIAL

Telangana Martyrs Memorial is situated near Public Gardens, Saifabad, Hyderabad. It is a 25-feet tall black and red granite monument. It was designed by professor of architecture Aekka Yadagiri Rao. It was commissioned in 1974. It commemorates the sacrifices of more than three hundred and sixty youngsters who faced bullets for the cause of separate Telangana in the 1969 movement. A white lily flower, atop the Memorial pays rich tributes to those martyrs. After Telangana became a separate state, it was decided to observe June 2nd every year as the Martyrs’ Memorial Day officially.

IV. DESCRIBING A PROCESS:

Presenting various steps involved in a relatively simple activity is called ‘Describing a Process’. It requires the knowledge of necessary items and the method of doing that ‘activity’, It also demands the capacity to put that knowl-edge into a neat and properly organised paragraph. Observing as many descriptions of processes as possible is definitely of great help. You come across a number of examples in books on chemistry and cookery, for instance. All the flow charts serve as good examples. After observing sufficient number of examples, start writing on your own.

Points to ponder over for the purpose of Public Examination?

• Use the Simple Present Tense, as the ‘actions’ are universal.
• Imperative sentences are commonly used,
  eg: Take a glass of water. Then put it.
• Passive sentences are also used,
  eg: A spoonful of salt is added.
• The use of linkers (markers) like first, then, after that is very important.A list of linkers is given here
  
• Give the list of ingredients, if necessary. (Generally given in the case of recipes).
• The given activity may require about 6 to 8 steps for completion. So your answer may have 6 to 8 sentences.
• The entire process can also be presented in the form of a FLOWCHART.
• Give a HEADING.

Examples:

1. Preparing potato fry

Given below is a description of how a beginner can make Aloo fry. Read the recipe and try it out at home! Observe the concise, numbered instructions, the list of ingredients, as also the preparation and cooking time, making it easy for a person to decide whether to attempt the dish or not.

Recipe for potato fry
Preparation time: 5 min, Cooking Time: 20 min, Serves: 2
Ingredients

• 4 Potatoes – 2, large, washed, peeled’ and cubed 4
• Oil – 2 tbsps 4
• Turmeric powder – 1- 2 tsp
• Chilli powder -1 tsp
• 4 Salt to taste Method

1. Heat oil in a heavy bottomed vessel.
2. Add the diced potatoes and cook on medium flame for 4 to 5 minutes, tossing the cubed potatoes frequently.
3. Reduce flame, add turmeric powder and mix. Place lid and cook on low flame for 10 minutes.
4. Remove lid, add salt. Allow the potatoes to reach a nice brown shade. It may take another 5-6 minutes.
5. Add chilli powder: and mix well. The potato fry is ready.

2. Sending a letter by Registered Post

When a letter is sent by registered post, the postman personally hands over the letter to the recipient and takes his signature, instead of simply leaving it in the recipient’s letter box. If the sender wants the proof of the letter having been delivered, he can attach an acknowledgement card to the letter. Registered post is, therefore, a safe way of sending important letters and documents. Given below is a paragraph about the process of sending a letter by registered post.

Observe the use of the passive voice.

Registered post is a safe way of sending important letters and documents. The letter is placed in an envelope and the words “By Registered Post” are inscribed on the top. The address of both the recipient as well as the sender must be written. The cover is then taken to the post office. It should be handed over the clerk at the Registration counter.

The clerk weighs it and informs us about the value of the stamps that should be affixed on the cover. The minimum charges are Rs. 25. The stamps should be bought and affixed accordingly and, once again, the cover is handed to the registration clerk. A receipt is given by the clerk and this should be kept safely till we come to know that the letter has reached the recipient.

3. Applying for a passport

A passport is an official document issued by the government, certifying the identity and citizenship of the person and entitling him to travel to foreign countries and return. The process of getting a passport was quite difficult till a few years ago but it has become considerably easy now. Given below is the process that must be followed * by Indian citizens to get a passport.

Applying for a passport online

Applying for a passport is difficult offline but is easy online. The applicant should go to the Passport Seva Website first. He/She should fill the User Registration Form carefully and click on ‘Register’. The applicant will then reach Applicant Home page. He should click on the link Apply for a Fresh Passport.

The form contains details of name; place and date of birth, gender, marital status, educational qualifications, identification marks, Aadhaar Card Number and so on. An ARN (Acquirer Reference Number) is generated when the form is submitted Next, the applicant must book an appointment at the Passport Seva Kendra depending on the availability of dates. A fee of Rs. 1500 is charged for booking an appointment.

On the appointed day the applicant must go to the Passport office with all original documents, self attested Photostat (Xerox) documents and the passport form with a photo pasted on it. The original documents are checked. A photograph is taken and biometric information taken. The person will be interviewed briefly. An acknowledgement slip is given. Within a day or two, police verification takes place at the address given. In a few days the applicant receives the passport.

Exercise

Question 1.
Describe how Sandbya drew Rs. 500 from an ATM.
Answer:
Drawing Money from ATM

Sandhya first went to the nearest ATM. After entering the ATM room, she inserted the card in the slot. Then from the displayed options she selected banking. Later she entered her 4 digit pin. After that she selected withdrawal from the choices. Next she clicked on Savings- account. And then she entered the amount. Then she opted for a printed advice by clicking on yes. As the machine then rolled out cash first and printed note later, she collected both, picked up her ATM card arid left the chamber.

Question 2.
Describe how you and your friends made arrangements for the College Day Function. (Hints: date, chief guest, programme, speeches, cultural items, invitations, stage arrangements, catering)
Answer:
How We Made Arrangements for the College Day

The moment our Principal announced proposals for college day, we Intermediate second year students) met in the students’ counsel room with our English lecturer in the chair. We discussed and decided upon the most convenient date. Later we thought of a guest who could really inspire us through his talk. We all decided-unanimously – our Collector to be our guest.

Then we chalked out the programme sheet. What first, what next, etc. We selected speakers to represent all sections, classes, groups, categories also. Then the talks dealt with cultural programmes. After that invitation cards, stage arrangement, food needs, etc. occupied our attention. Committees were formed for each activity. With the approval of our Principal and support and guidance of our lecturers we went ahead with our plans and made our College Day a grand success.

Question 3.
Describe how your elder brother got a Driving license.
(Hints : Learner’s License – Road signs, general rules – application – documents to be attached – Practice – Test driving after 30 days – Getting permanent license)
Answer:
GETTING A DRIVER’S LICENSE

First, my brother started learning how to drive. He also improved his knowledge of the symbols and signs on road sides and in driving. Then he filled in the prescribed form and submitted it in the Local RTA (Road Transport Authority) or MVI (Motor Vehicle Inspector) office along with the payment details of necessary fee. Then, he took a vision (eye- sight) test and a written test. He cleared both of them and got a Learner’s License. Then, he went on practising driving on roads. After a month, he took a test in actual driving. He got through it too. After that, he was issued the Driving License.

Telangana TSBIE TS Inter 2nd Year Chemistry Study Material Lesson 3(b) Chemical Kinetics Textbook Questions and Answers.

TS Inter 2nd Year Chemistry Study Material Lesson 3(b) Chemical Kinetics

Very Short Answer Questions (2 Marks)

Question 1.
Define the speed or rate of a reaction.
Answer:
The rate of a reaction is the change in con-centration of a reactant or product in unit time. It can be expressed in terms of:

1. the rate of decrease in concentration of any one of the reactants, or
2. the rate of increase in concentration of any one of the products.

Question 2.
Assuming that the volume of the system is constant, derive the expression for the average rate of the system R → P in terms of R and P. [time = t sec; R = reac-tant, P = product]
Answer:
Considering a hypothetical reaction assuming that the volume of the system remains constant.
R → P
One mole of the reactant R produces one mole of the product P. If [R], and [P], are the concentrations of R and P respectively at time t₁ and [R]₂ and [P]₂ are their concentrations at time t₂ then
Δt = t₂ – t₁ ; Δ[R] = [R]₂ – [R]₁
Δ[P] = [P]₂ – [P]₁
Rate of disappearance of R =

Rate of appearance of P =

These equations represent the average rate of a reaction.

Question 3.
What are the units of rate of reaction?
Answer:
The units of rate of reaction are concentration × time^-1. If concentration is mol L^-1 and time is in seconds then the units will be mol L^-1 s^-1. For gaseous reaction the units of rate equation are atm s^-1.

Question 4.
Draw the graphs that relate the concen-tration (C) of the reactants and the reac-tion times (t) and the concentrations of the products (C) and the reaction times (t) in chemical reactions.
Answer:

Question 5.
Write the equation for the rate of the reaction
5Br^– (aq) + \(\mathrm{BrO}_3{ }^{-}(\mathrm{aq})\) + 6H⁺(aq) → 3Br₂(aq) + 3H₂O(l)
Answer:

Question 6.
What is rate law? Illustrate with an example.
Answer:
The rate law is the expression in which rea-ction rate is given in terms of molar concen-tration of reactants. Each concentration term is raised to some power, which may or may not be same as the stoichiometric coefficient of the reacting species in balanced chemical equation. Example
2NO (g) + O₂(g) → 2NO₂(g)
Rate = k [NO]² [O₂]

Question 7.
Mention a reaction for which the exponents of concentration terms are not the same as their stochiometric coefficient in the rate equation.
Answer:
i) Formation of CCl₄ from CHCl₃?
CHCl₃ + Cl₂ → CCl₄ + HCl
Rate = k[CHCl₃] [Cl₂]^1/2

ii) CH₃COO C₂H₅ + H₂O → CH₃COOH
+ C₂H₅OH
Rate = k[CH₃COOC₂H₅]¹ [H₂O]⁰

Question 8.
Define order of a reaction. Illustrate your answer with an example. (TS 15)
Answer:
The sum of the powers of the concentration terms of the reactants in the rate law expre-ssion is called the order of that reaction.
Order of reaction can be 0, 1, 2, 3 and even fraction.

Example :
Rate = k[A]^1/2 [B]^3/2
Order = \(\frac{1}{2}\) + \(\frac{3}{2}\) = 2. i.e., second order

Question 9.
What are elementary reactions?
Answer:
The reactions taking place in one step are called elementary reactions.

Question 10.
What are complex reactions ? Name one complex reaction.
Answer:
If a reaction takes place in a sequence of elementary reactions called mechanism in which reactants convert into products, it is called complex reaction. These may be consecutive reactions, e.g. Oxidation of ethane to CO₂ and H₂O proceeds through a series of intermediate steps in which alcohol, aldehyde and acid are formed.

Question 11.
Give the units of rate constants for zero, first order and second order reactions.
Answer:

Question 12.
Define molecularity of a reaction. Illustrate with an example.
Answer:
The number of reacting species (atoms, ions, molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction is called molecularity of a reaction.

Example:

i) NH₄NO₂ → N₂ + 2H₂O
Since only one molecule is involved in the reaction it is unimolecular reaction.

ii) H₂ + I₂ → 2HI
Since two molecules are involved in the reaction it is a bimolecular reaction.

Question 13.
What is rate determining step in a complex reaction ?
Answer:
Suppose if a reaction proceeds in a sequence of elementary reactions, the overall rate of reaction is controlled by the slowest step and it is called as rate determining step.

Question 14.
Give the mechanism for the decomposition reaction of H₂O₂ in alkaline medium cata-lysed by I^– ion.
Answer:
Decomposition of H₂O₂ in alkaline medium catalysed by I^– ions proceeds in the following steps.

1. H₂O₂ + I^– → H₂O + IO^–
2. H₂O₂ + IO^– → H₂O + I^– + O₂

Question 15.
Write the equation relating [R], [R]₀ and reaction time ‘t’ for a zero order reaction [R] = concentration of reactant at time ‘t’ and [R]₀ = initial concentration of reactant.
Answer:
k = \(\frac{[\mathrm{R}]_0-[\mathrm{R}]}{\mathrm{t}}\) ; Rate = \(\frac{-\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}\) = k[R]₀
k = rate constant
[R]₀ = initial concentration
[R] = concentration of reactant at time t.

Question 16.
Draw the graph that relates the concentra-tion ‘R’ of the reactant and’t’ the reaction time for a zero order reaction.
Answer:

Question 17.
Give two examples for zero order reactions. (TS Mar. ‘ 19 )
Answer:

1. The decomposition of gaseous ammonia on a hot platinum surface is a zero order reaction at high pressure.
   
2. Thermal decomposition of HI on gold surface is zero order reactions.
   
3. Photochemical reactions are zero order

Question 18.
Write the Integrated equation for a first order reaction In terms of [R], [R]₀ and t.
Answer:
k = \(\frac{2.303}{t}\) log \(\frac{[\mathrm{R}]_0}{[\mathrm{R}]}\)
k = First order rate constant
[R] = Concentration of R at time t
[R]₀ = initial concentration R

Question 19.
Give two examples for gaseous first order reactions. (IPE 14)
Answer:

1. Hydrogenation of ethane is an example of first order reaction.
   C₂H₄(g) + H₂(g) → C₂H₆(g)
2. Decomposition of N₂O₅
   N₂O₅ (g) → 2NO₂ + \(\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\)

Question 20.
For the reaction: A(a) → B(g) + C(g), write the integrated rate equation in terms of total pressure p and the partial pressures P_AP_BP_C.
Answer:
Total pressure P_t = P_A + P_B + P_C
p_A, p_B and p_C are the initial pressures of A, B and C respectively.
p_i = initial pressure at time, t = 0.
k = \(\frac{2.303}{t}\)log \(\frac{p_i}{p_A}\) = \(\frac{2.303}{\mathrm{t}}\)log \(\frac{p_i}{2 p_i-p_t}\)

Question 21.
What is half-life of a reaction? Illustrate your answer with an example.
Answer:
The half – life of a reaction is the time in which the concentration of a reactant is reduced to one half of its initial concen-tration. It is represented by t_½.
Ex: Half life of C – 14 is 5730 years.
t_1/2 = \(\frac{0.693}{\mathrm{k}}\)
For a first order reaction half-period Is constant and is independent of initial con-cent ration.

Question 22.
Write the equation relating the half-life (t_1/2) of a reaction and the rate constant ‘K’ for first order reaction.
Answer:
For the first order reaction
k = \(\frac{2.303}{t}\)log \(\frac{[\mathrm{R}]_0}{[\mathrm{R}]}\) at t_1/2 [R] = \(\frac{[\mathrm{R}]_0}{2}\)
So the above equation becomes
k = \(\frac{2.303}{t_{1 / 2}}\) log \(\frac{[\mathrm{R}]_0}{[\mathrm{R}]_{0 / 2}}\)
or t_1/2 = \(\frac{2.303}{\mathrm{k}}\)log 2 or t_1/2 = k
For a first order reaction half-period is constant and is independent of initial con-centration.

Question 23.
Write the equation useful to calculate half-life (t_1/2) values for zero and first order reactions.
Answer:

1. Half-life for zero order reaction
   t_1/2 = \(\frac{[\mathrm{R}]_0}{2 \mathrm{k}}\)
2. Half-life for first order reaction
   t_1/2 = \(\frac{693}{\mathrm{k}}\) k is rate constant

Question 24.
What are Pseudo first order reactions? Give one example.
Answer:
The reactions which appear to be second order but follow the first order rate equation are called Pseudo first order reactions.
Example :

Since the concentration of water is large excess, the change in concentration of water is negligible. So its concentration is taken as constant. Then rate of reaction depend only on the concentration of CH₃COOC₂H₅.
Rate = k[CH₃COOC₂H₅]
Inversion of cane sugar is another example of pseudo first order reaction.

Question 25.
Write the Arrhenlus equation for the rate constant (k) of a reaction.
Answer:
Arrhenius equation for the rate constant (k) is
k = Ae^-Ea/RT
A = Arrhenlus frequency factor
R = Gas constant
E_a = Activation energy measured in Joules mol^-1
T = Absolute temperature

Question 26.
By how many times the rate constant increases for rise of reaction temperature by 10°C?
Answer:
For a chemical reaction with rise in temperature by 10° the rate constant is nearly doubled.

Question 27.
Explain the term ‘activation energy’ of a reaction with a suitable diagram.
Answer:

The energy required to form the intermediate called activated complex (C) is known as activatioñ energy E_a.

Question 28.
Write the equation which relates the rate constants k₁ and k₂ at temperatures T₁ and T₂ of a reaction.
Answer:
log \(\frac{\mathrm{k}_2}{\mathrm{k}_1}\) = \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left[\frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{~T}_1 \mathrm{~T}_2}\right]\)
k₁ and k₂ are the rate constants of a reaction at temperatures T₁ and T₂ respectively. E_a is activation energy.

Question 29.
What is collision frequency (Z) of a reac-tion? How is rate related to it for the reaction A + B → Products.
Answer:
The number of collisions per second per unit volume of the reactibn mixture is known as collision frequency (Z).
For a bimolecular elementary reaction
A + B → Products
Where Z_AB represents the collision frequency of reactants A and B. \(\mathrm{e}^{-\mathrm{Ea} / \mathrm{RT}}\) represent the fraction of molecules with energies equal to or greater than E_a.

Question 30.
Draw the graphs between potential energy – reaction coordinates for catalysed and uncatalysed reactions.
Answer:

Effect of catalyst on activation energy

Question 31.
What is the effect of temperature on the rate constant?
Answer:
The rate constant of reaction is almost doubled for every rise of 10° in temperature. The temperature dependence of rate of a
chemical reaction is given by Arrhenius equation
k = A. \(\mathrm{e}^{-\mathrm{Ea} / \mathrm{RT}}\)
A = Arrhenius frequency factor
T = Absolute temperature
R = Gas constant
E_a = Activation energy

Short Answer Questions (4 Marks)

Question 32.
Define average rate of a reaction. How is the rate of reaction expressed in terms of change in the concentration of reactants and products for the following reactions.
1) 2HI (g) → H₂(g) + I₂(g)
2) Hg(l) + Cl₂ (g) → HgCl₂(s)
3) 5Br,sup>- (aq) + \(\mathrm{BrO}_3^{-}(\mathrm{aq})\) + 6H⁺(aq) → 3Br₂(aq) + 3H₂O(l)
Answer:
The average rate of a reaction is defined as the rate of change of concentration per unit time.

Question 33.
What is rate equation? How is it obtained? Write the rate equations for
1) 2NO(g) + O₂(g) → 2NO₂(g)
2) CHCl₃ + Cl₂ → CCl₄ + HCl
3) CH₃COOC₂H₅(l) + H₂O(l) → CH₃COOH(aq) + C₂H₅OH(l)
Answer:
The mathematical expression written in terms of concentration of reactants, which actually influence the rate is called rate law or rate equation.

1) 2NO(g) + O₂(g) → 2NO₂(g)
Rate = k [NO]² [O₂]
2) CHCl₃ + Cl₂ → CCl₄ + HCI
Rate = k [CHCl₃] [Cl₂]^1/2
3) CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH
Rate = k [CH₃COOC₂H₅]¹ [H₂O]⁰

Experimental data shows that the exponents of the concentration terms may be same as their stoichiometric coefficients in the balance chemical equation as in the example (1) or may not be same as in examples (2) and (3).

Question 34.
Define and explain the order of a reaction. How is it obtained experimentally?
Answer:
The sum of the powers of exponents to which the concentration terms are raised in the rate law expression is called order of reaction.
For a hypothetical reaction
aA + bB → Products
The rate equation is Rate = k[A]^x [B]^y
Where x + y = n = order of the reaction. The decrease in the concentration of reactants or the increase in the concentration of products are measured experimentally. These values are substituted in rate constant equations of different order of reactions. The equation that gives constant value gives the order of reaction.

Question 35.
What is “molecularity” of a reaction? How is it different from the ‘order’ of a reaction ? Name one bimolecular and one tri- molecular gaseous reactions. (IPE 14)(Mar. 2018 AP)
Answer:
The number of reacting species (atoms, ions or molecules) taking part in an elementary reaction which must collide simultaneously in order to bring about a chemical reaction is called molecularity of a reaction.

1. Order of a reaction is an experimental quality. It can be zero and even a fraction but molecularity cannot be zero or non – integer.
2. Order is applicable to elementary as well as complex reactions whereas molecularity is applicable only for elementary reactions. For complex reaction molecularity has no meaning.
3. For complex reaction, order is given by the slowest step and molecularity of the slowest step is taken as the order of the overall reaction.

Dissociation of hydrogen iodide into H₂ and I₂ is a bimolecular reaction.
2HI → H₂ + I₂
Formation of NO₂ from NO and O₂ is a trimolecular reaction.
2NO + O₂ → 2NO₂

Question 36.
Derive the integrated rate equation for a zero order reaction.
Answer:
Zero order reaction means that the rate of the reaction is proportional to zero power of the concentration of reactants. Consider the reaction
R → P
Rate = –\(\frac{\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}\) = k[R]°
or Rate = –\(\frac{\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}\) = k × 1( ∵ [R]° = 1)
d[R] = – kdt
Integrating both sides
[R] = – kt +1 (I = integration constant) … (1)
At t = 0, the concentration of the reactant
R₀ = [R]₀, When [R]₀ is initial concentration of the reactant.
Substituting in equation (1)
[R]₀ = -k × 0 + 1
[R]₀ = I
Substituting the value of I in (1)
[R] = – kt + [R]₀ ….. (2)
Further simplifying equation (2) we get the rate constant k as
k = \(\frac{[\mathrm{R}]_0-[\mathrm{R}]}{\mathrm{t}}\)

Question 37.
Derive an Integrated rate equation for a first order reaction.
Answer:
In this type of reactions, the rate of the reaction is proportional to the first power of the concentration of the reactant R. For
example
Rate = –\(\frac{\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}\) = k[R] or \(-\frac{\mathrm{d}[\mathrm{R}]}{\mathrm{R}}\) = kdt
Integrating tuis equation we get
In [R] = -kt + I (I = integration constant) ….. (1)
When t = 0, [R] = [R]₀
∴ ln [R]₀ = -k × 0 + I
ln [R]₀ = I
Substituting the value I in equation (1)
ln[R] = -kt + ln[R]₀ or ln \(\frac{[\mathrm{R}]}{\left[\mathrm{R}_0\right]}\) = -kt
or k = \(\frac{1}{\mathrm{t}} \ln \frac{[\mathrm{R}]_0}{[\mathrm{R}]}\) or k = \(\frac{2.303}{t} \log \frac{[R]_0}{[R]}\)
This is the integrated rate equation for first order reaction.

Question 38.
Derive an integrated rate equation in terms of total pressure [P] and the partial pressures P_A, P_B, P_C for the gaseous reaction
A(g) → B(g) + C(g).
Answer:
The gaseous reaction A(g) → B(g) + C(g)
Let P₁ be the initial pressure of A and b_t the total pressure at time t. Integrated rate equation for such a react Ion can be derived as
Total pressure P_t = P_A + P_B + P_C
P_A, P_B and P_C are the partial pressures of A, B and C respectively.
If x atm be the decrease in pressure of A at time t and one mole each of B and C is being formed, the increase in pressure of B and C will also be x atm each.

When p_i is the intial pressure at time t = 0
P_t = (p_i – x) + x + x = p_i + x
x = p_t – p_i
Where P_A = p_i – x
= p_i – (p_t – p_i)
= 2p_i – p_t
k = \(\frac{2.303}{t}\)log\(\frac{p_i}{p_A}\)
= \(\frac{2.303}{t}\)log\(\frac{p_i}{2 p_i-p_t}\)

Question 39.
What is half-life (t_1/2) of a reaction? Derive the equations for the ‘half-life’ value of zero and first order reactions.
Answer:
The half – life of a reaction is the time in which the concentration of a reactant is reduced to one half of its initial concentra-tion. It is represented as t_½.
For the first order reaction
k = \(\frac{2.303}{t}\) log\(\frac{[\mathrm{R}]_0}{[\mathrm{R}]}\) at t_1/2[R] = \(\frac{[\mathrm{R}]_0}{2}\)
So the above equation becomes

Question 40.
What is Arrhenius equation? Derive an equation which describes the effect of rise of temperature (T) on the rate constant (k) of a reaction.
Answer:
Arrhenius equation explains the temperature dependence of the rate of a chemical reac-tion. Arrhenius equation is
k = Ae^-E_a/RT
Where A is the Arrhenius factor, or the frequency factor. R is gas constant and E_a is activation energy in J mol^-1.
Taking natural logarithms of both sides
In k = –\(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}\) + ln A …… (1)
At temperature T₁, the equation (1) is
In k₁ = –\(\frac{E_a}{R T}\) + ln A …… (2)
At temperature T₂, the equation (1) is
In k₂ = –\(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}\) + ln A …… (3)
Subtracting equation (2) from (3) we get

Question 41.
Discuss the effect of catalyst on the kinetics of a chemical reaction with a suitable diagram.
Answer:
A catalyst is a substance which increase the rate of a reaction without itself undergoing any permanent chemical change.

It is believed that the catalyst provides an alternate pathway or reaction mechanism by reducing the activation energy between reactants and products and hence lowering the potential energy barrier as shown in the figure. Lower the value of activation energy faster will be the rate of reaction.

Question 42.
Describe the salient features of the collision theory of reaction rates of bimolecular reactions. (Mar. 2018-TS)
Answer:
Collision theory is based on kinetic theory of gases. According to this theory

1. The reactant molecules are assumed to be hard spheres and
2. Reaction is postulated to occur when molecules collide with eách other.
3. A reaction occurs on collision of two molecules only if they possess a certain minimum amount of energy in excess of the normal energy of molecules.
4. The minimum energy which molecules must possess before collision should be equal to or greater than the activation energy.
5. The collisions in which the reactant molecules have proper orientation only leads to the formation of products.
6. Whereas Improper orientation makes them bounce back and no products are formed.
7. The collisions in which reactant molecules convert into product molecules are called effective or fruitful collisions.

Question 43.
Explain the terms
a) Activation energy (E_a)
b) ColIsIon frequency (Z)
c) Probability factor (P) with respect to Arrhenius equation.
Answer:
a) Activation energy : According to Arrhenius, a reaction takes place only when reactant molecules collide and form an unstable intermediate which have higher potential energy than reactant or product molecules. The energy required
for the formation of this intermediate or activated complex is called activation energy.

b) Collision frequency : According to collision theory the reactant molecules are assumed to be hard spheres and reaction occurs when molecules collide with each other. The number of collisions per second per unit volume of the reaction mixture is known as collision frequency (Z). For the bimolecular elementary
A + B → Products
Rate = Z_AB\(\mathrm{e}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}\)
Z_AB represents the collision frequency of reactants A and B.

c) Probability factor: To account for effective collisions, a factor known as probability factor or steric factor is introducted. It takes into account the fact that in a collision molecules must be properly oriented.
(ie.,) Rate = P × Z_ABe^Ea/RT

Long Answer Questions

Question 44.
Explain the following terms with suitable examples.
(a) Average rate of a reaction
(b) Slow and fast reactions
(c) Order of a reaction
(d) Molecularity of a reaction
(e) Activation energy of a reaction.
Answer:
a) Average rate of a reaction : The average rate of a reaction is defined as the rate of change of concentration per unit time.

b) Slow and fast reactIons : The reactions which takes place instantaneously are called fast reactions. Ionic reactions are fast reactions e.g. Precipitation of silver chloride occurs instantaneously by mixing aqueous solutions of silver nitrate and sodium cholride.
Some reactions takes place slowly e.g. rusting of iron, invension of sugar, hydrolysis of starch. These reactions are called slow reactions.

c) Order of ReactIon : The sum of the powers or exponents to which the concentration terms are raised in the rate law expression is called order of reaction.
For a hypothetical reaction
aA + bB → Products
The rate equation is Rate = k[A]^x [B]^y
Where x + y = n = order of the reaction.

d) Molecularity of the reaction : The number of reacting species (atoms or ions or molecules) taking part in an elementary reaction which must collide simultaneously in order to bring about a chemical reaction is called molecularity of a reaction.

e) Activation energy of a reaction: Accor-ding to Arrhenius, a reaction takes place only when reactant molecules collide and form an unstable intermediate which have higher potential energy than reac-tant or product molecules. The energy required for the formation of this inter-mediate or activated complex is called activation energy.
E_a = E_T – E_R

Question 45.
Give two examples for each of zero order and first order reactions. Write the equa-tions for the rate of a reaction interms of concentration changes of reactants and products for the following reactions.
1) A(g) + B(g) → C(g) + D(g)
2) A(g) → B(g) + C(g)
3) A(g) + B(g) → C(g)
Answer:
Examples for zero order reaction :

i) The decomposition of gaseous ammonia on a hot platinum surface is a zero order reaction at higher pressure.

At high pressure, the surface of the metal is completely covered with gas molecules, so unable to alter the ammonia on the surface of the catalyst making the rate of reaction independent of its concentration.

ii) Thermal decomposition of HI on gold surface is another example of zero order reaction.

iii) Some enzyme catalysed reactions are zero order reactions.

Examples for first order reactions:

i) Decomposition of N₂O₅.
N₂O₅ → 2NO₂ + \(\frac{1}{2}\)O₂

ii) All natural and artificial radioactive decay of unstable nuclei take place by first order kinetics.
\({ }_{88}^{226} \mathrm{Ra}\) → \({ }_2^4 \mathrm{He}\) + \({ }_{86}^{224} \mathrm{Rn}\)

Rate of reactions:

Question 46.
Discuss the effect of temperature on the rate of a reaction. Derive necessary equations in this context.
Answer:
The rate constant of a reaction increases with increase of temperature. This increase is generally two fold to five-fold for 10° rise in temperature. This is explained on the basis of collision theory. The main points of collision theory are as follows.

1. For a reaction to occur, there must be collisions between the reacting species.
2. Only a certain fraction of total collisions are effective in forming the products.
3. For effective collisions the molecules must possess the sufficient energy equal to or greater than activation energy as well as proper orientation.

On the basis of above conclusions, rate of reaction is given by
Rate = f × z where f is the effective collisions and z is total number of collisions per unit volume per second.

Quatitatively the effect of temperature on the rate of reaction and hence on the rate constant k was proposed by Arrhenlus.
k = Ae^-E_a/RT —– (1)

Where A is a constant called frequency factor, E_a is the energy of activation R is gas constant and T is the absolute temperature.
The factor e^-E_a/RT gives the fraction of molecules having energy equal to or greater than the activation energy, E_a.
Taking logarithms on both sides of equation (1) we get
In k = ln A – \(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}\)
The value of rate constant at temperatures T₁ and T₂ are k₁ and k₂ respectively, then we have
ln k₁ = ln A – \(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}_1}\) …. (2)
ln k₂ = ln A – \(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}_2}\) …. (3)
Subtracting equation (2) from (3) we get

Question 47.
Give a detailed account of the collision theory of reaction rates of bimolecular gaseous reactions. (TS 16: IPE 14)
(TS Mar. 19; (Mar.2018.TS)
Answer:
According to collision theory

1. The reactant molecules are assumed to be hard spheres.
2. Reaction occur only when molecules collide each other.
3. All collisions do not lead to the forma-tion of products.
4. The collisions in which the molecules having threshold energy and proper orientation leads to the formation of products.
5. Such collisions are called effective collisions.
6. The number of collisions per second per unit volume of the reaction mixture is known as collision frequency (Z).
7. Activation energy also effects the rate chemical reactions. For a bimolecular elementary reaction.
   A + B → Products Rate of reaction is
   Rate = Z_AB e^-E_a/RT
   Where Z_AB represents the collision frequency of reactants. A and B and e^-E_a/RT represents the fraction of molecules with energies equal to or greater than E_a.
8. If the colliding molecules have improper orientation they bounce back and no reaction takes place.
9. To account for effective collisions a new factor ‘p called probability or steric factor is introduced
   Rate = PZ_ABe^-E_a/RT
   
10. Thus, in collision theory activation energy and proper orientation of the molecules together determine the criteria for
    an effective collision and hence the rate of a chemical reaction.

Numerical Data Based And Concept Oriented Questions

Question 48.
A reaction is 50% completed in 2 hours and 75% completed In 4 hours. What is the order of the reaction?
Answer:
50% completed means half – life.
So for first order reaction t_1/2 = 0.
or k = \(\frac{0.693}{k}\) = 0.3465 mol hr^-1
In the second experiment 75% completed in 4 hours.
k = \(\frac{2.303}{t} \log \frac{[R]_0}{[R]}\)
[R]₀ = 100, [R] = 25
∴ k = \(\frac{2.303}{4}\)log\(\frac{100}{25}\) = \(\frac{2.303}{4}\) log 4
= \(\frac{2.303}{4}\) × 0.6021 = 0.3466
Since k is constant in both experiments, it is first order reaction.

Question 49.
A reaction has a half-life of 10 minutes. Calculate the rate constant for the first order reaction. (TS ’16)
Answer:
For first order reaction
k = \(\frac{0.693}{t_{1 / 2}}\) = \(\frac{0.693}{10}\) = 0.0693 min^-1

Question 50.
In a first order reaction, the concentration of the reactant is reduced from 0.6 mol/L to 0.2 mol/L m 5 min. Calculate the rate constant (k).
Answer:
Initial concentration [R]₀ = 0.6 moL/L
Concentration after 5mn [R] = 0.2 mol /L
First order rate equation

Question 51.
The rate conant for a zero order reaction in A is 0.0030 mol L^-1 s^-1. How long it will take for the initial concentration of A to fall from 0.10M to 0.075M.
Answer:
For zero order reaction
k = \(\frac{[\mathrm{R}]_0-[\mathrm{R}]}{\mathrm{t}}\) or t = \(\frac{[\mathrm{R}]_0-[\mathrm{R}]}{\mathrm{t}}\)
k = 0.0030 mol L^-1 s^-1
[R]₀ = 0.10M
[R] = 0.075
∴ t = \(\frac{0.10-0.075}{0.0030}\) = 8.33 sec

Question 52.
A first order decomposition reaction takes 40mm for 30% decomposition. Calculate its t_1/2 value.
Answer:

Question 53.
Calculate the half-lIfe of first order reaction whose rate constant is 200 s^-1.
Answer:
For first order reaction
t_1/2 = \(\frac{0.693}{\mathrm{k}}\) = \(\frac{0.693}{200}\) = 0.003465
= 3.465 × 10^-3 s

Question 54.
The thermal decomposition of HCOOH is a first order reaction. The rate constant is 2.4 × 10^-3 s^-1 at a certain temperature. Calculate how long will It take for 3/4 of initial quantity of HCOOH to decompose.
Answer:
Rate constant k = 2.4 × 10^-3s^-1
[R]₀ = 1; [R] = 0.25 ;

Question 55.
The decomposition of a compound is found to follow first order rate law. If it takes 15 minutes for 20% of original material to
react, calculate the rate constant.
Answer:
For a first order reaction
k = \(\frac{2.303}{t} \log \frac{[R]_0}{[R]}\)
[R]₀ = 100; [R] =80; t = 15
k = \(\frac{2.303}{15}\)log\(\frac{100}{80}\) = \(\frac{2.303}{15}\) log 1.25
= \(\frac{2.303}{15}\) × 0.0963 = 0.0148 min^-1

Question 56.
In a pseudo first order hydrolysis of ester in water, the following results are obtained

Calculate the average rate of reaction between the time Interval 30 to 60 s.
Answer:
i) Average rate of reaction between interval time 30 to 60 seconds is given by
Average rate = \(\frac{\Delta \mathrm{x}}{\Delta \mathrm{t}}\) = \(\frac{\mathrm{C}_2-\mathrm{C}_1}{\Delta \mathrm{t}}\)
= \(\frac{0.17-0.31}{60}\) = \(-\frac{0.14}{30}\)
= -0.00467 = -4.67 × 10^-3 Ms^-1
Minus sign indicate that the rate of reaction is decreasing with time as concentration of ester is decreasing with time.

ii)

Question 57.
The half – life for a first order reaction is 5 × 10^-6 s. What percentage of the Initial reactant will react in 2 hours?
Answer:
Given that
Half-life for the first order reaction = 5 × 10^-6 s
First order rate constant, k = \(\frac{0.693}{5 \times 10^{-6} \mathrm{~s}}\)

Question 58.
H₂O₂ (aq) decomposes to H₂O(l) and O₂(g) in a first order reaction w.r.t. H₂O₂. The rate constant is k = 1.06 × 10^-3 min^-1. How long it will take 15% of the sample to de-compose ?
Answer:
Given that
Rate constant k = 1.06 × 10^-3 min^-1
Initial concentration of reactant [R]₀ = 100
Concentration of reactant at time
t[R]_t = 100 – 15 = 85
First order rate constant
k = \(\frac{2.303}{t} \log \frac{\left[R_o\right]}{[R]}\);
t = \(\frac{2.303}{1.06 \times 10^{-3}} \log \frac{100}{85}\) ; t = 153 min

Question 59.
Show that in the case of first order reaction, the time required for 9999% completion of the reaction is 10 times that required for 50% completion. (log 2 = 3010)
Answer:
When the reaction has completed 99.9%

Question 60.
The rate constant of a reaction is doubled when the temperature is raised from 298K to 308K. Calculate the activation energy.
Answer:

Question 61.
The first order rate constant k for the reaction C₆H₅I(g) → C₂H₄(g) + HI(g) at 600K is 1.60 × 10^-5s^-1. The energy of activation is 209 kJ mol. Calculate k at 700 K.
Answer:
Given that
Rate constant at 600 K = 1.60 × 10^-5 s^-1
Rate constant at 700 K =?
R = 8.314 J kJ mol^-1
T₁ = 600K T₂ = 700K
Activation energy E_a = 209 kJ mol^-1 = 209000 J

Question 62.
The activation energy for the reaction 2HI(g) → H₂(g) + I₂(g) at 581K is 209.5 kJ/mol. Calculate the fraction of molecules having energy equal to or greater than activation energy. (R = 8.31 JK^-1 mol^-1]
Answer:
Temperature T = 581K
Activation energy E_a = 209.5 JK mol^-1
R = 8.314 JK^-1 mol^-1

Question 63.
For the reaction R → p, the concentration of a reactant changes from 0.03M to 0.02M in 25 minutes. Calculate the average
rate of reaction using units seconds.
Answer:
Given that
[R]₀ = 0.03M
[R] = 0.02M
[R]₀ – [R] = 0.03 – 0.02 = 0.01M
Time = 25 mm = 25 × 60 sec.
Average rate of reaction = \(\frac{[R]_0-[R]}{t}\)
= \(\frac{0.01}{25 \times 60}\) = 6.66 × 10^-6 Ms^-1

Question 64.
In a reaction 2A → Products, the concentration of A decreases from 0.5 mol L^-1 to 0.4 moL^-1 in 10 minutes. Calculate the rate during this Interval.
Answer:
2A → Products
Rate \(\frac{\mathrm{dx}}{\mathrm{dt}}\) = \(-\frac{1}{2} \frac{\Delta[\mathrm{A}]}{\Delta \mathrm{t}}\)
Δ[A] = 0.5 – 0.4 = 0.1 mol^-1
Rate = \(-\frac{1}{2} \times \frac{0.1}{10}\) = 5 × 10^-3 mol^-1 L^-1 min^-1

Question 65.
For a reaction, A + B → product : the rate law is given by r = k[A]^1/2[B]². What is the order of reaction?
Answer:
Order of reaction Is the sum of the powers of concentration terms in rate equation.
Rate equation r = k[A]^1/2 [B]²
∴ order = 0.5 + 2 = 2.5

Question 66.
The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased by three times, how will it affect the rate of formation of Y.
Answer:
Since the reaction follows second order kinetics r = k(x)²
x increases by 3 times
∴ r = k(3)²
So rate increases by nine times.

Question 67.
A first order reaction has a rate constant 1.15 × 10^-3s^-1. How long will 5g of this reactant take to reduce to 3g ?
Answer:
Rate constant k = 1.15 × 10^-3s^-1
Initial cone. [R]₀ = 5
Conc. at time t [R] = 3

Question 68.
Time required to decompose SO₂Cl₂ to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.
Answer:
For first order reaction
k = \(\frac{0.693}{t_{1 / 2}}\) = \(\frac{0.693}{60 \times 60}\) = 1.925 × 10^-4 sec^-1
(or) 1.155 × 10^-2 min^-1

Question 69.
From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
i) 3NO(g) → N₂O(g)
Rate = k[NO]²
ii) H₂O₂(aq) + 3I^–(aq) + 2H⁺ → 2H₂O(l) + \(I_3^{-}\)
Rate = k[H₂O₂] [I^–]
iii) CH₃CHO(g) → CH₄(g) + CO(g)
Rae = k[CH₃CHO]^3/2
iv) C₂H₅Cl (g) → C₂H₄ (g) + HCl(g)
Rate = k[C₂H₅Cl]
Answer:
i) 3NO(g) → N₂O (g)
Rate = k[NO]²
Order w.r.t NO = 2
Overall order = 2
Units of k = (mol L^-1)^1-2 s^-1
= mol^-1 L s^-1

ii) H₂O₂ + 3l^– + 2H⁺ → 2H₂O + \(I_3^{-}(\mathrm{aq})\)
rate = k[H₂O₂] [I^–]
Order w.r.t. H₂O₂ = 1
Order w.r.t. I^– = 1
Overall order =2
Units of k = (mol L^-1)^1-2 s^-1
= mol^-1 L s^-1

iii) CH₃ CHO(g) → CH₄(g) + CO(g)
rate = k[CH₃CHO]^3/2
Order w.r.t. CH₃CHO = 1.5
Overall order = 1.5
Units of k = (mol L^-1)^1-1.5s^-1
= mol^1/2 L^1/2 s^-1

iv) C₂H₅Cl(g) → C₂H₄(g) + HCl (g)
rate = k[C₂H₅Cl]
order w.r.t. C₂H₅Cl = 1
overall order = 1
Units of k = (mol L^-1)^1-1 s^-1
= s^-1

Question 70.
For the reaction 2A + B → AB, the rate = k[A][B]² with k = 2.0 × 10⁶ v mol^-2 L²s^-1. Calculate the Initial rate of the reaction when [A] = 0.1mol L^-1, [B] = 0.2 mol L^-1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L^-1.
Answer:
Given that
k = 2.0 × 10^-6 mol^-2 L² s^-1
[A] = 0.1 mol L^-1
[B] = 0.2 mol L^-1

i) Initial rate = k[A] [B]²
= 2 × 10^-6 × 0.1 × (0.2)²
∴ Rate of reaction = 8 × 10^-9 mol L^-1sec^-1

ii) 2A + B → AB

r = 2 × 10^-6 × 0.06 × (0.18)²
∴ Rate of reaction = 3.89 × 10^-9 mol L^-1sec^-1

Question 71.
The decomposition of NH₃ on platinum surface is zero order reaction. What are the rates of production of N₂ and H₂ if k = 2.5 × 10^-4 mol^-1 Ls^-1.
Answer:
Since the decomposition of NH3 on platinum surface is zero order.
So rate of reaction k = 2.5 × 10^-4 mol L^-1 sec^-1.
2NH₃ → N₂ + 3H₂
\(\frac{\mathrm{dx}}{\mathrm{dt}}\) = \(\frac{\mathrm{dN}_2}{\mathrm{dt}}\) = \(\frac{1}{3} \frac{\mathrm{dH}_2}{\mathrm{dt}}\)
Rate production of N₂ = \(\frac{\mathrm{dN}_2}{\mathrm{dt}}\) = \(\frac{\mathrm{dx}}{\mathrm{dt}}\)
= 2.5 × 10^-4 mol^-1 L^-1 sec^-1
Rate production of H₂ = \(\frac{\mathrm{dH}_2}{\mathrm{dt}}\) = \(\frac{3 \mathrm{dx}}{\mathrm{dt}}\)
= 7.5 × 10^-4 mol L^-1 sec^-1

Question 72.
The rate expression for the decomposition of dimethyl ether in terms of partial pressures is given as Rate = k(pCH₃ O CH₃)^3/2. If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constant ?
Answer:
Units of rate = bar min^-1
r = k p^3/2
bar min^-1 = k (bar)^3/2
∴ k = bar^-1/2 min^-1

Question 73.
A reaction is second order with respect to a reactant. How is the rate of reaction is affected if the concentration of the reactant is
(i) doubled
(ii) reduced to half.
Answer:
Ratio of second order reaction = k[Reactant]²
When the concentration is doubled Rate = k[2]²
∴ Rate increases by 4 times.
When concentration is reduced to half.
rate = k\(\left[\frac{1}{2}\right]^2\)
∴ Rate decreases to \(\frac{1}{4}\) times.

Question 74.
A reaction is first order in A and second order in B.
i) Write the differential rate equation,
ii) How is the rate affected on increasing the concentration of B three times ?
iii) How is the rate affected when the concentration of both A and B are doubled ?
Answer:
The reaction is first order in A and second order in B.
∴ The differential rate equation r = k [A]¹ [B]²
When the concentration of B is increased three times r = k[A][B]²
∴ rate increases by 9 times
When the concentration of both A and B are doubled
r = k[2][2]²
∴ Rate increases by 8 times.

Question 75.
In a reaction between A and B, the initial rate of reaction (r₀) was measured for different initial concentrations of A and B as given below :

What is the order of the reaction with respect to A and B ?
Answer:
Rate of reaction r = kA^m Bⁿ
5.07 × 10^-5 = k(0.2)^m (0.3)ⁿ
5.07 × 10^-5 = k(0.2)^m (0.1)ⁿ
1.43 × 10^-5 = k(0.4)^m (0.05)ⁿ
∴ order of reaction with respect to A = 1.5
∴ r = k A^1.5 B⁰
order of reaction with respect to B = 0

Question 76.
The following results have been obtained during the kinetic studies of the reaction:

Determine the rate law and rate constant for the reaction.
Answer:
When [A] = [B] the intial rate of formation of D is 6.0 × 10^-3 mol L^-1 min^-1
From the experiment no.4 when the concentration of A is increased four times keeping the concentration B. Constant the rate increases by 4 times i.e.
6.0 × 10^-3 × 4 = 2.4 × 10^-2
So the order of reaction with respect to A is 1
From the experiments 2 and 3, when the concentration of B is doubled keeping the concentration of A constant the rate increases by 4 times i.e.,
7.2 × 10^-2 × 4 = 2.88 × 10
∴ The order of reaction w.r.t B = 2
Rate law = k[A][B]²
Rate = k[A][B]²
6.0 × 10^-3 = k[0.1] [0.1]²
6.0 × 10^-3 = k × 1 × 10^-3
k = \(\frac{6.0 \times 10^{-3}}{10^{-3}}\) = 6.0 Mol^-1 L² min^-1

Question 77.
The rate constant for a first order is 60 s^-1. How much time will It take to reduce the initial concentration of the reactant to its 1/16th value.
Answer:
Rate constant = 60 s^-1
For first order reaction
t_1/2 = \(\frac{0.693}{k}\) = \(\frac{0.693}{60}\) = 0.01155 sec.
t = 4t_1/2 = 4 × 0.01155 = 4.62 × 10^-2 sec.

Question 78.
For a first order reaction, show that the time required for 99% completion is twice the time required for completion of 90% of reaction.
Answer:

∴ The time required for 99% completion is twice the time required for completion of 90% reaction.

Question 79.
For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.

Answer:

Question 80.
The following data were obtained during the first order thermal decomposition of SO₂Cl₂ at a constant volume.
SO₂Cl₂(g) → SO₂ (g) + Cl₂ (g)

Calculate the rate of reaction when total pressure is 0.65 atm.
Answer:

Pressure of SO₂Cl₂ = 0.35
∴ r = k[SO₂Cl₂] = (2.23 × 10^-3) (0.35)
= 7.8 × 10^-4 atm sec^-1

Question 81.
The rate constant for the decomposition of hydrocarbons is 2.418 × 10^-5 s^-1 at 546K. If the energy of activation is 179.9 kJ / mol. What will be the value of pre – exponential factor?
Answer:
Rate constant for the decomposition of hydrocarbon = 2.418 × 10^-5s^-1
Temperature T = 546 K
Energy of activation E_a = 179.9 kJ / mol.
The value of pre-exponential factor ?
k = Ae^-E_a/RT
2.418 × 10^-5 = A.e^{-179.103/8.314 × 546}
∴ A = 3.9 × 10¹² sec^-1.

Question 82.
Consider a certain reaction A → Products with k = 2.0 × 10^-2 s^-1. Calculate the concentration of A remaining after 100s if the initial concentration of A is 1.0 mol L^-1.
Answer:
Rate constant k = 2.0 × 10^-2 sec^-1
Initial concentration [R]₀ = 1.0 mol L^-1
Time t = 100 s
Concentration after time T = [R] =?
k = \(\frac{2.303}{t} \log \frac{[R]_0}{[R]}\)
2 × 10² = \(\frac{2.303}{100} \log \frac{1}{1-x}\)
log \(\frac{1}{1-x}\) = 0.868
∴ 1 – x = 0.135M

Question 83.
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t_1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours ?
Answer:
Decomposition of sucrose in acid solution follows first order kinetics
Half – life t_1/2 = 3.00 hours
For first order
k = \(\frac{0.693}{3} \mathrm{hr}^{-1}\)
∴ \(\frac{0.693}{3}\) = \(\frac{2.303}{8} \log \frac{1}{1-x}\)
1 – x = 0.157

Question 84.
The decomposition of hydrocarbon follows the equation
K = (4.5 × 10¹¹ s^-1) e^-28000k/T. Calculate E_a.
Answer:
K = Ae^-E_a/RT
\(\frac{E_a}{R}\) = 28000
E_a = 232.79 kJ/mole

Question 85.
The rate constant for the first order de-composition of H₂O₂ is given by the following equation: log k = 14.34 – 1.25 × 10⁴ K/T. Calculate E_a for this reaction and at what a temperature will its half – life period be 256 minutes ?
Answer:

Question 86.
The decomposition of A into product has value of k as 4.5 × 10³ S^-1 at 10°C and energy of activation 60 kJ mol^-1. At what temperature would k be 1.5 × 10⁴ s^-1?
Answer:

Question 87.
The time required for 10% completion of a first order reaction at 298 K is equal to that required for Its 25% completion at 308K. If the value of A is 4 × 10¹⁰ s^-1, cal-culate k at 318 K and E_a.
Answer:

Question 88.
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation
of the reaction assuming that it does not change with temperature.
Answer:

Intext Questions – Answerš

Question 1.
For the reaction R → P, the concentration of a reactant changes from 003M to 002M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
Answer:
The average rate of reaction will be
= \(-\frac{\Delta R}{\Delta t}\) = –\(\frac{[R]_2-[R]_1}{t_2-t_1}\)
Now [R]₂ = 0.02M, [R]₁ = 0.03M, Δt = 25mm
∴Average rate = –\(\frac{0.02-0.03}{25}\) = –\(\frac{(-0.01)}{25}\)
= 4 × 10^-4M min^-1
= \(-\frac{(0.01)}{25 \times 60}\) = 6.66 × 10^-6Ms^-1

Question 2.
In a reaction 2A → products, the concentration of A decreases from 0.5 mol L^-1 to 0.4 mol L^-1 in 10 minutes. Calculate the rate during this Interval.
Answer:
Average rate = \(-\frac{1}{2} \frac{0.4-0.5}{10}\) = \(-\frac{1}{2} \frac{(-0.1)}{10}\)
= 5 × 10^-3 M min^-1

Question 3.
For a reaction A + B → product; the rate law is given by, r = k[A]^1/2 [B]². What is the order of the reaction?
Answer:
The order of reaction ¡s sum of powers of concentration terms.
Order = \(\frac{1}{2}\) + 2 = 2.5

Question 4.
The conversion of molecules x to y follows second order kinetics. If concentration of x is increased to threetlmes, how will it
effect the rate of formation of y?
Answer:
The reaction x → y follows second order Kinetics. Therefore, the rate equation for this reaction will be Rate = K[x]²
Let [x] = a mol L^-1, then equation (1) can be written as
Rate = K. (a)² = Ka²
If the concentration of x is Increased to three times then [x] = 3a mol L^–
Now the rate equation will be
Rate = K(3a)² = 9(Ka)²
Hence the rate of formation will increase by 9 times.

Question 5.
A first order reaction has a rate constant 1.15 × 10^-3s^-1. How long will 5gm of this reactant take to reduce 3g?
Answer:
Given, Initial amount = 5g
Rate constant = 1.15 × 10^-3 s^-1
We know that for a 1st order reaction

Question 6.
Time required to decompose SO₂Cl₂ to half of its initial amount is 60 minutes. If the decomposition is a first order reaction. Calculate the rate constant.
Answer:
We know that for a ¡st order reaction

Question 7.
What will be the effect of temperature on rate constant?
Answer:
The rate constant of reaction is almost doubled for every rise of 100 in temperature. The temperature dependence of rate of a chemical reaçtion is given by Arrhenius equation
k = A.e^-Ea/RT
A = Arrhenius frequency factor
T = Absolute temperature
R = Gas constant
E_a = Activation energy

Question 8.
The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298 K. Calculate E_a.
Answer:
Given that T₁ = 298 K
T₂ = 298 + 10 = 308K
The rate of reaction will be doubled when temperature is raised by 100.
∴If the value of k₁ = k; k₂ = 2k
Substituting these values in Arhenious equation

Question 9.
The activation energy for the reaction
2HI (g) → H₂(g) + I₂(g) is 209.5kJ mol^-1 at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.
Answer:
Here
E_a = 209.5kJ mol^-1 = 209500 J mol^-1
T = 581K; R = 8.314 JK^-1 mol^-1
The fraction of molecules having energy equal or more than activation energy is given by

Now x = Anti log of 18.8323
= Anti log 19.1677 = 1.471 × 10^-19

Telangana TSBIE TS Inter 2nd Year English Study Material Grammar Curriculum Vitae Exercise Questions and Answers.

TS Inter 2nd Year English Grammar Curriculum Vitae

Q.No.15  (1 ×4 = 4 Marks)

Curriculum Vitae (CV) presents the personal and academic details of an individual to the prospective employer in a systematically organised form. Slight variations in the format entitle this account to be called ‘Resume’ / ‘rezjumei /( ‘Biodata’ or ‘Academic Vitae’. Whatever be the title, presenting it properly paves the way for a prosperous future, for it initiates the process of selection of candidates. Hence, it is very important, from the real life point of view’, for the students to learn to prepare the CV as impressively as possible.

It is equally important from the examination point of view too, for the student can score all the allotted five marks with a little bit of care and practice. The question may be asked in two different ways. You may be asked to respond to a job notification. Then, you will have to ‘imagine’ the information required for the notified job and prepare the CV with that information. This is a little demanding.

The other way is to ask you to prepare the CV with the information provided byway of hints. This is relatively easier. Either way, you must know the format well. Any format, written correctly, fetches you. full marks. Choose the one you feel comfortable with.

Examine the following model carefully.
RESUME OF VASUDEVARAO MANNEM
VASUDEVARAO MANNEM
1-1-32, New World Colony,
Happy Society, Gandhi marg,
Hyderabad 512345
Mobile: 91-9999223444
[email protected]

OBJECTIVE
A position as Project Manager in the information technology Industry.

2006-2010
B.Tech in Electronics and Communication Engineering ABC Engineering College, Karimnagar, 78% aggregate.

2004-2006
Intermediate Board of Intermediate Education.
DFE College, Karimnagar. 73.4% aggregate.

2004
SSC. Board of Secondary Education.
GIH School, Karimnagar,76% aggregate.

TECHNICAL SKILLS
Language : C, C++
DBMS Packages : Oracle 8i
Operating System : MS DOS, Windows 98, NT, 2000, XP, 7
Web Designing : JAVA, HTML, XML

PROFILE

• Higly energetic, growth oriented individual seeking to establisha career in the Information Technology industry.
• Ability to take on challenges, work under pressure dedication towards work.

STRENGTHS

• Good Communicator
• Team Player
• Problem solver

ADDITION AI INFORMATION

Language Known : English, Telugu and Hindi
Interest : Listening to Music, Playing Chess
Age : 22 years
Father’s name : Sri. Mannem Ramarao
References : Available on Request

I. Here is a Sample Resume written by Sunil Kumar Yemula.

SUNIL KUMAR VEMULA
ADDRESS : H.No. 2-1-446/2A, Sharma Apartments,
New Nallakunta, Hyderabad – 500044
Ph: 9999955555
E-mail: [email protected]

CAREER OBJECTIVE
To be a professional who can make a qualitative difference with an esteemed organization, where high competence and skill is rewarded.

EDUCATION

• Discontinued Bachelor of Commerce from Masterminds College affiliated to Nagarjuna University, Guntur, in March 2021 due to Covid pandemic
• Discontinued C A from Masterminds College affiliated to Nagarjuna University, Guntur, in March 2021 due to Covid pandemic
• Intermediate (Mathematics, Economics, Commerce), Board of Intermediate Education, GJC, Kacheguda, TS in March 2020
• SSC, Board of Secondary Education from GNR Model School, Malakpet, 2018

TECHNICAL. OTHER KEY SKILLS

• Working knowledge of computers
• Quick learner – Learn things very quickly and adopt it in understandable manners
• Quick Driven – Focus more on quality of work and strive very hard to achieve it.
• Logical thinker – Always solve the problem with a best logical solution
• Communication Skills – Fluency in English and Telugu logical solution

HOBBIES :

• Singing – Won awards in singing competition during schooling
• Modern arts – Earn money by modern arts (by using discarded material)

LANGUAGES KNOWN : English and Telugu
PERSONAL PROFILE : Will be provided on request
REFERENCES : Will be provided on request

Place : HYDERABAD
Sunil Kumar Vemula

II. Look at the call for News Anchor for a television channel advertised in job search portal https://in.indeed.com. Prepare a CV in response.

NEWS ANCHOR
Nakshatra Media
Hyderabad, Telangana.
₹ 9,678 – ₹ 27,947 a month

A. Apply securely with Resume

• Responsive employer
• Organizing the breaking news in a way that it presentsd the most interesting aspects first.
• Meeting up the reporters, news directors and fellow news anchors to..

Here is a CV prepared by Manoja in response to the advertisement.

III. Here is a CV prepared by Manoja in response to the advertisement.

MANOJA KSSP
ADDRESS : Flat no. 402, Nagarjuna Apartments, Ramalayam Road,
Ramgiri, Nalgonda. Telangana- 508001
New Nallakunta, Hyderabad – 500044
Ph: 9XXXXXXXX8
E-mail: [email protected]

CAREER OBJECTIVE
To be a professional who can make a qualitative difference with an esteemed organization, where high competence and skill is rewarded.

EDUCATION:

TECHNICAL SKILLS

• Telugu Typing, Computer skills
• Communication Skills – Fluency in English and Telugu.

WORK EXPERIENCE:

• Worked as News presenter for City Cable Nalgonda, Summer 2021
• Covered Covid hit areas and interviewed officialsand did live coverage

HOBBIES:

• Singing – Won awards in singing competition during schooling
• Making short films – on mobile

Languages Known : Telugu, Hindi and English
Personal Profile : Will be provided on request
References : 1. Dr. Ghanshyam, Principal, GDC(W) Nalgonda
2. Venu Madhavi, Program Executive, City Cable, Nalgonda.

Place : HYDERABAD
KSSP Manoja

IV. Let’s look at a job advertisement that was published in a newspaper.

Join Swiggy family as a Part-Time or Full-Time Food Delivery Executive Who can be an ideal Rider/Driver/Delivery Boy/Logistics/Admin/Delivery Executive for Swiggy?

Urgently hiring-Multiple vacancies – Weekends/Weekday options
Freshers also can apply. Prefer candidate above 18 years of age and can ridea two¬wheeler, 10th, 10th Pass, 12th Pass, students/candidates looking to work part-time, in shifts can visit our website and post your resume for the following positions:

• Delivery rider, Delivery Boys / Delivery Executive, * Back Office Executive, * Loading, Unloading, * Delivery Operations,
• Call Centre Executive, Collection agents, * Office Assistants, * Data Entry, * Field Sales Job, * Logistics *Helper * Night Shift Partime Helper

Here is a Resume prepared by Senthil Mammutty in response to the advertisement.

Senthil Mammutty
134 Swagath Hotel Street
S.D road, Secunderabad, T.S
(040)2xxxxx42
[email protected]

OBJECTIVE
To obtain an operations executive position in food retail industry

EDUCATION
NIFT, B.Des (Bachelor of Design), Madapur, Discontinued in fisal year.

EXPERIENCE
Food Service Worker, McDonald’s Restaurant 2020-Present, Madapur, Hyderabad.
Responsibilities: To provide excellent customer service, operate cash registers, maintain the work area clean and neat, prepare food and refill lobby items as necessary

AWARDS
Best Delivery executive 4.5 star ratings 2021

ACTIVITIES
Debates, Theatre at Junior College 2018-19
Participated in local and state level debate/drama competitions Won ThirdPrize in the District level debate conference, November 2018

VOLUNTEER
Red Ribbon Club, Red Cross Centre- Summer 2021 Banjara Hills, Hyderabad Performed office work such as data processing, filling, and answering phones; provided child-care services and general help as necessary.

REFERENCES
Available upon request

Place: SECUNDERABAD
Senthil Mammutty

V. Look at this advetisement on the job portal https://naukri.com.

Wanted Social Media Manager for “Swecchha” NGO. We are hiring Full time / Permanent / Part time social media managers, who will be managing all social media platforms of the organization Facebook, Instagram, twitter, Iinkedln, YouTube etc. The person will be responsible for all end to end social media actioities. Any Graduate can apply.

Key Skills required: Social Media, Social Networking, Social Media Content Marketing NGO Management, Digital Marketing, Social Media Marketing, Marketing, Social Service.
Here is a functional CV that Vaishnav has prepared.

Gottumukkala Vaishnav
ADDRESS: Flat #108, VImala Nivas, Opp. Community Hall,
Huzurabad, Karimnagar
Ph : +91 70 XXXXXXX
E-mail id : [email protected]

CAREER OBJECTS
To work in an organization that provides challenges and provides learning opportunities to hone amy skills and gives scope for serving society.

EDUCATION
Pursuing Masters in Social Work (Distance Mode)
Year of completion of Degree 2024 completed BBA from GDC, Warangal, KU with CGPA 7.5, Grade A 2022

TECHINAL SKILLS
‘Networking-Multimedia’ course, Bengaluru 2023 Web Programming, web designing, multimedia tools

WORK EXPERIENCE
Handled GDC Warangal official website, social media admin of Facebook, Instagram, Twitter, Telegram, WhatsApp You-Tube Channel, Blog sites.

WORK EXPERIENCE
Workship Resource person to Tharuni’ NGO to train girl students in developing self confidence 2019-20. Conducted workshops for Ru jroo’s Agaaz-e-Baatcheet, an Interfaith programme for College students sponsored by The British Deputy High Commission, India 2018 -2020 Associated with MAVA (Men Against Violence Against Women) for their awareness campaigns in colleges 2018 – 2019 Have been an admin of social media campaigns of Sneha, Mitr.

Additonal Information

Languages known : English, Telugu, Hindi and French
Interests : Rock Climbing. Rappelling, Photography
References :

1. Dr. Mamatha Raghveer, Founder Tharuni Warangal
2. Principal, GDC, Warangal

Excercise

Preparea Curriculum Vitae/Resume/Bio data in response to the advertisements published in Posters. News papers and College groups.

Question 1.

Answer:
For the post of Asst. Professor: Global Institute of Engineering & Technology Chilkur
CV of Ramarao Kokkiragadda

OBJECTIVE
Looking for a position that provides opportunities to mould budding geniuses into flowering personalities of Mother India.

DETAILS

NAME : Ramarao Kokkiragadda
FATHER : Vijay Kumar K
DATE OF BIRTH : 17. 08.1996
ADDRESS : 3-2-19-A, Masjid Street, Vaheb Nagar Armoor – 503224, Nizamabad (dt)
Mobile: 9986786324
Mail Id: [email protected]

EDUCATIONAL QUALIFICATIONS:

Technical Skills
PPT Designer Certificate

Work Experience
Lab (English Language) Incharge – One year JNTU Campus, Hyderabad

Languages Known : Urdu, Hindi, Telugu, English

References

1. Prof. Isaac Saqueria
   – Dept, of ELT. OU Campus
2. Prof. Anwar Pasha
   – Dept, of English JNTU, Hyderabad

10.06.2021
Armoor
Signature

Question 2.

Answer:
BIO-DATA of Gongati Deepak
Name  :  CHARAN GANTA
Father’s Name  : VENKATESWARA RAO GANTA
Date of Birth and Age  :  17.11.1995, Age 27 years
Full Address  :  11/C, Sri Ramakrishna Complex,
Vidyanagar Colony, Adilabad – 504001
+919848546666
//[email protected]//

Academic Qualifications

Experience
5 years as Co-ordinator, Sales and Marketing Dept., Solar Industries limited

Language known
English, Hindi, Urdu, Telugu and French

Place: Adilabad
Date : 14 May 2022

Signature

Question 3.

Answer:
A. BIO DATA of SRIMUKHI B

Name  :  SRI LAKSHMI BADETI
Father’s Name  :  KRISHNA RAO
Date of Birth and Age  :  26.11.1995
Full Address  : H.No. 3114/5/2-A CHRUCH ROAD
Near Post Office, Siddipet, Medak (Dist)
+91 9848486854
email id: [email protected]

Academic Qualifications

Experience : Three years part-time – a private school.

Language known : Teiugu, English, Hindi

Place: Siddipet
Date : 14 May 2022

Signature

Question 4.

Answer:
For the post of Graphic Designer
ATMAKURI RADHAKRISHNA
F/o A. Srinivas
DOB : 16.08.1998
H.No.: 14-11-A-Z
Main Road
MAHABUBABAD – 506101
MOBILE : 91 – 8323435678
Mail ID : [email protected]

Objective
A POSITION that opens greener avenues for enchanting graphics for ad and entertainment industry

Experience
One year – Graphic Designer Creative Commercials

Education Qualifications

Technical Skills
C++, PYTHON, JAVA

Languages Known
Telugu, English, Hindi

Place : Mahabubabad
Date: 29-06-2021

Signature

Question 5.

Answer:
For the post of Radio Jockey in an eminent Entertainment Service Provider Firm
Resume of ALTAF MD

ALTAF MD
F/o. Md. Jani
DOB: 06.07.1998
H.No.: 13-14-27
Muzaphar Street
Tirumalagiri – 508202
Suryapet (Dt)

Objective
Aspiring to join a firm that enhances my confidence and skills to be able to sell icecream on Mount Everest and water heaters in That Desert.

Experience
Part time PRO of Corporate colleges 3 years with excellent record for my communication skills

Educational Qualifications :

Technical Skills
Capable of creating educational games software, entertainment with musical voice

Languages Known
Urdu, Hindi, Telugu, English

Place: Tirumalagiri
Date: 24-04-2021

Signature

Question 6.

Answer:
Bio- data of M. Praveen

Name : M PRAVEEN
Father’s Name : VENKATA PRASAD
Date of Birth and Age : 13. 06.1998, Age 24 years
Full Address : 12-83, Kali Temple Road
Near Market Yard, Narayanpet
Mahabubnagar Dt.
7842614856
[email protected]

Academic Qualifications

Experience
2 years Interior Designer Dhruva Architects Allied Book Services, Mahabubnagar

Language known
Telugu, English,Hindi Urdu, Kannada

Place: Narayanapet
Date: 13.06.2022

Signature

Question 7.

Answer:
CV of Jani Mohammad
Objective: A POSITION IN IT
DETAILS:

Name : JANI MOHAMMAD
Father’s Name : IBRAHIM MOHAMMAD
Date of Birth and Age : 22.11.1998,
Full Address : H.No. 22/14-2/A
Bakal wadi, Old Town,
NIZAMABAD ‘
Mobile: 9785631257
e-mail:[email protected]

Academic Qualifications

Experience
Two years – IT Professional (wireless) Softech Solutions – Hyderabad

Language known
Telugu, English,Hindi

Place: Nizamabad
Date : 05.07.2022

Signature

Students must practice these Maths 2B Important Questions TS Inter Second Year Maths 2B Differential Equations Important Questions Long Answer Type to help strengthen their preparations for exams.

TS Inter Second Year Maths 2B Differential Equations Important Questions Long Answer Type

Model III – Problems on non-homogeneous D.E

Question 1.
Solve (2x + y + 1) dx + (4x + 2y – 1) dy = 0. [(TS) Mar. ’15]
Solution:

Question 2.
Solve the differential equation \(\frac{d y}{d x}=\frac{x+2 y+1}{2 x+4 y+3}\). [Mar. ’19 (TS)]
Solution:
Given differential equation is \(\frac{d y}{d x}=\frac{x+2 y+1}{2 x+4 y+3}\) ……(1)

4v + log (4v + 5) = 8x + 8c
4(x + 2y) – 8x + log [4(x + 2y) + 5] = c
4x + 8y – 8x + log(4x + 8y + 5) = c
8y – 4x + log (4x + 8y + 5) = c

Question 3.
Solve \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}-\mathrm{y}+3}{2 \mathrm{x}-2 \mathrm{y}+5}\). [(AP) May ’15]
Solution:

Question 4.
Solve \(\frac{d y}{d x}=\frac{4 x+6 y+5}{3 y+2 x+4}\). [(TS) May ’19]
Solution:



8z + 9 log|8z + 23| = 64x + c
8(2x + 3y) + 9 log|8(2x + 3y) + 23| = 64x + c
24y + 9 log|16x + 24y + 23| = 48x + c
Which is the required solution.

Question 5.
Solve \(\frac{d y}{d x}=\frac{3 y-7 x+7}{3 x-7 y-3}\)
Solution:

Question 6.
Solve \(\frac{d y}{d x}=\frac{x+2 y+3}{2 x+3 y+4}\)
Solution:

Question 7.
Solve the differential equation (x – y) dy = (x + y + 1) dx. [Mar. ’17 (TS)]
Solution:

Question 8.
Solve (2x + y + 3) dx = (2y + x + 1) dy
Solution:

Question 9.
Solve \(\frac{d y}{d x}=\frac{6 x+5 y-7}{2 x+18 y-14}\)
Solution:

Question 10.
Find the solution of the equation x(x – 2) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 2(x – 1)y = x³(x – 2), which satisfies the condition that y = 9, when x = 3. [(TS) May ’17]
Solution:

Question 11.
Form the differential equation corresponding to the family of circles of radius r given by (x – a)² + (y – b)² = r², where ‘a’ and ‘b’ are parameters.
Solution:
Given equation is (x – a)² + (y – b)² = r² ………(1)
a, b are parameters.
differentiating (1) w.r.t. ‘x’ on both sides we get
2(x – a)(1 – 0) + 2(y – b)(\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – 0) = 0
(x – a) + (y – b)(\(\frac{\mathrm{dy}}{\mathrm{dx}}\)) = 0 ……..(2)
Again differentiating (2) w.r.t ‘x’ on both sides, we get



Which is the required differential equation.

Question 12.
Form the differential equation from y = c(x – c)², where ‘c’ is an arbitrary constant. [(TS) Mar. ’16]
Solution:
Given y = c(x – c)² ………(1)
(where ‘c’ is an arbitrary constant)
Diff. with respect to x on both sides

Telangana TSBIE TS Inter 2nd Year English Study Material Grammar Filling in Forms Exercise Questions and Answers.

TS Inter 2nd Year English Grammar Filling in Forms

Q.No. 14 (8 × 1/2 = 4 Marks)

In our day to day life, we need to fill in various kinds of forms. Before filling in a particular form, one must go through the relevant instructions carefully. Forms should be filled in very carefully in legible hand writing. To fill in forms correctly, one must first know what information is required to be filled in a given blank. To understand this one must carefully read what is given before a blank, and sometimes, what is given after a blank too. In the Public Examination question paper, the blanks in the given form are numbered. The student can write the relevant information against those numbers. It is not necessary to copy down the entire form

Now study the following examples carefully

I. Depositing Money in a Bank:

When we deposit money in the Savings Bank Account, we fill in the Savings Bank Account Pay-in-SUp and enter the details. The slip has two parts. The longer one is retained with the bank while the smaller one, called counter foil, is returned to us after the required particulars are entered. First observe how the form was filled in and then attempt the exercise that follows.

Exercise

Question 1.
Mr. L. Srinivas wants to deposit an amount of Rs. 9600/- on 22.05.2022 in his Savings Account with the SBI, Asifabad branch. His account number is 63217569982. He carries the amount in the denomination of Rs. 500 notes 18 and the remaining are Rs. 200 notes. His mobile number is 9848123456. Fill the details in the following voucher to deposit the amount.

Answer:
1) Asifabad
2) 22.05.2022
3) 63217569982
4) L. Srinivas
5) 9848123456
6) 9600/-
7) Nine thousand and six hundred
8) 500 × 8 18 ; 200 × 3
9) 9600
10) L. Srinivas

Question 2.
You are M. Venkateswarlu Your Savings Bank Account number is 12545635489 in the SBi Branch of Shamsnabad, Your mobile number is 7502012543. Deposit an amount of Rs. 8000/- into your account today. And the donornlnation is 3 notes of Rs. 2000 and the remaining are Rs. 500 notes.

Answer:
1) Shamshabad
2) 20.04.2022
3) 12545635489
4) M. Venkateswrarlu
5) 7502012543
6) 8000/-
7) Eight thousand only
8) 2000 × 3; 500 × 4
9) 8000
10) M. Venkateswarlu

Question 3.
You are Ch. Krishna. Your Account Number is 56218561932. Deposit an amount of Rs. 9,300/- into your Savings Bank Account with the SBI, Punjagutta Branch. Your mobile number is 8374373612. The amount is the denomination of Rs. 500 notes 18 and the remaining are Rs. 100 notes.

Answers:
1) Panjagutta
2) 14.06.2022
3) 56218561932
4) Ch. Krishna
5) 8374373612
6) 9300/-
7) Nine thousand and three hundred only
8) 500 × 18 ; 100 × 3
9) 9300
10) Ch. Krishna

II. Depositing Money in a Bank:

Observe how the Withdrawal form was filled before you attempt the exercise.

Exercise

Question 1.
You are Laxmi. You want to withdraw an amount of Rs. 14500/- from your account in the SBI, Mancherial Branch on 02.06.2022. Your account number is 52178999999 and PAN is CNPCKI234N. Fill the following withdrawal form.

Answer:
1) Macherial
2) 02.06.2022
3) Laksmi
4) 52178999999
5) fourteen thousand and five hundred only
6) 14500/-
7) CNPCK1234N
8) 6384725109
9) Mancherial
10) Lakshmi

Question 2.
You are Ch. Gopi Krishna. You have a Savings Bank Account in the SBI Hanumakonda Main Branch, your account number is 52738963762 and PAN is BCKPC2356P. Withdraw an amount of Rs. 25000 – today by filling the Savings Bank Withdrawal Form .

Answer:
1) Hanumakonda Main
2) 06.05.2022
3) Ch. Gopi Krishna
4) 52738963762
5) Twenty five thousand only
6) 25000/-
7) BCKPC2356P
8) 8436271590
9) Hanumakonda Main
10) Ch. Gopi Krishna (Ch. Gopi Krishna)

III. Application for a Demand Draft / Banker’s Cheque

Demand draft is drawn in favour of a person or a firm. The particulars are to be filled in with utmost care. We should also mention in which branch the DD is to be realized. The commission charged by the bank for ‘th,e service is called exchange.

Now carefully go through the sample DD form and do the exercise which follows.

Exercise

Question 1.
You are K. Gangadhar residing at Warangal. You want to apply for EAMCET – 2O2. Apply for a DD’for Rs. 800/- in favour of The Convener, TS EAMCET-2022, JNTU, Hyderabad, in SB!, War an gal Main Branch. It should be payable on SBI, JNTU Branch, Hyderabad. The exchange to be paid is Rs. 35/-.

Answer:
1) Warangal Main
2) 05.06.2022
3) K. Gangadhar
4) The Convener, TS EAMCET-2022
5) Rs 800/-
6) Rs 35/-
7) Eight hundred and thirty five only
8) Not applicable
9) SBI JNTU Branch Hyderabad
10) 8642103579
11) K. Gangadhar

Question 2.
You are S, Srinivas, staying at Vidyanagar, Adilabad. You want to purchase some books from ‘Assorted Book Store’, Hanumakonda. Apply for a Demand Draft of Rs. 3,600/ today in favour of the book store payable at the SBI, Hanumakonda Main Branch. The exchange to be paid is Rs. 40/-

Answer:
1) Vidyanagar
2) 04.05.2022
3) S. Srinivas
4) Assorted Bookstore Hanumakonda
5) Rs 3600/-
6) Rs 40/-
7) Three thousand six hundred and forty only
8) Not applicable
9) Hanumakonda Main Branch
10) 9361485710
11) S. Srinivas

Question 3.
You are Murali Mohan. You want to get a Bankers Cheque for Rs.6500/- on Hanumakonda Main Branch of the State Bank of India in favour of ‘Nava Telangana Publishing House’. The exchangeis Rs. 45/-. Fill in the form given below. You are taking the Banker’s cheque at Jankapur Branch, Asifabad.

Answer:
1) Jankapur
2) 03.05.2022
3) Murali Mohan
4) Nava Telangana Publishing House
5) Rs 6500/-
6) Rs 45/-
7) Six thousand five hundred and forty five only
8) Not applicable
9) Hanumakonda Main Branch
10) 9210364587
11) Murali Mohan

IV. Depositing Cash in a Post Office Account

We can also avail ourselves of the banking services at various branches of Post offices. We can save our money at any branch of a post office. Given below is the Post Office Savings Bank form.

Study the form carefully and attempt the exercise that follows.

Exercise

Question 1.
You are K. Madhubala. Your account number Is 18975 in Bellampally Post office, Mancherial district. Deposit an amount of Rs. 25000/- into your account by filling in the Post Office Savings Account form.

Answer:
1) Bellampalli
2) 18975
3) 06.04.2022
4) K. Madhubala
5) Twenty five thousand
6) Rs 25000/-
7) Not applicable
8) K. Gangadhar

Question 2.
You are P. Neelaveni. Your Account Number is 1981 in Mahboobnagar Post Office. Deposit an amount of Rs. 6000/- into your account by filling in the Post Office Savings Bank form.

Answer:
1) Mahaboobnagar
2) 1981
3) 08.05.2022
4) P. Neelaveni
5) Six thousand
6) Rs 6000/-
7) Not applicable
8) P.Neealaveni

V. Withdrawing Cash from a Post Office Account

Withdrawing money from a Post Office is a simple transaction. Just fill in the Post Office Savings Bank Withdrawal form and submit it to the official concerned. The passbook must accompany the form. Now look at the Withdrawal form and do the exercise that follows.

Exercise

Question 1.
You are M. Dinakar having a savings account in the post office Narsampet, Warangal district. Withdraw an amount of Rs. 12000/- today by filling in the withdraw! form.

Answer:
1) Narsampet
2) 03.06.2022
3) Savings (SB)
4) 9816
5) 12000/-
6) Twelve thousand only
7) Not applicable
8) Dinakar.M

Question 2.
You are S. Swetha, Your Post Office Savings Bank Account Number is 56847 at Post Office of Godavarikhani, Peddapalli district. Withdraw an amount of Rs. 1200/- today by filling in the withdrawal form.

Answer:
1) Godavarikhani
2) 13.07.2022
3) Savings (SB)
4) 56847
5) 1200/-
6) Twelve hundred only
7) Not applicable
8) S. Suuetha

VI. Withdrawing Cash from a Post Office Account

If we plan to undertake a long railway journey it is advisable to reserve a berth in advance in a particular train and class so that we can travel comfortably. Look at the filled-in- form below and then fill in the blank form given after.

Exercise

Question 1.
You are Mr. D. Venkateswarlu, You want to go to Sirpur Kaghaznagar on 24.05.2022 from Warangal along with your father, Rama Rao, aged 55 and your mother, Janaki, aged 50 by the train number 12656, Navajeevan Express. Reserve three seats in advance for 2nd seating (2S) class by filling in the form given below

Answer:
1) 12656
2) Navjeevan Express
3) 24.5.2022
4) 2nd Seating (2S)
5) three (3)
6) Warangal
7) Sirpur Kaghaznagar
8) Warangal
9) Sirpur Kaghaznagar
10) 9848943545
11) D. Venkateswarlu : Ramarao Janaki
12) D. Venkateswarlu
13) 8-13, Gandhi Nagar, Warangal
14) D. Venkateswarlu
15) 9848943545
16) 15.04.2022
17) 4.15 PM

Telangana TSBIE TS Inter 2nd Year English Study Material Grammar Idioms and Phrases Exercise Questions and Answers.

TS Inter 2nd Year English Grammar Idioms and Phrases

Q.No. 12 (4 Marks)

Do you have any idea of what an idiom is?
Let us read this conversation.

Anitha  :  Hi, Akhila… seem to be over the moon!
Akhila  :  Yeah…Don’t you know? The Inter results are music to my ears. What about you?
Anitha  :  Ah! I had to eat humble pie.
Akhila  :  Why… ? You burnt the midnight oil… I think.
Anitha  :  Nevertheless, I failed in one subject.
Akhila  :  Don’t cry your eyes out. Pull up your socks. Prepare we for the next exam.
Anitha  :  It’s OK. But my father doesn’t have belief in my ability, often says, “Don’t tell me cock and bull stories”.
Akhila  :  Come on Anitha. Don’t cry over spilt milk. Keep bad company at bay. You will succeed. Cheer up…
Anitha  :  Thank you Akhila. You have been a pillar of support.
Akhila  :  It’s alright. See you

Have you gone through the groups of words in italics ? What do they mean? What can you get from the words in isolation? Don’t they sound strang Yes, they are called idioms.

What is an idiom ?
An idiom is an artistic expression whose meaning is unpredictable from the usual meaning of its component words. Use of idioms adds glamour to the language, and finds a place in everyday use of language as well.

Now, let’s look at some idiomatic expressions and their usage.

1. once in a blue moon: happen very rarely
He attends classes once in a blue moon.

2. an arm and a leg: very expensive
It cost me an arm and a leg to study in the USA.

3. a piece of the cake: very easy
Batting is a piece of cake for Virat these days.

4. a drop in the ocean: a tiny part of something much bigger
The small donation was just a drop in the ocean.

5. bite one’s tongue: want to say something but stop oneself
Sitara wanted to speak out about the unfair decision. But she bit her tongue.

6. go the extra mile: doing much more than required
My father always goes the extra mile to help the needy.

7. get ducks in a row: to get one’s affairs in order or organized
I can’t hope to do well at my job until I get my ducks in a row.

8. let the cat out of the bag: reveal a secret accidentally
Joel let the cat out of the bag about my surprise birthday party.

9. working against the clock: not having enough time to do something
I am really working against the clock now. I must hurry.

10. flogging a dead horse: attempting to continue with something that is over We are flogging a dead horse. Our present business is making no money.
So, let’s do something else.

11. on cloud nine: very happy
Rahul is on cloud nine since he got a good job.

12. bolt from the blue: something happened unexpectedly
The results of the recent polls were a bolt from the blue to many parties.

13. storm in a tea cup: making unnecessary fuss/ getting excited about something unimportant
There was a storm in a tea cup over who should be the Chief Guest on Hostel Day.

14. make hay while the sun shines : to take advantage of a good situation that may not last long
Our boss is on vacation. Let’s make hay while the sun shines and relax.

15. beat black and blue: covered with bruise marks caused by being hit
Kiran was beaten black and blue by the violent crowd.

16. fight an uphill battle: struggle against very unfavourable circumstances
The Kings XI Punjab must fight an uphill battle to win the IPL title.

17. donkey’s years: a long time .
I have been teaching grammar for donkey’s years.

18. at the eleventh hour: at the last possible moment
If you want to do your best, don’t do things at the eleventh hour.

19. bee in one’s bonnet: an idea that constantly occupies one’s thoughts
Our English teacher has a bee in her bonnet about correct pronunciation.

20. cook someone’s goose: spoil someone’s plans or chances of success
This year’s severe drought cooked Ramaiah’s goose.

21. feather in one’s cap: the achievement of which one can be proud of
Bahubali’s success is a feather in Rajamouli’s cap.

22. rags to riches: start offbeing very poor and become very rich and successful
Dhirubhai Ambani’s life is a story of rags to riches.

23. in the blink of an eye: happen fast and instantaneously
The announcement of intermediate results reached every come” in the blink of an eye.

24. in/by leaps and bounds: make rapid or spectacular progress
Our business flourished in leaps and bounds.

25. risk life and limb: in danger of death or severe injury
Don’t risk your life and limb by participating in reality shows.

26. save one’s neck/skin: escape from death, punishment, etc. especially by leaving others in an extremely difficult situation
Satish got his friends into trouble to save his skin.

27. birds of a feather flock together: similar in many ways, so spend time together
Arun and Varan are sports enthusiasts and are often found together. So, our friends usually speak of them as birds of a feather (flock together).

29. bite off more than you can chew: try to manage something too difficult
Actor Shireesh bit off more than he could chew in his debut movie.

30. take the bull by the horns: act decisively to deal with a complex problem
I want to take the bull by the horns by deciding to swim across the fiver.

31. leave no stone unturned: try everything possible to achieve something
President Kalam left no stone unturned to motivate the youth.

32. no spring chicken: one is quite old or well past one’s youth
I am no spring chicken, you know. How can I dance?

33. beyond wildest dreams: better than you imagined for
Last year’s rainfall was beyond our wildest dreams.

34. keep your nose to the grindstone: concentrate on working or studying hard (informal)
In the Intermediate class, my daughter has to keep her nose to the grindstone. No time for games or music.

35. paddle your own canoe: do something without the help of others
Sonu refused all help as he believed in paddling his owm canoe.

36. have a bone to pick: annoyed with somebody and talk to them about it
Lakshmi has a bone to pick with Revati as she was not invited to the marriage.

37. give a tongue-lashing: scold someone severely
The teacher gave the lazy boy a tongue-lashing when he called Rani a lazy girl.

38. dressed up to the nines: wearing bright or glamorous clothes
Actors often dress up to the nines for public functions.

39. make one’s ears burn: embarrassed by hearing something being said about you
The discussion about my childhood pranks made my ears burn.

40. turn a deaf ear: refuse to listen to somebody
Our boss turned a deaf ear to our request to change the meeting venue.

41. back to the salt mines: returning to work with some reluctance
After the vacation, the students had to go back to the salt mines.

42. nuts and bolts: detailed facts and the practical aspects (informal)
Unless I get to know the nuts and bolts of the business, I can’t venture into it.

43. step into someone’s shoes: take over a job / a position held by someone before you
When the manager retires, I’ll step into his shoes.

44. catch-22: a frustrating situation that will lead to further frustration
In big cities, if you don’t have a place to live in, you can’t get a job, and with no job, you can’t get a place to live in! Thus, it’s a catch-22 situation.

45. at the drop of a hat: do it immediately without hesitation
The wealthy socialite throws parties at the drop of a hat.

46. wild goose chase: a search for something that is impossible for you to find or that does not exist, that makes you waste a lot of time
Without a proper address or phone number, it would be a wild goose chase to locate someone in Hyderabad.

47. taste your own medicine: have the same bad treatment that you have given to others
The evil-minded will taste their own medicine sooner or later.

48. blow one’s horn: to praise oneself; to boast
People avoid her as she constantly blows her horn.

49. burn the midnight oil: to work very late into the night
I have to present this report by tomorrow. So, I must bum the midnight oil tonight.

50. thrilled to bits: extremely pleased about something
I was thrilled to bits when I received a fancy gift.

51. a bed of roses: something which is easy
Life is not always a bed of roses.

52. a white elephant: expensive but not that useful
His new car has become a white elephant.

53. hit the nail on the head: say something correctly
She hit the nail on the head with her response.

54. spill the beans: to reveal a secret
They were afraid he would spill the beans.

55. hot cakes: fastselling
The new model cars are selling like hot cakes.

56. face the music: accept unpleasant consequences
Having lost his character, he has to face the music.

57. judge a book by its cover: judge something primarily on appearance
You can’t judge a book by its cover. Just because he looks strange, that doesn’t mean . he is not a nice person.

58. Achilles heel: a weak (vulnerable) spot
Maths has always been my Achilles heel.

59. by a whisker: by a very small amount/margin
Finally, our team lost by a whisker and I was disappointed.

60. every cloud has a silver lining: every misfortune has some positive aspect
If we hadn’t missed the plane, we wouldn’t have met you. It is rightly said every cloud has a silver lining.

61. part and parcel: an essential or fundamental element
Don’t get disheartened at losing your form. It’s part and parcel of being a professional.

62. a thick skin: an ability to not be upset by criticism
A politician needs a thick skin.

63. a sea change: a complete change in someone’s attitude or behavior
There is a sea change in the behavior of the culprit after his release from the prison.

64. in a nutshell: briefly; in essence
Let me explain the proceedings in a nutshell.

65. get the nod: to receive permission from someone to start something
Rahul got the nod after a lengthy discussion among the members.

66. take the rap: to be blamed or punished, especially for something you have not done
She was prepared to take the rap for the shoplifting, though it had been her sister’s idea.

67. dark horse: one who is previously unknown and is now prominent
The Gujarat Titans has proved to be a dark horse in the recent IPL.

68. a hot potato: controversial and sensitive issue
Racism is currently a hot potato in the international cricket.

69. a square meal: a large, filling, nutritious meal
The soldiers are very tired. They haven’t had a square eat for four days.

70. by fair means or foul: by any possible method
They never gave up trying to recover their property by fair means or foul.

71. a labour of love: a task that you do for pleasure without expecting payment
Preparing this book is clearly a labour of love.

72. keep the pot boiling: keep going on actively
I threw in a question just to keep the pot boiling while my brain caught up.

73. tie the knot: to get married
The couple tied the knot last year.

74. from top to bottom: very thoroughly
I would clean my room from top to bottom every Sunday.

Exercises

I. Fill in the blanks ITI the following sentences with suitable idiomatic expressions given below. Make necessary changes in the idioms if needed.

take the rap
in a nut shell
a sea change
by a whisker
get the nod
a thick skin

1. Finally, our team lost _____________ and I was disappointed.
2. A politician needs _____________
3. There is _____________ in the behavior of the culprit after his release from the prison.
4. Let me explain the proceedings _____________
5. Rahul _____________ after a lengthy discussion among the members.
6. She was prepared to _____________ for the shoplifting, though it had been her sister’s idea.
Answers:
1. by a whisker
2. a thick skin
3. a sea change
4. in a nutshell
5. go the nod
6. take the rap

II. Fill in the blanks in the following sentences with suitable idiomatic expressions given below. Make necessary changes in the idioms if needed.

on cloud nine
tie the knot
go the extra mile
apiece of cake
turn a deaf ear
hot cakes

1. The new model cars are selling like _____________
2. My father always _____________ since he got a good job.
3. Rahul is _____________ for Virat these days.
4. The couple _____________ last year.
5. Our boss _____________ to our request to change the meeting venue.
Answers:
1. hot cakes
2. goes the extra mile
3. on cloud nine
4. a piece of cake
5. tied the knot
6. turned a deaf ear

Telangana TSBIE TS Inter 2nd Year Accountancy Study Material 4th Lesson Partnership Accounts Textbook Questions and Answers.

TS Inter 2nd Year Accountancy Study Material 4th Lesson Partnership Accounts

Very Short Answer Questions

Question 1.
What is partnership?
Answer:

1. According to section 4, the Partnership Act 1932, partnership means “The relationship between persons who have agreed to share profits of the business carried on by all or any one of them acting for all”.
2. The persons who have entered into partnership business are called individually as “partners” and collectively as “a firm” and the name under which their business is carried on is called the name of the firm.

Question 2.
What is a partnership deed?
Answer:

1. It is a document that defines the rights and liabilities of partners of the firm besides containing other matters pertaining to the conduct and management of the firm.
2. Partnership deed is signed by all the partners.

Question 3.
State the methods of preparing partners’ capital accounts.
Answer:
There are methods of preparing partners’ capital accounts.
They are: A) Fixed capital method; and B) Fluctuating capital method.

Question 4.
What is the fixed capital method?
Answer:

1. Under fixed capital method, we prepare two accounts partner capital account and partners current account.
2. All the adjustments such as drawings, interest on capital, salary or commission, share of profits will be recorded in a partners current account, and the partners capital account is remains unchanged. Because of this, it is called fixed capital method.

Question 5.
What is fluctuating capital method?
Answer:

1. Under this method, all transactions relating to a partner such as capital contributed by partner, interest on capital, drawings, interest on drawings, salary, commission to partner, share of profit or loss are recorded in capital account only.
2. Because of this, the balance in capital account keeps on changing every year. Hence, it is called fluctuating capital method.

Question 6.
What is partnership? State its features.
Answer: Definition: According to section 4, Partnership Act 1932, partnership means “The relationship between persons who have agreed to share profits of the business carried on by all or any one of them acting for all”.
Features:

• Agreement: The partnership is formed to make profits. It is an agreement entered between/among all the partners.
• Business: It must be of a lawful business. It includes trade, vocation and profession.
• Profit sharing: The partners share the profit and losses of the business as per the
  agreement.
• Management: Partnership business may be managed or carried on by all the partners or any one of the partners acting on behalf of all the other partners, f) Unlimited liability: The liability of a partner in partnership firm is not only unlimited, but also joint and several.

Question 7.
What is partnership deed? State its contents.
Answer:
Partnership deed; It is a document which defines the rights and liabilities of partners of the firm besides containing other matters pertaining to the conduct and management of the firm.
Contents of partnership deed: The written agreement can contain as much, or as little, as partners want. The law does not say what it must contain. The usual contents of partnership deed are listed as here under.

1. Name of the partnership firm.
2. Names, addresses and occupation of all the partners.
3. Nature, object and duration of business.
4. Amount of capital to be contributed by each partner.
5. Drawings if any, allowed for private purposes.
6. Sharing of profits and losses.
7. Rate of interest on capital and drawings.
8. Rights, duties and liabilities of each partner.
9. Method of keeping books of accounts and audit.
10. Mode of payment to retired partner’s share.

Question 8.
What is Profit and Loss appropriation account?
Answer:

1. Profit and Loss appropriation account is prepared after ascertaining the net profit (or) loss through profit and loss account.
2. Profit and Loss appropriation account is a normal account. It is an extension of profit and loss account.
   This account is debited with interest on capital of partners, salary, commission remuneration to the partners, if allowed and profit transfer to general reserve and credited with the net profit, interest on drawings.
3. The balance if any, may be net profit or loss and will be distributed to the partners as per their profit sharing ratio.

Question 9.
State the provisions of partnership act in the absence of partnership deed,
Answer:
It is to be noted that if there is no partnership deed all the partners should abide by the provisions of partnership Act 1932.
In the absence of partnership deed, the partnership Act states that:

• All the profits and losses are to be shared equally.
• No interest is to be provided on capital.
• No interest is to be charged on drawings.
• No salary or commission is payable to managing partner.
• The partners who provided loan to firm are entitled to get interest @ 6% p.a. on the loan.
• Every partner shall take part in management of the business.

Textual Problems

Question 1.
A, B and C are partners sharing profits and loss equally with share capital of Rs. 50,000, Rs. 40,000 and t 30,000 respectively. The profit for the year ended 31st March 2015 amounted to Rs. 56,000 before allowing interest on capital @ 8% p.a. and salary to A Rs. 6,000 and commission to C Rs. 4,000.
Prepare profit and loss appropriation account.
Answer:
Profit & Loss appropriation a/c

Question 2.
X,Y and Z are partners sharing profits and losses in the ratio of 2: 2: 1. Each partner withdraws the fixed sum of Rs. 4,000 per month for their personal use.
Mr. X, withdraws on 1st day of every month, Mr. Y withdraws on the last day of every month and Mr. Z withdraws on middle of every month. Calculate interest on drawings. Rate of interest is 6% p.a.
Answer:
Calculation of Interest on Drawings:
Total drawings 4000 x 12 = 48,000
Rate of interest 6%

Question 3.
A and B are partners sharing profits and loss in the ratio of 3: 2 respectively. A contributed Rs. 2,00,000 and B contributed Rs. 1,50,000 towards their capital at the beginning of the year. Their drawings during the jfear amounted to Rs. 10,000 and 12,000 respectively. Mr. A is entitled to a salary of Rs. 8,000 and Mr. B to be given a commission of Rs. 10,000. Interest on capital to be provided @ 10% p.a. Interest on drawings amounted to t 500 for A and 1600 for B. Net profit before above adjustments amounted to Rs. 62,000.
Prepare profit or loss appropriation account and partners capital accounts under fluctuating capital method.
Answer:
Profit & Loss appropriation a/c

Fluctuation capital method

Question 4.
Sitha and Githa are partners sharing profits and losses in the ratio of 2: 3. Their capital accounts showed a balance of Rs. 1,00,000 and Rs. 1,50,000 respectively as on 1st April 2014. During the year, Sitha and Githa withdraw Rs. 5,000 and Rs. 7,000 respectively for personal use. Interest on capital to be provided @ 8% p.a. Interest on drawings amounted to Rs. 500 and Rs. 700 respectively. Profit for the year ended 31st March 2015, before making above adjustments amounted to Rs. 75,000.
Prepare: (a) Profit & Loss Appropriation account, (b) Partners capital accounts under (i) Fixed capital method, (ii) Fluctuation capital method.
Answer:
Profit & Loss appropriation a/c

Fixed capital method:

Fluctuations capital method:

Question 5.
Ramu and Robert are partners sharing profits and losses in the ratio of 5: 3. Give journal entries for the following and give capital accounts of Ramu and Robert.
(a) Capital introduced by Ramu Rs. 50,000 and Robert Rs. 80,000.
(b) Drawings by Ramu Rs. 5,000 and Robert Rs. 6,000.
(c) Interest on capital: Ramu Rs. 2,500 and Robert Rs. 4,000.
(d) Interest on drawings Ramu Rs. 300 and Robert Rs. 400.
(e) Profit transferred to General Reserve Rs. 9,000.
(f) Salary to Ramu Rs. 10,000 and Commission to Robert Rs. 7,000.
Answer:
Journal entries in the books of Ram and Robert

Question 6.
X and Y are partners sharing profits and losses in the ratio of 3: 2. Their trial balance as on 31 March, 2020 was as under.

You are required to prepare Trading, Profit and loss account, Profit and loss appropration account for the year ended 31 March 2020 and balance sheet as on that date after considering the following.
a) Closing stock value at t 48,000.
b) Interest on capital allowed at 6% p.a.
c) X is allowed a salary of Rs. 600 per month.
d) Depreciate machinery by 5% and Furniture by 8%.
e) Provide for bad debts @ 5% on debtors.
f) Transfer of general Reserve Rs. 5,000.
Answer:
Trading & Profit and loss a/c of X & Y for the year ended 31.03.2020

Balance Sheet of X & Y as on 31-03-2020

Question 7.
The following is the trial balance of Rarnesh and Suresh as on 31 March 2019. They share profit and losses in the ratio of 1: 3 respectively.


Prepare final accounts after taking into account the following.
a) Stock on March 2019 valued at Rs. 13,000.
b) Wages due Rs. 1,000
c) Taxes paid in advance Rs. 500
d) Provide for bad debts @ 6% on debtors.
e) Interst on capital allowed at 8%.
f) Interest on drawings @ 6% p.a.
g) Ramesh is allowed a salary of Rs. 7,000.
h) Depriciate buildings by Rs. 3,000, write off goodwill Rs. 2,000 and furniture by Rs. 800.
Answer:
Trading & Profit and loss a/c of Ramesh & Suresh for the year ended 31.03.2019

Balance Sheet of Ramesh & Suresh as on 31-03-2019

Question 8.
Ganga and Yamuna carrying on partnership business sharing profits and losses equally. Their trial balance as on 31 March 2020 was as under.

Additional information:
a) Closing Stock valued at Rs. 17,000.
b) Salaries payable Rs. 2,000.
c) Provide for bad debts @ 6% on debtors.
d) Interest on Loan to be provided.
e) Depreciate Fixed assets by 3% and furniture by Rs. 500.
f) Interest allowed on capital at 6%. No interest required on current accounts.
g) Interest on drawings Ganga Rs. 250, Yamuna Rs. 150.
h) Transfer to general Reserve Rs. 6,000.
Answer:



Balance Sheet of Ganga & Yamuna as on 31-03-2020

Textual Examples

Question 1.
Mr. Ram withdraws Rs. 1000 per month for his personal use. The rate of interest islO % per annum. Calculate interest on drawings in the following cases;
(a) When the amount (drawings) is taken on 1st day of every month.
(b) When the amount is drawn on last day of every month.
(c) When the amount is drawn in the middle of every month.
Answer:
Ram’s total drawings for the year Rs. 12,000
(12 months @ Rs. 1000 per month)
Rate of interest 10% per annum.

Question 2.
Mr. Ravi and Mohan are partners in a firm on 1st April 2019 with a capital of Rs. 1,50,000 and Rs. 2,00,000 respectively. On 1st July 2014 Mr. Ravi introduced Rs. 50,000 as an additional capital. Rate of it interest on capital is 10% p.a. Calculate interest on capital. Books are closed on 31st March, 2020.
Answer:
Calculation of interest on Capital On Ram’s capital
On Rs. 1,50,000 for one year @ 10% p.a.
(1 – 4-2019 to 31 – 3-2020)

On Mohan’s capital
On Rs. 2,00,000 @ 10% p.a. for one year
(1-4-2019 to 31 -3 – 2020)

Question 3.
Ram, Laxman and Hanuman are partners in a firm sharing profits and loss in the ratio of 2: 2: 1 respectively. Ram withdraw Rs. 5,000 per month on 1st day of every month. Laxman withdraw Rs. 4,000 per month on last day of every month. Hanuman withdraw^ Rs. 3,000 per month at the middie of every month. Rate of interest on drawings is 6% per annum. Calculate interest on drawings.
Answer:
Calculation of interest on Drawings:
Total drawings for the year;
Ram: 12 months @ t 5,000 per month = Rs. 60,000
Laxman: 12 months @ Rs. 4,000 per month = Rs. 48,000
Hanuman: 12 months @ Rs. 3,000 per month = Rs. 36,000
Interest on drawings = Total drawings x Rate of interest x Time

Note: (Student may follow product method. It take more time to calculate. For example in case of Ram’s Drawings interest is to calculated for 12 times, i.e., on 1st Rs. 5,000 drawn, interest is for 12 months and 2nd Rs. 5,000 drawn, interest is for 11 months so on. If we take average of this we get 6.5 months (time). So, calculate interest for 6.5 months on total drawings for the year. Therefore, student is advised to remember the above formula for each mode of drawings).

Question 4.
Ram and Rahim commenced a business on 1st April 2019. Ram brings in Rs. 20,000 in cash and Furniture worth Rs. 5,000 towards his capital. Mr. Rahim brings Rs. 10,000 in cash and Rs. 15,000 worth of building and creditors Rs. 5,000 towards his capital. Give journal entries and capital account of Ram and Rahim.
Answer:
Journal Entries


Ledger Accounts:

Question 5.
Mr. A and B are partners sharing profits and losses in the ratio of 2:1 have contributed Rs.1,00,000 and Rs. 60,000 respectively towards their capital. The partnership agreement provides that;
(a) Interest is allowed on capital @ 8% per annum.
(b) Mr. A is allowed a salary of Rs. 20,000 p.a.
(c) During the year A takes t 10,000 and B takes Rs. 5,000 towards personal use.
(d) Interest on drawings amounted to Rs. 1,000 for A and Rs. 500 for B.
(e) Partners decided to transfer Rs. 8,000 to general reserve a/c.
Net profit of the firm before above adjustments amounted to Rs. 45,000 for the year ended 31st March 2020.
You are required to give journal entries and profit and loss appropriation account and partners capital accounts (Fluctuating capital method).
Answer:
Journal entries

Note: Capital accounts may be prepared separately for A and B.

Question 6.
P, Q and R are partners sharing profits and losses in the ratio of 4: 3: 3. On 1st April 2019, their capital accounts showed a balance Rs. 80,000, Rs. 60,000 and Rs. 60,000 respectively. P is entitled to a salary of Rs.15,000 and Q is to get a commission of Rs. 8,000. Interest on capital is allowed @ 5% per annum. Their drawings during the year ended 31st March 2020, amounted to Rs. 4,000, Rs. 3,000 and Rs. 3,000 respectively for P,Q and R.

The interest on drawing amounted to Rs. 200, Rs. 150 and Rs. 150 respectively They decided to transfer Rs. 10,000 to general reserve. The profit earned for the year ended 31 March, 2020 amounted to Rs. 60,000 before making above adjustments.
Give profit & loss appropriation account and partners capital accounts under
(a) Fixed capital method
(b) Fluctuation capital method,
Answer:

Note: The balance of profit available is Rs. 17,500 i.e., Rs. 60,500 – Rs. 43,000. It- is distributed among P, Q and R in the ratio of 4: 3: 3. P gets Rs. 7,000: Q gets Rs. 5,250 and R gets Rs. 5,250.
(a) Fixed capital method:

(b) Fluctuating capital method:

Question 7.
Ram and Rahim are partners sharing profits and losses in the ratio of 3: 2 respectively. The following trail balance is extracted from their books as on 31 March 2020. You are required to prepare Trading, profit and loss account, profit and loss appropriation account and balance sheet as on that date.

The following and additional Informations is provided:
a) Closing stock valued at T 50,000.
b) Depreciate Machinery by 10%, Furniture by 8%, Buildings by 5%.
c) Provide for outstanding wages Rs. 6,000 and salaries Rs. 8,000.
d) Insurance paid in advance Rs. 500.
e) Provide for doubtful debts at 5% on debtors.
f) Interest on capital is allowed @ 6% p.a.
g) Interest on drawings: Ram Rs. 300, Rahim Rs. 180.
h) Transfer to general Reserve t 10,000.
i) Salary allowed to Ram Rs. 8,000 and commission allowed to Rahim Rs. 6,000.
Answer:
Trading & Profit and loss a/c of Ram and Rahim for the year ended 31 March, 2020

Note: Profit is Rs. 1,00,500 (ie., 1,30,500 – 30,000) transferred to capital accounts of Ram 3/5 of 1,00,500 ie., Rs. 60,300 and to Rahim 2/5 of 1,00,500 ie., 40,200.

Balance Sheet of Ram and Rahim as on 31st March, 2020


Note: Capital accounts age maintained under fluctuating capital method. Hence, all adjustments relating to interest on capital, drawings, salary and commission to partners, drawings, general reserve are recorded in capital accounts of partners concerned.

Question 8.
A and B are partners sharing profits and losses in the ratio of their capital. Their trial balance as on 31 March, 2020 is as follows.

You are required to prepare financial statements for the year ending 31 March 2020 after taking into account the following adjustments.
a) Closing stock valued at Rs. 28,000.
b) Salaries due Rs. 4,000.
c) Depreciate buildings by 2%, Machinery by 10%.
d) Interest due on investments Rs. 2,500.
e) Provide for bad debts at 5% on debtors.
f) Interest on capital is allowed at 8% p.a. No interest allowed on current account balances.
g) Interest on drawings: A – Rs. 80 and B – Rs. 40.
h) Salary allowed to B * Rs. 6,000.
i) Transfer to general Reserve Rs. 15,000.
Answer:





Balance Sheet of A and B as on 31 March, 2020

Telangana TSBIE TS Inter 2nd Year English Study Material Grammar One-word Substitutes Exercise Questions and Answers.

TS Inter 2nd Year English Grammar One-word Substitutes

Q.No.11 (4 Marks)

A single word that can be used in the place of many words is usually referred to as a one-word substitute. Knowing the meanings and usage of as many one-word substitutes as possible is very helpful in note-making and precis writing. It also improves one’s insight into the functioning of the language.

One h Ipful way to understand and remember the meaning of longer words is to know the r anings of their parts. ‘Telephone’, for example, is a combination of two parts ire. t ie (= distance) and ph?ne (= sound). Now think of the meanings of telescope (vision), telegram (writing), pbonedcs etc. ‘Autobiography’ is made up of three parts i. e. auto (= self), bio (=life) and graphy (= writing). Now you can easily understand and remember the meanings of words like automatic, automobile, biology, biometric, geography. Another way of acquiring quick command of one-word substitutes is to study them in groups, for example, of words pertaining to various fields of knowledge, referring to persons, etc. But the time – tested method is to have constant practice in knowing the meanings first arid then using them in sentences.

One – word substitutes
We can master our vocabulary by learning one-word subsdtutes. A one-word substitute, as its name indicates, is a word that replaces a group of words. The knowledge of one-word substitutes not only saves time while writing but also helps you in scoring good marks in competitive examinations.
To have a clearer understanding of one-word substitutes, read the following two passages. Pay special attention to the words or groups of words in bold.

(a) Here is my cousin Saraswathi. She is a person who loves books very much. Whatever book she S:il:n lay h~r hands on, she loves it and reads it. Recently shehas developed a keen interest in books dealing with the study of human mind. Yesterday, at a seminar, she explained how some people suffer from a morbid fear of water. She also discussed how some . people exhibit obsessive devotion to their own people. Later, she talked about the study of gods. Participants in the seminar appreciated her as a person with total knowledge in the topic she discusses. But, she humbly admitted that she was just like anyone of them there,

(b) Here is my cousin Saraswathi. She is a blbUophlle. Whatever ~ book she can lay her hands on, she loves it and reads it. Recently she bas developed a keen interest in books dealing with psychology. Yesterday, at a seminar, she explained how some people suffer from hydrophobia or aquaphobia. She also discussed how some people exhibit ethnomanla. Later, she talked about theology. Participants in the seminar appreciated her as omniscient in the topic she discusses. But she humbly admitted that she was just like anyone of them there.

Have you noticed how one-word substitutes (in Passage (b)) replaced groups of words (in Passage (a))? Between passages (a) and (b), which do you like better? Yes, you are right! One-word substitutes lend brevity (the soul of wit), variety (the spice of life), beauty (a joy forever) and clarity (the soul of communication) to language. They help one enhance one’s understanding of the mechanism that governs the formation of words. Keeping in view the multiple advantages of learning as many one-word substitutes as possible, keep on acquiring a few one-word substitutes a week like a hobby (culture).

Activity

Fill in the blanks in the following sentences with one word substitutes. Choose the right word from the list given in the box below

anaemia
benefactor
chronicle
debut
ecology
germicide
infanticide
misanthrope
optimist
panacea

1. Repeated and reckless use of a ______________ results in irreparable damage because of the side effects of the chemicals.
2. The sage says that his view of life is his ______________ to every problem one encounters in one’s life.
3. The present pandemic has very effectively emphasized the importance of preserving and protecting ______________
4. Severe fever for a few months left my mother with ______________
5. My uncle is such an ______________ that he sees opportunities even in dreaded difficulties!
6. Swami Vivekananda said that one who extends spiritual knowledge to others is the best ______________
7. Srilekha invited loud applause with her melodies in her very ______________
8. Some people resort to ______________ when a girl child is born in their homes, which is inhuman.
9. Our teachers always encourage us to stay updated. For that, they advise us to go through a ______________ a day and be in touch with all major historical events.
10. You can never find a miserable ______________ as this gentleman. Introduce to him
Answer:
1. germicide
2. panacea
3. ecology
4. anaemia
5. optimist
6. benefactor
7. debut
8. infanticide
9. chronicle
10. misanthrope

Given below is a list of words to enrich your vocabulary.

1. agends  :  a list of things to be discussed at a meeting
2. agnostic  :  a person who claims neither faith nor dis-belief in the existence of God
3. altruism  :  unselfish interest in the welfare of others
4. amateur  :  one who engages in an activity as a pastime rather than as a profession
5. ambidextrous  :  able to use both hands equally well
6. ambiguous  :  having more than one meaning and so, is unclear
7. amphibious  :  living on land as well as in water
8. anarchist  :  one who rebels against authority or established order
9. anarchy  :  the absence of government or control in a society
10. annihilation  :  complete destruction of something
13. anonymous  :  (a person) not identified by name, of unknown name
12. anthology  :  a collection of poems or stories
13. antidote  :  a substance that’can act against the effect of poison
14. antiseptic  :  a medicine that prevents infection
15. archaeology  :  study of life and culture of ancient people through the excavation of sites
16. atheist  :  a person who does not believe in the existence of god.
17. audience  :  a number of people listening to a lecture or a concert
18. autobiography  :  the life history of a person written by him self/herself
19. autonomous  :  (a person or an insurute) that can take decisions independently
20. bacteriology  :  a scientific study of bacteria
21. benefactor  :  one who gives money or help to another person or cause
22. bibliophile  :  a person who loves reading and keeping books
23. biography  :  a story of someone’s life written by another person
24. biosphere  :  the totality of living organisms and their environemnt
25. bouquet  :  a bunch of flowers tied together to be given as a present or to welcome someone

26. brunch  :  a late morning meal eaten instead of breakfast and lunch
27. calligraphy  :  the art of good handwriting
28. cannibal  :  a person who eats human flesh
29. cantonment  :  a permanent station for soldiers, garrison
30. cardiologist  :  a doctor who treats heart diseases
31. celibacy  :  the state of remaining unmarried
32. centennial  :  (centenary) the hundredth anniversary of an event
33. chef  :  a professional cook, typically the head cook in a restaurant
34. chronicle  :  a record of historical events
35. colleagues  :  people who work in the same organization
36. contemporary  :  living or occurring at the same time
37. cosmopolitan  :  an outlook that is influenced by people from all over the world
38. den  : the home of lions
39. dermatologist  :  a doctor who treats skin diseases
40. drought  :  prolonged period of abnormally low rain- fall
41. edible  :  fit to be eaten as food by humans
42. egoist (egotist)  :  a selfish person who talks and thinks of himself/herself
42. egoist (egotist)  :  a selfish person who talks and thinks of himself/herself
43. encyclopedia  :  a book of information covering all subjects (disease) regularly found in a particular
44. endemic  :  area or among particular people
45. ephemeral  :  that which has a short life
46. epidemic  :  the spread of an infectious disease in a very short time in a place
47. epitaph  : a short text written on a tombstone
48. epitome  :  the perfect example of something
49. etiquette  :  the rules of accepted polite behaviour in a society
50. etymology  :  the study of the origin and history of words
51. geocentric  :  having the earth at the centre
52. glutton  :  one who eats excessively

53. graphophobia  :  fear of writing
54. gregarious  :  (of people) who love the company of others
55. gymnasium  :  a room that has equipment for physical exercises
56. hematophobia  :  fear of blood
57. herbarium  :  a collection of dried plants
58. hydrosphere  :  all the water of the earth
59. iconoclast  :  one who attacks established and cherished beliefs; idol breaker
60. imminent  :  about to happen in the immediate future
61. immune  :  resistant to a particular disease or toxin
62. incorrigible  :  (of people) who cannot be corrected or changed
63. indefatigable  :  able to work for a long time without becoming tired
64. indelible  :  (a mark) that cannot be erased easily
65. inevitable  :  that which will happen and cannot be avoided
66. infallible  :  incapable of making mistakes
67. inflammable  :  catching tire quickly
68. insolvent  :  unable to pay debts
69. invincible  :  too strong to be defeated
70. invisible  :  that which cannot be seen
71. irrevocable  :  something that cannot be changed
72. kennel  :  a house or shelter for a dog
73. lethal  :  designed to cause death
74. loquacious  :  talking a lot or too much
75. maxim  :  a short statement expressing the rule of conduct
76. mercenary  :  concerned with making money at the expense of ethics
77. misogynist  :  a man who hates women
78. mortuary  :  a place where dead bodies are kept until cremation
79. notorious  :  well known for some bad quality
80. novice  :  one who is inexperienced or new to a job

81. nuance  :  a slight difference in meaning that is difficult to detect
82. obsolete  :  something which is out of date
83. omnipotent  :  having unlimited power
84. omniscient  :  having complete or unlimited knowledge
85. opaque  :  that which cannot be seen through, not transparent
86. optician  :  a person whose job is to examine people’s eyes and to recommend and sell glasses
87. optimist  :  one who looks at the bright side of things
88. ornithology  :  a scientific study of birds
89. orthopaedician  :  one who treats conditions involving the musculo-skeletal system / bone specialist
90. paediatrician  :  a doctor who treats diseases of children
91. palindrome  :  a word or phrase that reads the same backwards or forwards Example: madam
92. pantheism  :  the belief that God is present in all things
93. patent  :  sole right to produce or sell an invention
94. pedestrian  :  a person walking on a street
95. penchant  :  strong taste or liking for something
96. perennial  :  lasting for a long time, continually recurring
97. peregrination  :  a long slow journey, especially on foot
98. pessimist  :  one who looks at the dark side of things
99. philanthropist  :  a person who helps the needy
100. physician  :  one who attends to sick people and prescribes medicines
101. pilgrimage  :  a journey to a holy place for religious reasons
102. polyglot  :  one who can speak many languages
103. professional  :  a person with proven practical knowledge in a field
104. psychology  :  the study of human mind and behaviour
105. quarantine  :  confinement to one place to prevent the spread of infection
106. seismograph  :  an instrument for detecting earthquakes
107. somnambulism  :  the habit or activity of walking in sleep
108. spendthrift  :  a person who wastes money
109. stalwart  :  a loyal supporter of an organization

110. stoic  :  one who is indifferent to pleasure or pain
111. teetotaller  :  one who never takes alcoholic drinks
112. theist  :  a person who believes in the existence of God
113. verbose  :  using more words than required
114. veteran  :  someone who has a lot of experience in a field
115. web  :  the home of spiders

Exercises

Match the following word in Column A with their meanings / definitions in Column B.

Question 1.

Answer:
i) d
ii) e
iii) a
iv) g
v) b
vi) c

Question 2.

Answer:
i) f
ii) g
iii) e
iv) c
v) b
vi) a

Students must practice these Maths 2B Important Questions TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type to help strengthen their preparations for exams.

TS Inter Second Year Maths 2B Differential Equations Important Questions Short Answer Type

Model I – Problems on Variables Separable Method

Question 1.
Solve \(\frac{\mathbf{d y}}{\mathbf{d x}}\) = e^x-y + x² (e^-y). [(AP) May ’18, (TS) ’17]
Solution:

Question 2.
Solve (e^x + 1) y dy + (y + 1) dx = 0. [May ’10, mar. ’04]
Solution:

Question 3.
Solve tan y sec²x dx + tan x sec²y dy = 0.
Solution:
Given D.E is tan y sec²x dx + tan x sec²y dy = 0
⇒ \(\frac{\sec ^2 x d x}{\tan x}+\frac{\sec ^2 y d y}{\tan y}=0\)
⇒ \(\int \frac{\sec ^2 x}{\tan x} d x+\int \frac{\sec ^2 y}{\tan y} d y=c\)
⇒ log|tan x| + log|tan y| = c
⇒ log|tan x| + log|tan y| = log c
⇒ log|tan x tan y| = log c
⇒ tan x tan y = c

Question 4.
Solve (xy² + x) dx + (yx² + y) dy = 0. [(TS) Mar. ’20, (AP) ’15; May ’13]
Solution:

Question 5.
Solve \(\frac{d y}{d x}=\frac{x(2 \log x+1)}{\sin y+y \cos y}\). [Mar. ’05]
Solution:

Question 6.
Solve \(\frac{d y}{d x}=\frac{-\left(y^2+y+1\right)}{\left(x^2+x+1\right)}\). [May ’08, Mar. ’03]
Solution:


which is the required general solution.
(where k = √3c = constant)

Question 7.
Solve \(\sin ^{-1}\left(\frac{d y}{d x}\right)\) = x + y. [(TS) Mar. ’20; May ’07]
Solution:

⇒ ∫sec²t dt – ∫tan t sec t dt = x + c
⇒ tan t – sec t = x + c
⇒ tan(x + y) – sec(x + y) = x + c
which is the required general solution.

Question 8.
Solve \(\frac{d y}{d x}\) = tan²(x + y).
Solution:

Question 9.
Solve \(\frac{d y}{d x}\) – x tan(y – x) = 1. [(TS) May ’15]
Solution:
Put y – x = z, then
\(\frac{d y}{d x}-1=\frac{d z}{d x}\)
⇒ \(\frac{d y}{d x}=1+\frac{d z}{d x}\)

∴ The solution is sin(y – x) = \(c \mathrm{e}^{\mathrm{x}^2 / 2}\), where c is an arbitrary constant.

Question 10.
Solve (x + y + 1) \(\frac{d y}{d x}\) = 1
Solution:

⇒ x + y + 1 – log(1 + x + y + 1) = x + c
⇒ y + 1 – log(x + y + 2) = c
⇒ y – log(x + y + 2) = c
Which is the required solution.

Question 11.
Solve \(\sqrt{1+x^2} d x+\sqrt{1+y^2} d y=0\). [Mar. ’09]
Solution:

Question 12.
Solve y(1 + x) dx + x(1 + y) dy = 0. [(AP) Mar. ’20]
Solution:
Given D.E. is y(1 + x) dx + x(1 + y) dy = 0
y(1 + x) dx = -x(1 + y) dy
\(\frac{1+x}{x} d x=\frac{-(1+y)}{y} d y\)
\(\left(\frac{1}{x}+1\right) d x=\left(\frac{-1}{y}-1\right) d y\)
Now, Integrating on both sides, we get
\(\int\left(\frac{1}{x}+1\right) d x=\int\left(\frac{-1}{y}-1\right) d y\)
\(\int \frac{1}{x} d x+\int 1 d x=-\int \frac{1}{y} d y-\int 1 d y\)
log x + x = -log y – y + c
log x + x + log y + y = c
x + y + log xy = c
Which is the required solution.

Question 13.
Solve \(\frac{d y}{d x}=\frac{x y+y}{x y+x}\). [(TS) May ’16]
Solution:

Question 14.
Solve \(\frac{d y}{d x}\) = sin(x + y) + cos(x + y)
Solution:
Given D.E. is \(\frac{d y}{d x}\) = sin(x + y) + cos(x + y)
Put x + y = z
differential w.r.t ‘x’
\(1+\frac{d y}{d x}=\frac{d z}{d x}\)
⇒ \(\frac{d y}{d x}=\frac{d z}{d x}-1\)
(1) ⇒ \(\frac{\mathrm{dz}}{\mathrm{d} \mathrm{x}}\) – 1 = sin z + cos z

Question 15.
Solve \(\frac{d y}{d x}\) = (3x + y + 4)²
Solution:
Given D.E. is \(\frac{d y}{d x}\) = (3x + y + 4)² …….(1)
Put 3x + y + 4 = z
differential w.r.t ‘x’
\(3+\frac{d y}{d x}=\frac{d z}{d x}\)
⇒ \(\frac{d y}{d x}=\frac{d z}{d x}-3\)

Question 16.
Find the equation of the curve whose slope, at any point (x, y) is \(\frac{y}{x^2}\) and which satisfies the condition y = 1, when x = 3.
Solution:
We know that the slope at any point (x, y) on the curve is m = \(\frac{\mathrm{dy}}{\mathrm{dx}}\)
Given that, slope of at any point (x, y) on the curve is m = \(\frac{y}{x^2}\)

Question 17.
Solve (y² – 2xy) dx + (2xy – x²) dy = 0. [May ’01]
Solution:

Question 18.
Solve \(\frac{d y}{d x}=\frac{y^2-2 x y}{x^2-x y}\). [Mar. ’19 (AP)]
Solution:
Given D.E is \(\frac{d y}{d x}=\frac{y^2-2 x y}{x^2-x y}\) ……..(1)
The given equation is a homogeneous equation since both the numerator and denominator are homogeneous functions of degree 2.
Now, put y = vx
Diff. w.r. to x, we get



This is the required general solution of the given equation.

Question 19.
Find the equation of a curve whose gradient is \(\frac{d y}{d x}=\frac{y}{x}-\cos ^2 \frac{y}{x}\), where x > 0, y > 0 and which passes through the point (1, \(\frac{\pi}{4}\)). [(AP) May ’19, ’16, (TS) ’16]
Solution:
Given D.E. is \(\frac{d y}{d x}=\frac{y}{x}-\cos ^2 \frac{y}{x}\) ………(1)
The given equation is a homogeneous equation since both the numerator and denominator are homogeneous functions of degree 1.
Put y = vx
diff. w.r.t ‘x’ on both sides

Question 20.
Give the solution of \(x \sin ^2\left(\frac{y}{x}\right) d x=y d x-x d y\) which passes through the point (1, \(\frac{\pi}{4}\)). [Mar. ’14]
Solution:
Given differential equation is

The given equation is a homogeneous equation since both the numerator and denominator are homogeneous functions of degree 1.
Put y = vx
Diff. w.r. to ‘x’ on both sides


which is the required particular solution.

Model II – Problems on Homogeneous D.E.

Question 21.
Solve (x² – y²) dx – xy dy = 0. [(AP) May ’17; Mar. ’09]
Solution:

Question 22.
Solve \(\frac{d y}{d x}=\frac{(x+y)^2}{2 x^2}\). [Mar. ’05]
Solution:

Question 23.
Solve (x² – y²) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = xy
Solution:
Given D.E. is (x – y ) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = xy
\(\frac{d y}{d x}=\frac{x y}{x^2-y^2}\) ………(1)
This is a homogeneous equation since both the nr and dr are homogeneous functions of degree 2.
Let y = vx

⇒ -x² = 2y² log(cy)
⇒ x² + 2y² log(cy) = 0
Which is the required general solution.

Question 24.
Solve (x² + y²) dx = 2xy dy. [(AP) Mar. ’20, Mar. ’17, ’16]
Solution:
Given diff. equation is \(\frac{d y}{d x}=\left(\frac{2 x y}{x^2+y^2}\right)^{-1}\) …….(1)
Clearly, it is a homogeneous diff. equation of degree zero.

which is the required general solution.

Question 25.
Solve y² dx + (x² – xy + y²) dy = 0.
Solution:
Given D.E is y² dx + (x² – xy + y²) dy = 0
(x² – xy + y²) dy = -y² dx


Which is the required general solution.

Question 26.
Solve \(\frac{d y}{d x}+\frac{y}{x}=\frac{y^2}{x^2}\)
Solution:

Question 27.
Solve x dy = \(\left(y+x \cos ^2 \frac{y}{x}\right) d x\). [Mar. ’13, ’11]
Solution:
Given diff. equation is
x dy = \(\left(y+x \cos ^2 \frac{y}{x}\right) d x\)
\(\frac{d y}{d x}=\frac{y+x \cos ^2 \frac{y}{x}}{x}=\frac{y}{x}+\cos ^2 \frac{y}{x}\) …….(1)
Clearly, it is a homogeneous differential equation of degree zero.
Let y = vx then v + x \(\frac{d v}{d x}=\frac{d y}{d x}\)
From (1), x \(\frac{\mathrm{dv}}{\mathrm{dx}}\) = v + cos²v – v = cos²v
sec²v dv = \(\frac{\mathrm{dx}}{\mathrm{x}}\)
Integrating on both sides
∫sec²v dv = ∫\(\frac{\mathrm{dx}}{\mathrm{x}}\)
tan v = log x + c
⇒ tan(\(\frac{y}{x}\)) = log x + c
Which is the required general solution.

Question 28.
Solve (2x – y) dy = (2y – x) dx. [Mar. ’12]
Solution:
Given diff. equation is \(\frac{d y}{d x}=\frac{2 y-x}{2 x-y}\) …….(1)
Clearly, it is a homogeneous diff. equation of degree zero.

Which is the required general solution.

Question 29.
Solve xy² dy – (x³ + y³) dx = 0. [(TS) May ’18]
Solution:
Given D.E. is xy² dy – (x³ + y³) dx = 0
xy² dy = (x³ + y³) dx
\(\frac{d y}{d x}=\frac{x^3+y^3}{x y^2}\) ……(1)
The given equation is a homogeneous equation since both the numerator and denominator are homogeneous functions of degree 3.
Put y = vx
diff. w.r.t ‘x’ on both sides

Which is the general solution of the given equations.

Question 30.
Solve (x³ – 3xy²) dx + (3x²y – y³) dy = 0. [(AP) May ’18; May ’14]
Solution:
Given D.E is (x³ – 3xy²) dx + (3x²y – y³) dy = 0
(3x²y – y³) dy = -(x³ – 3xy²) dx
\(\frac{d y}{d x}=\frac{-\left(x^3-3 x y^2\right)}{3 x^2 y-y^3}\)
\(\frac{d y}{d x}=\frac{3 x y^2-x^3}{3 x^2 y-y^3}\) ……..(1)
It is a homogeneous equation since both nr and dr are homogeneous functions of degree 3.
Put y = vx
Then diff. w.r. t. x



Which is the required solution of the given equation.

Question 31.
Solve (x²y – 2xy²) dx = (x³ – 3x²y) dy. [Mar. ’18 (AP)]
Solution:

Question 32.
Solve \(\frac{d y}{d x}=\frac{2 x-y+1}{x+2 y-3}\)
Solution:
Given D.E. is \(\frac{d y}{d x}=\frac{2 x-y+1}{x+2 y-3}\) …….(1)
Comparing (1) with \(\frac{d y}{d x}=\frac{a x+b y+c}{a^1 x+b^1 y+c^1}\)
We get a = 2, b = -1, c = 1
a’ = 1, b’ = 2, c’ = -3
Now b = -1 = -(1) = -a’
∴ b = -a’
Now (x + 2y – 3) dy = (2x – y + 1) dx
⇒ x dy + 2y dy – 3dy = 2x dy – y dx + dx
⇒ x dy + 2y dy – 3 dy – 2x dx + y dx – dx = 0
⇒ (x dy + y dx) + 2y dy – 3 dy + 2x dx – dx = 0
⇒ d(xy) + 2y dy – 3 dy + 2x dx – dx = 0
By integrating, we get
∫d(xy) + 2∫y dy – 3 ∫1 dy + 2 ∫x dx – ∫dx = c
xy + 2 . \(\frac{y^2}{2}\) – 3y + 2 . \(\frac{x^2}{2}\) – x = c
xy + y² – 3y + x² – x = c
Which is the required solution.

Model III – Problems on non-homogeneous D.E.

Question 33.
Solve \(\frac{d y}{d x}=\frac{-3 x-2 y+5}{2 x+3 y+5}\)
Solution:
Given diff. equation is \(\frac{d y}{d x}=\frac{-3 x-2 y+5}{2 x+3 y+5}\) …….(1)
Comparing it with \(\frac{d y}{d x}=\frac{a x+b y+c}{a^{\prime} x+b^{\prime} y+c^{\prime}}\)
We get a = -3, b = -2, c = 5, a’ = 2, b’ = 3, c’ = 5
Then, we can solve equation (1) by the case (i)
∴ b = -a’
\(\frac{d y}{d x}=\frac{-3 x-2 y+5}{2 x+3 y+5}\)
⇒ dy (2x + 3y + 5) = dx (-3x – 2y + 5)
⇒ 2x dy + 3y dy + 5 dy + 3x dx + 2y dx – 5 dx = 0
⇒ 2(x dy + y dy) + 3y dy + 5 dy + 3x dx – 5 dx = 0
⇒ 2d(xy) + 3y dy + 5 dy + 3x dx – 5 dx = 0
Integrating on both sides, we get
2∫d(xy) + 3∫y dy + 5∫dy + 3∫x dx – 5∫dx = c
⇒ 2xy + 3 \(\frac{y^2}{2}\) + 5y + 3 \(\frac{x^2}{2}\) – 5x = c
⇒ 4xy + 3y² + 10y + 3x² – 10x = 2c = k
(k = constant)
Which is the required general solution.

Question 34.
Solve 2(x – 3y + 1) \(\frac{d y}{d x}\) = 4x – 2y + 1
Solution:
Given D.E is 2(x – 3y + 1) \(\frac{d y}{d x}\) = 4x – 2y + 1
\(\frac{d y}{d x}=\frac{4 x-2 y+1}{2 x-6 y+2}\) …….(1)
Comparing (1) with \(\frac{d y}{d x}=\frac{a x+b y+c}{a^{\prime} x+b^{\prime} y+c^{\prime}}\)
We get a = 4, b = -2, c = 1
a’ = 2, b’ = -6, c’ = 2
Now, b = -2 = -(2) = -a’
∴ b = -a’
Hence we can solve by the case (1).
(2x – 6y + 2) dy = (4x – 2y + 1) dx
⇒ 2x dy – 6y dy + 2 dy = 4x dx – 2y dx + 1 dx
⇒ 2x dy – 6y dy + 2 dy – 4x dx + 2y dx – 1 dx = 0
⇒ 2(x dy + y dx) – 6y dy – 4x dx + 2dy – 1 dx = 0
⇒ 2d(xy) – 6y dy – 4x dx + 2 dy – dx = 0
By integrating we get,
2∫d(xy) – 6∫y dy – 4∫x dx + 2∫dy – ∫dx = c
⇒ 2(xy) – 6(\(\frac{y^2}{2}\)) – 4(\(\frac{x^2}{2}\)) + 2y – x = 0
⇒ 2xy – 3y² – 2x² + 2y – x = c
Which is the required solution.

Question 35.
Solve x \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + 2y = log x
Solution:
Given D.E. is x \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + 2y = log x
dividing on both sides by ‘x’
\(\frac{d y}{d x}+\frac{2 y}{x}=\frac{\log x}{x}\)
Which is in the form \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + Py = Q
It is linear in y.

Model IV(a) – Problems on L.D.E in y

Question 36.
Solve x log x \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y = 2 log x. [(AP) Mar. ’20; May ’14]
Solution:

Question 37.
Solve (1 + x²) \(\frac{d y}{d x}\) + y = \(e^{\tan ^{-1} x}\). [(AP) Mar. ’18, ’16; May ’16, (TS) Mar. ’15 (TS); May ’13]
Solution:

Question 38.
Solve \(\frac{d y}{d x}\) = y tan x + e^x sec x. [(TS) May ’19; Mar. ’09]
Solution:
Given diff. equation is
\(\frac{d y}{d x}\) – y tan x = e^x sec x ……..(1)
It is a linear diff. equation in y.
Comparing it with \(\frac{d y}{d x}\) + Py = Q
We get P = -tan x, Q = e^x sec x
I.F. = e^2∫Pdx
= e^{-∫tan x dx}
= e^{log|cos x|}
= cos x
The solution of (1) is,
y (I.F) = ∫Q(I.F) dx + c
y (cos x) = ∫e^x sec x (cos x) dx + c = ∫e^x + c
y (cos x) = e^x + c
Which is the required solution.

Question 39.
Solve \(\frac{d y}{d x}\) + y sec x = tan x. [May ’10]
Solution:
Given diff. equation is
\(\frac{d y}{d x}\) + y sec x = tan x
It is a linear diff. equation in y.
It compared with \(\frac{d y}{d x}\) + Py = Q
We get P = sec x, Q = tan x
I.F. = e^∫Pdx
= e^{∫sec x dx}
= e^{log|sec x + tan x|}
= sec x + tan x
The solution of (1) is,
y (I.F) = ∫Q(I.F) dx + c
y(sec x + tan x) = ∫tan x (sec x + tan x) dx + c
= ∫sec x tan x dx + ∫tan²x dx + c
= ∫sec x tan x dx + ∫(sec²x – 1) dx + c
= sec x + tan x – x + c
Which is the required solution.

Question 40.
Solve \(\frac{d y}{d x}\) + y tan x = sin x. [(TS) Mar. ’16, ’12]
Solution:
Given diff. equation is
\(\frac{d y}{d x}\) + y tan x = sin x ……..(1)
It is a linear diff. equation in y.
Comparing it with \(\frac{d y}{d x}\) + Py = Q
Where P = tan x, Q = sin x
I.F = e^∫Pdx
= e^{∫tan x dx}
= sec x
The solution of (1) is,
y(I.F) = ∫Q(I.F) dx + c
y sec x = ∫sin x . sec x dx + c
y sec x = ∫tan x dx + c
y sec x = log|sec x| + c
Which is the required solution.

Question 41.
Solve \(\frac{d y}{d x}\) + y tan x = cos³x. [(AP) May ’19, ’18; Mar. ’17]
Solution:
Given diff. equation is \(\frac{d y}{d x}\) + + y tan x = cos³x
It is a linear diff. equation in y of the first order.
Comparing it with \(\frac{d y}{d x}\) + Py = Q
we get P = tan x, Q = cos³x

Which is the required solution.

Question 42.
Solve cos x . \(\frac{d y}{d x}\) + y sin x = sec²x. [Mar. ’19 (TS); Mar. ’14]
Solution:
Given differential equation is
cos x . \(\frac{d y}{d x}\) + y sin x = sec²x
Dividing on both sides by cos x, we get
\(\frac{d y}{d x}\) + y tan x = sec³x
Which is in the form of \(\frac{d y}{d x}\) + Py = Q
It is linear in ‘y’
Here, P = tan x, Q = sec³x
IF = e^∫Pdx
= e^{∫tan x dx}
= e^{log|sec x|}
= sec x
∴ Solution is y(I. F) = ∫Q(I. F) dx + c
y sec x = ∫sec²x . sec²x dx + c
= ∫sec⁴x dx + c
= ∫sec²x . sec²x dx + c
= ∫(1 + tan²x) sec²x dx + c
= ∫(1 + t²) dt + c
Put tan x = t ⇒ sec²x dx = dt
= ∫1 dt + ∫t² dt + c
= t + \(\frac{\mathrm{t}^3}{3}\) + c
y sec x = tan x + \(\frac{1}{3}\) tan³x + c

Question 43.
Solve (1 + x²) \(\frac{d y}{d x}\) + y = tan^-1x. [(AP) May ’16]
Solution:

Question 44.
Solve (1 + x²) \(\frac{d y}{d x}\) + 2xy – 4x² = 0. [Mar. ’06]
Solution:
Given diff. equation is
\(\frac{d y}{d x}+\frac{2 x y}{1+x^2}=\frac{4 x^2}{1+x^2}\) …….(1)
It is a linear diff. equation in y.
Comparing it with \(\frac{d y}{d x}\) + Py = Q

Which is the required solution.

Question 45.
Solve \(\frac{d y}{d x}+y\left(\frac{4 x}{1+x^2}\right)=\frac{1}{\left(1+x^2\right)^2}\). [(AP) May ’17]
Solution:

Question 46.
Solve \(\frac{d y}{d x}+\frac{3 x^2}{1+x^3} y=\frac{1+x^2}{1+x^3}\)
Solution:

Question 47.
Solve \(\frac{1}{x} \frac{d y}{d x}+y e^x=e^{(1-x) e^x}\). [(AP) May ’18]
Solution:

Question 48.
Solve x(x – 1) \(\frac{d y}{d x}\) – y = x³(x – 1)³. [(AP) Mar. ’19]
Solution:

Model IV(b) – Problems on L.D.E in x

Question 49.
Solve (1 + y²) dx = (tan^-1y – x) dy. [Mar. ’18 (TS); May; Mar. ’15 (AP); May ’15 (TS)]
Solution:

Question 50.
Solve (x + 2y³) \(\frac{d y}{d x}\) = y. [May ’12]
Solution:

Question 51.
Solve (x + y + 1) \(\frac{d y}{d x}\) = 1. [Mar. ’17 (TS); Mar. ’13]
Solution:
Given diff. equation is \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{x}+\mathrm{y}+1}\)
\(\frac{d x}{d y}\) – x = y + 1 ……..(1)
It is a linear diff. equation in x.
Comparing it with \(\frac{d x}{d y}\) + Px = Q
Where P = -1, Q = y + 1
I.F. = e^∫Pdy = e^{∫(-1) dy} = e^-y
The solution of (1) is
x(I.F) = ∫Q(I.F) dy + c

Which is the required general solution.

Model V – Problems on Bernoulli’s D.E.

Question 52.
Solve \(\frac{d y}{d x}\) (x²y³ + xy) = 1. [Mar. ’11]
Solution:

Telangana TSBIE TS Inter 2nd Year English Study Material Grammar Punctuation Exercise Questions and Answers.

TS Inter 2nd Year English Grammar Punctuation

(Symbols- .,,) to be used out of 10 necessary

Q.No. 10 (8 × 1/2 = 4 Marks)

Communication is not simply uttering words. It’s just one aspect of it. Communication is an art that only a few people truly understand. It includes signs we may convey indirectly, such as body language. Many people believe it is a more effective medium of communication. Body language is influenced by where we stop, the words we focus on, and our facial expressions.

However, you cannot use body language when writing. Despite this, many people are able to express themselves beautifully through this medium. Punctuation serves as body language in written form, empowering our emotions. This is made up of wordplay, pauses, exclamations, and introspection.

Here, we’ll become acquainted with the various types of punctuation marks used in the English language. They all serve a different purpose, and their proper use can improve the meaning of a sentence. They not only give your words more substance, but also the kind of depth that can only be conveyed verbally. We’ll take a quick look at all of these punctuation marks and how they’re used in written communication.

We’ll break them down for you so you don’t have any doubts.

Punctuations are an important part of written communication, and learning about them will help you understand how to use them correctly in sentences. To grasp Punctuation knowledge, re-read the content you write and try to check whether you are using Punctuation correctly or not, as this is the most common mistake we all make at times. You’ll be surprised by the results. It’s a fascinating experiment that everyone should try.

Play with punctuation

Let us understand and relish the following sets of sentences.

1. Stop, not hang him.
Stop not, hang him.
(A comma kills a man. Punctuation matters!)

2. Woman, without her man, is nothing.
(Woman is nothing unless there is a man.)
Woman! Without her, man is nothing.
(Man is nothing unless there is a woman.)

3. Let’s eat Grandpa.
(Grandpa is going to be eaten!)
Let’s eat, Grandpa.
(inviting grandpa to eat)

4. Akbar said Ashoka was a great warrior.
(Ashoka was a great warrior.)
“Akbar”, said Ashoka, “was a great warrior.”
(Akbar was a great warrior.)

5. This section consists of seven- year-old children.
(The children are aged seven.)
This section consists of seven year-old children.
(There are seven children who are one-year old.)

6. These are my employees.
(These people are my employees.)
These are my employee’s.
(These things belong to my employee.)
These are my employees’.
(These things belong to my employees;)

The above sentences clearly indicate how punctuation marks are helpful in conveying the correct meaning.

Punctuation marks, like traffic signals, are visual indicators used in written/ printed texts to make texts meaningful.

Types of punctuation: There are three types of punctuation. They are:

Now, let us look at their specific functions in detail.

A. End punctuation
1. Full stop (.)
It indicates the most extended pause and is always placed at the end of a sentence. It is used
i) at the end of declarative and imperative sentences.
Naveen is a university student.
Come here.

ii) after most abbreviations and initials.
M.A. (Master of Arts), B.Sc. (Bachelor of Science),
a.m. (ante meridian)

iii) to separate hour from minute and date from month and year.
The class begins at 7.45 a.m. daily.
Vanitha was born on 15.12.2015.

Note: Full stops are omitted (NOT used)
i) in acronyms (abbreviations pronounced as complete words).
NATO, UNESCO, VAT

ii) when the capital letter in the abbreviation doesn’t stand for a complete word.
TV (Television), TB (Tuberculosis)

iii) in newspaper headlines.
Art is more than unbroken lines
Heat wave condition persists until May

2. Question mark (?)
It is used
i) at the end of an interrogative sentence.
Do you like sweets?

ii) after question tags or similar words.
He likes music, doesn’t he? This is your car, right?

iii) after el I iptical questions. .
Doing well? In trouble?
Note: A question mark is not used at the end of an indirect question.
The principal asked me where I had gone

3. Exclamation mark (!)
It is used .
i) after an emotional expression of joy/sorrow/surprise/shock/anger, etc.
How fabulous the movie is!
What an awful experience it was!

ii) after an interjection or one-word exclamation.
Pity! She lost her father.
Hurrah! We won the match.

iii) after an imperative sentence when it is charged with feeling (a strong : command).
Shut up! Go and bring your notebook.
Get lost! I need no explanation.

Excercise 

I. Use appropriate end punctuation marks (full stop question mark / exclamation mark) with the following sentences.

Questions:
1) Will you show me the book
2) How intelligent you are
3) You like English, don’t you
4) Stop the bus for me
5) It is raining now
6) What an idea
7) Was the meal nourishing
8) Hold me I am going to faint
9) The dam burst Run for your lives
10) Watch out A car is coming
Answers:
1) Will you show me the book?
2) How intelligent you are!
3) You like English, don’t you?
4) Stop the bus for me.
5) It is raining now.
6) What an idea!
7) Was the meal nourishing ?
8) Hold me! I am going to faint.
9) The dam burst! Run for your lives.
10) Watch out! A car is coming.

II. Read the following paragraph and use appropriate end punctuation marks (full stop/ question mark/exclamation mark) wherever necessary.

Once upon a time there lived a duck and a kangaroo They were friends One day the duck asked the Kangaroo, ‘Dear friend, “How do you jump” The Kangaroo replied, “Ah it is very easy Do you love it” The duck said, “Oh is it I love to jump like you then The Kangaroo said, “Sit on my back I will take you round the world”
Answer:
Once upon a time there lived a duck and a kangaroo. They were friends. One day the duck asked the kangaroo, “Dear friend, How do you jump?” The Kangaroo replied, “Ah! It is very easy. Do you love it ?” The duck said, “Oh! Is it ? I love to jump like you then. The kangaroo said, “Sit on my back. I will take you round the world.”

Exercise

1. Use appropriate internal punctuation marks (semicolon / colon /comma/ quotation marks / ellipses) in the following sentences.

Questions:
1. The monsoon failed this year too and the country is in the grip of a famine.
2. I wasn’t just annoyed I was absolutely furious
3. Sumit said, Where are you going?
4. She worked hard she failed.
5. August 151947. It was the day on which we won independence.
6. The man said I wasn’t wounded.
7. He was not stupid just passive.
8. The grand tour usually included Paris France Vienna Austria Rome and Italy
Answers:
1. The monsoon failed this year too: and the country is in the grip of a famine.
2. I wasn’t just annoyed; I was absolutely furious
3. Sumit said, “Where are you going?
4. She worked hard: she failed.
5. August 151947-It was the day on which we won independence.
6. The man said, “I wasn’t wounded.
7. He was not stupid just passive.
8. The grand tour usually included Paris France,Vienna, Austria, Rome, and Italy.

Excercise – A

I. Use appropriate word punctuation marks (apostrophe / hyphen/capitals) in the following sentences.ellipses) in the following sentences.

Questions:
1. this is my cousins car,
2. gandhiji led the non violent movement
3. there are forty-six boys in the class.
4. these are my father in laws clothes.
5. we have semi-skilled workers
6. iam a member of the officers club.
7. he wants an up to date account
8. they prefer sugar free sweet.
Answers:
1. This is my cousins car,
2. Gandhiji led the non violent movement
3. There are forty-six boys in the class.
4. These are my father in laws clothes.
5. We have semi-skilled workers
6. I am a member of the officers club.
7. He wants an up to date account
8. They prefer sugar free sweet.

II. Read the following paragraph and use appropriate word punctuation marks (apostrophe / hyphen / capitals) wherever necessary.

there was an old owl. everyday he used to see some incidents happening around him. yesterday he saw a boy helping his mother-in-law. today he saw him shouting at her. the boys father in law was kind and gentle, the boy shouted at his father in law was kind and gentle, the boy shouted at his father in law too. the owls curiosity grew more and more to know about the boy.
Answer:
There was an old owl. Everyday he used to see some incidents happening around him. Yesterday he saw a boy helping his mother-in-law. today he saw him shouting at her. The boys father in law was kind and gentle. The boy shouted at his father in law too. The owls curiosity grew more and more to know about the boy.

Exercises – B

I. Punctuate the following letter.

24-7/A
bank street
hyderabad
27 October 2015,

To
the editor
box no 128
the hindu
Hyderabad

dear sir,
with reference to your advertisement in todays newspaper for the post of an incharge of ads section i would like to apply for it before that let me be known of the details of the interview like date time and venue.

yours faithfully,
manisharma,
Answer:
24-7/A
Bank street,
Hyderabad
27 October 2015,

To
The editor
box no 128
The hindu
Hyderabad

Dear sir,

with reference to your advertisement in today’s newspaper for the post of an incharge of ads section i would like to apply for it before that let me be known of the details of the interview like date time and venue.

yours faithfully,
manisharma.

II. Punctuate the following dialogue.

Iasya ………… hai kavya how are you
kavya ………… fine what about you not seen for a week
Iasya ………… ive been to my grandmas village for vacation
kavya ………… oh how did you feel there
Iasya ………… fabulous what a pleasant life it was greenery cool breeze everywhere
kavya ………… you are right but we are living in towns noting but concrete
Iasya ………… but why dont we concentrate on planting
kavya ………… good idea why dent we start first
Iasya ………… ok lets meet here tomorrow again
kavya ………… ok bye see you
Answer:
Iasya ………… Hai Kavya. How are you?
kavya ………… Fine. What about you? Not seen for a week?
Iasya ………… I’ve been to my grandma’s village for vacation.
kavya ………… Oh! How did you feel there?
Iasya ………… Fabulous! What a pleasant life it was! Greenery, cool breeze everywhere
kavya ………… You are right. But we are living in towns. Nothing but concrete jungles!
Iasya ………… But, why shouldn’t we concentrate on planting?
kavya ………… Good idea! Why shouldn’t we start first?
Iasya ………… OK. Let’s meet here tomorrow again,
kavya ………… OK. Bye! See you.

III. There should be ten punctuation marks in the following paragraph. Try to insert at least eight of them.

two weeks ago i was amused when a friend who couldnt bear to sleep alone, woke me up close to midnight at the hotel into which we had checked in. can we hire a double room im
totally spooked, she said
Answer:
Two weeks ago, I was amused when a lriend who couldn’t bear to sleep alone, woke me up close to midnight, at the hotel into which we had checked in. “Can we hire a double room? I’m totally spooked,” she said.

Exercise(From the prescribed prose Lesssons)

Rewrite the following sentence / sentences using punctuation marks where necessary.

Question 1.
aristotle could have aooided the mistake of thinking that women have fewer teeth than men by the simple device ot asking mrs aristotle to keep her mouth open while he counted
Answer:
Aristotle could have avoided the mistake of thinking that women have fewer teeth than men by the simple device of asking Mrs. Aristotle to keep her mouth open while he counted.

Question 2.
if someone maintains that two and two are five or that Iceland is on the equator you feel pity rather than anger unless you know so little of arithmetic or geography that his opinion shakes your own contrary conviction
Answer:
If someone maintains that two and two are five or that Iceland is on the equator, you feel pity rather than anger unless you know so little of arithmetic or geography that his opinion shakes your own contrary conviction.

Question 3.
When i was young I lived much outside my own country in france germany italy and the united states.
Answer:
When I was young, I lived much outside my own country in France, Germany, Italy, and the United States.

Question 4.
mahatma gandhi deplores railways and steam boats and machinery he would like to. undo the whole of the Industrial Revolution you may never have an opportunity of actually meeting anyone who holds this opinion because in western countries most people take advantage of modern techniques forgranted
Answer:
Mahatma Gandhi deplores railways and steamboats, and machinery; he would like to undo the whole of the Industrial Revolution. You may never have an opportunity of actually meeting anyone who holds this opinion because, in Western countries, most people take advantage of modem techniques for granted.

Question 5.
be very wary of opinions thatflatteryourselfesteem both men and women, nine times out of ten are firmly convinced of the superior excellence of their own sex
Answer:
Be very wary of opinions that flatter your self-streern. Both men and women, nine times out of ten, are firmly convinced of the superior excellence of their own sex.

Question 6.
from the earliest days there had been many notable women in India poets scholars capable administrators and leaders of religious movements
Answer:
From the earliest days there had been many notable women in India – poets, scholars, capable administrators and leaders of religious movements.

Question 7.
women all over india came forward defying all social taboos sacrificing physical comforts and denying the validity of all restrictions which had been enforced against them to take up every kind of work connected with national movement
Answer:
Women all over India came forward, defying all social taboos, sacrificing physical comforts, and denying the validity of all restrictions which had been enforced against them, to take up every kind of work connected with national movement.

Question 8.
It was a matter of surprise to the outside world that independent india should have appointed women to the highest posts so freely as members of the cabinet as governors of provinces as ambassadors and as leaders of delegations to international conferences for in an oriental country such as india women are presumed to be held in subjection and therefore all this seemed to be unnatural.
Answer:
It was a matter of surprise to the outside world that independent India should have appointed women to the highest posts so freely, as members of the Cabinet, as Governors ofProvinces, as Ambassadors and as leaders of delegations to international conferences, for in an oriental country such as India, women are presumed to be held in subjection and therefore all this seemed to be unnatural.

Question 9.
originally he seems to have been uncertain of the response or at least of the kind of work that women could do in the national movement for t hough he was a passionate believer in the equality of women he seems to have been doubtful whether the women of india
Answer:
Originally, he seems to have been uncertain of the response, or at least of the kind of work that women could do in the national movement, for though he was a passionate believer in the equality of women, he seems to have been doubtful whether the women of India….

Question 10.
what the gandhian movement did was to release women from the social bandages that custom had imposed and conservatism had upheld
Answer:
What the Gandhian movement did was to release women from the social bond ages that custom had imposed and conservatism had upheld.

Question 11.
after the citys rain fed potholes the transition to smooth roads within the campus of the thiagarajar college of engineering is more than a treat the man behind the tar topped tracks is known as Madurai’s Plastic Road Man.
Answer:
After the city’s rain-fed potholes, the transition to smooth roads within the campus of the Thiagarajar College of Engineering (fCE) is more than a treat. T.he man behind the tar-topped tracks is known as Madurai’s Plastic Road Man.

Question 12.
The day we met Dr Vasudevan he was waving a special gazette notification of the ministry of environment & forests dated 4th February 2011 directing all municipal authorities across the country to encourage use of plastic waste by adopting suitable technology” such as in road construction.
Answer:
The day we met Dr. Vasudevan, he was waving a special gazette notification of the Ministry of Environment & Forests, dated 4th February, 2011, directing all municipal authorities across the country to “encourage the use of plastic waste by adopting suitable technology such as in road construction…”

Question 13.
dr kalams words proved prophetic with full support from the college correspondent Dr.oasudeoan laid the first 60 feet long plastic road within the campus
Answer:
Dr. Kalam’s words proved prophetic. With full support from the College Correspondent, Dr. Vasudevan laid the first 60 feet long plastic road within the campus.

Question 14.
the jamshedpur utilities and services company a tata enterprise approached Dr. Vasudevan last november for using plastic waste in laying roads in jamshedpur.
Answer:
The Jamshedpur Utilities and Services Company, a Tata enterprise approached Dr. Vasudevan for using plastic in laying roads in Jamshedpur,

Question 15.
the view of this world which india has taken is summed up in one compound Sanskrit word sacchidananda the meaning is that reality which is essentially one has three phases.
Answer:
The view of this world which India has taken is summed up in one compound Sanskrit word, Sacchidananda. The meaning is that Reality, which is essentially one, has three phases.

Question 16.
The first is sat it is the simple fact that things are the fact which relates us to all things through the relationship of common. existence The second is chit it is the fact that we know which relates us to all things, through the relation ship of knowledge the third is ananda it is the fact that we enjoy which unites us with all things through the relationship of love.
Answer:
The first is sat: it is the simple fact that things are, the fact which relates us to all things through the relationship of common existence. The second is chit: it is the fact that we know, which relates us to all things through the relationship of knowledge. The third is ananda: it is the fact that we enjoy, which unites us with all things through the relationship of love.

Question 17.
In kalidasas drama shakuntala the hermitage which dominates the play overshadowing the kings palace has the same idea running through it – the
recognition of the kinship of man with conscious and unconscious creation alike
Answer:
In Kalidasa’s drama, Stiakuntala, the hermitage, which dominates the play, ” overshadowing the king’s palace, has the same idea running through it – – the recognition of the kinship of man with conscious and unconscious creation alike.

Question 18.
in the west era dramas human characters drown our attention in the vortex of their passions nature occasionally peeps out but she is almost always a trespasser who has to offer excuses or bow apologetically and depart
Answer:
In the Western dramas, human characters drown our attention in the vortex of their passions. Nature occasionally peeps out, but she is almost always a trespasser, who has to offer excuses, or bow apologetically and depart.

Question 19.
but in all our dramas which still retain their fame such as Mrit- Shakatika Shakuntala Uttara ramachatita nature stands on her own right proving that she has her great function to impart, the peace of the eternal to human emotions
Answer:
But in all our dramas which still retain their fame, such as Mrit-Shakatika, Shakuntala, Uttara-Ramacharita, Nature stands on her own right, proving that she has her great function, to impart the peace of the eternal to human emotions.

Question 20.
yes mrs moore (opens door and talks loudly) come right in
Answer:
Yes, Mrs. Moore! (Opens door and talks lougly.) Come right in.

Question 21.
yes it was months and months before he had any sort of a job i worried a bit you know i was afraid he would get so discouraged he would
Answer:
Yes, it was months and months before he had any sort of a job. I worried a bit, you know, 1 was afraid he would get so discouraged…. he would….

Question 22.
no firm is going to smile and say forget it my boy no they dont do things that way these days
Answer:
No firm is going to smile .and say, ‘Forget it, my boy’ no, they don’t do things that way these days.

Question 23.
mr van king had brought it down to his office intending to have it reset – and put it in his office safe and forgot it imagine forgetting a diamond heirLoom worth a fabuLous fortune
Answer:
Mr Van King had broughtit down to his office, intending to have it re-set – – and put it in his office safe, and -forgot it! Imagine forgetting a diamond heirloom worth a fabulous fortune!

Question 24.
dont worry not a souL saw us wait ill get it (10)
Answer:
Don’t worry. Not a soul saw us! Wait, I’ll get it.

Question 25.
are you surejim that you put itthere oh oh… i see (laughs) you have it
Answer:
Are you sure Jim, that you put it there?… Oh, oh…. I see… You have it!

Question 26.
yes but you keep out of this youll never get the diamond jim poor Larry wiLson was sentenced to ten years in jail today… I” stealing.
Answer:
Yes! But you keep out of this, you’ll never get the diamond! Jim, poor Larry Wilson was sentenced to ten years in jail today… stealing.

Question 27.
JIM: ma you dont think i stoLe that diamond
MA: what eLse am i to think
Answer:
Jim: Ma! You don’t think I stole that diamond?
Ma: What else am I to think?

Question 28.
just what are you going to do with the diamond no jim keep quiet a moment mrs ryan what are you going to do with that diamond
Answer:
Just what are you going to do with the diamond? No. Jim, keep quiet a moment. Mrs. Ryan, what are you going to do with that diamond?

Question 29.
I promise but jitn i i (tries not to cry) ill never forget this night never(J0)
Answer:
I promise. But Jim, I…. I…. (tries not to cry) I’ll never forget this night, never.

Question 30.
a) King dilipa with Queen sudakshina has entered upon the the life of the forest the great monarch is busy tending the cattle of the hermitage [Model Question Paper]
Answer:
King Dilipa, with Queen Sudakshina, has entered upon the life of the forest. The great monarch is busy tending the cattle of the hermitage.

b) Sum it said where are you going [Model Question Paper] .
Answer:
Sumit said, “Where are you going ?”

Question 31.
It is more difficult to deal with the self esteem of man as a man because we cannot argue out the matter unth some non-human mind, the only way i know of dealing with this general human conceit is to remind ourselves that man is a brief episode in the life of a small planet in a little corner of the universe and that for aught we know other parts of the cosmos may contain beings as superior to ourselves as we are to jellyfish. [REVISION TEST-I]
Answer:
It is more difficult to deal with the self-esteem of man as a man because we cannot argue out the matter with some non- human mind. The only way I know of dealing with this general human conceit is to remind ourselves that man is a brief episode in the life of a small planet in a little corner of the universe and that, for aught we know, other parts of the cosmos may contain beings as superior to ourselves as we are to jellyfish.

Question 32.
The brahmo samaj led the movement for emancipation the ancient rules of purdah were broken and brahmo women moved freely in society: but this was but a false dawn as it was far in advance of popular opinion. [REVISION TEST-II]
Answer:
The Brahmo Samaj led the movement for emancipation. The ancient rules of purdah – were broken and Brahmo women moved freely in society: but this was but a false dawn as it was far in advance of popular opinion.

Question 33.
after a decades hard work and persistent efforts’ his simple invention of a technology ” to use – plastic waste to lay roads patented by TCE finally got a shot in the arm last month with the centre approving its wider application. [REVISION TEST-III]
Answer:
After a decade’s hard work and persistent efforts, his invention of a simple technology to use plastic waste to lay roads, patented by TCE, finally got a shot in the arm with the Centre approving its wider application.

Question 34.
a) King dilipa with Queen sudakshina has entered upon the the life of the forest the great monarch is busy tending the cattle of the hermitage [Model Question Paper]
Answer:
King Dilipa, with Queen Sudakshina, has entered upon the life of the forest. The great monarch is busy tending the cattle of the hermitage.

b) Sum it said where are you going [Model Question Paper]
Answer:
Sumit said, “Where are you going ?”

Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A Measures of Dispersion Important Questions Very Short Answer Type to help strengthen their preparations for exams.

TS Inter Second Year Maths 2A Measures of Dispersion Important Questions Very Short Answer Type

Question 1.
Find the mean deviation from the mean of the discrete data 6, 7, 10, 12, 13, 4, 12, 16. [TS – May 2016, Mar. ’14]
Solution:
Given data is 6, 7, 10, 12, 13, 4, 12, 16
The arithmetic mean of the given data is
\(\overline{\mathrm{x}}=\frac{\text { Sum of the observations }}{\text { No. of observations }}\)
= \(\frac{\Sigma x}{n}=\frac{6+7+10+12+13+4+12+16}{8}\)
\(\bar{x}=\frac{80}{8}\) = 10
The deviations of the respective observations from the mean are 6 – 10, 7 – 10, 10 – 10, 12 – 10, 13 – 10, 4 – 10, 12 – 10, 16 – 10 = – 4, – 3, 0, 2, 3, – 6, 2, 6
The absolute values of the deviations are 4, 3, 0, 2, 3, 6, 2, 6.
∴ The required mean deviation about mean is
M.D = \(=\sum_{i=1}^n \frac{\left|x_i-\bar{x}\right|}{n}=\sum_{i=1}^8 \frac{\left|x_i-\bar{x}\right|}{n}\)
= \(\frac{4+3+0+2+3+6+2+6}{8}\)
= \(\frac{26}{8}=\frac{13}{4}\) = 3.25.

Question 2.
Find the mean deviation about mean for the data 38, 70, 48, 40, 42, 55, 63, 46, 54, 44. [AP – May 2015].
Solution:
Given data is 38, 70, 48, 40, 42, 35, 63, 46, 54, 44
The arithmetic mean of the given data is
\(\overline{\mathrm{x}}=\frac{\text { Sum of the observations }}{\text { No. of observations }}\)
= \(\frac{\Sigma \mathrm{x}}{\mathrm{n}}\)
= \(\frac{38+70+48+40+42+55+63+46+54+44}{10}\)
= \(\frac{500}{10}\) = 50
The deviations of the respective observations from the mean are 38 – 50, 70 – 50, 48 – 50, 40 – 50, 42 – 50, 55 – 50, 63 – 50, 46 – 50, 54 – 50, 44 – 50 = – 12, 20, – 2, – 10, – 8, 5, 13, – 4, 4, – 6
The absolute values of the deviations are 12, 20, 2, 10,8, 5, 13, 4, 4, 6.
∴ The required mean deviation about mean is
M.D. = \(\sum_{i=1}^n \frac{\left|x_i-\bar{x}\right|}{n}=\sum_{i=1}^{10} \frac{\left|x_i-\bar{x}\right|}{n}\)
= \(\frac{12+20+2+10+8+5+13+4+4+6}{10}\)
= \(\frac{84}{10}\) = 8.4.

Question 3.
Find the mean deviation about mean for the data 3, 6, 10, 4, 9, 10.
[AP – Mar. ’18, May ’16; TS – Mar. ’17; May ’14]
Solution:
Given data is 3, 6, 10, 4, 9, 10
The arithmetic mean of the given data is
\(\overline{\mathrm{x}}=\frac{\text { Sum of the observations }}{\text { No. of observations }}\)
= \(\frac{\Sigma \mathrm{x}}{\mathrm{n}}=\frac{3+6+10+4+9+10}{6}\)
= \(\frac{42}{6}\) = 7
The deviations of the respective observations from the mean are 3 – 7, 6 – 7, 10 – 7, 4 – 7, 9 – 7, 10 – 7
= – 4, – 1, 3, – 3, 2, 3
The absolute values of the deviations are 4, 1, 3, 3, 2, 3.
The required mean deviation about mean is
MD = \(\sum_{i=1}^n \frac{\left|x_i-\bar{x}\right|}{n}=\sum_{i=1}^6 \frac{\left|x_i-\bar{x}\right|}{n}\)
= \(\frac{4+1+3+3+2+3}{6}\)
= \(\frac{16}{6}\) = 2.666 = 2.7.

Question 4.
Find the mean deviation about median for the data 13, 17, 16, 11, 13, 10, 16, 11, 18, 12, 17. [AP – Mar. 2016]
Sol.
Given data is 13, 17, 16, 11, 13, 10, 16, 11, 18, 12, 17
The ascending order of the observations in the given data is
10, 11, 11, 12, 13,13, 16,16, 17, 17, 18
Median of one given data is M = 13.
The deviations of the respective observations from the median i.e., x_i – M are
13 – 13, 17 – 13, 16 – 13, 11 – 13, 13 – 13, 10 – 13, 16 – 13, 11 – 13, 18 – 13, 12 – 13, 17 – 13
= 0, 4.3, – 2, 0, – 3, 3, – 2, 5, – 1, 4
The absolute values of the deviations are
0, 4, 3, 2, 0, 3, 3, 2, 5, 14.
∴ The required mean deviation about median is
M.D = \(\sum_{i=1}^n \frac{\left|x_i-M\right|}{n}=\sum_{i=1}^{11} \frac{\left|x_i-M\right|}{n}\)
= \(\frac{0+4+3+2+0+3+3+2+5+1+4}{11}\)
= \(\frac{27}{11}\) = 2.45.

Question 5.
Find the mean deviation about median for the data 4,6, 9, 3, 10,13,2. [TS – Mar. ‘18, ’15, May ’15; AP-Mar. ‘19, ’17]
Solution:
Given data is 4, 6, 9, 3, 10, 13, 2
The ascending order of the observations in given data is
2, 3, 4. 6, 9, 10, 13
Median of given data is M = 6
The deviations of the respective observations from the median i.e., x_i – M are
2 – 6, 3 – 6, 4 – 6, 6 – 6, 9 – 6, 10 – 6, 13 – 6 = – 4, – 3, – 2, 0, 3, 4, 7
The absolute values of the deviations are 4, 3, 2, 0, 3, 4, 7.
∴ The required mean deviation about median is
M.D = \(\sum_{i=1}^n \frac{\left|x_i-M\right|}{n}=\sum_{i=1}^7 \frac{\left|x_i-M\right|}{n}\)
= \(\frac{4+3+2+0+3+4+7}{7}\)
= \(\frac{23}{7}\) = 3.29.

Question 6.
Find the mean deviation about median for the data 6, 7, 10, 12, 13, 4, 12, 16.
Solution:
Given data is 6, 7, 10, 12, 13, 4, 12, 16
The ascending order of the observations in the given data is
4, 6, 7, 10, 12, 12, 13, 16
Median of one given data is M = \(\frac{10+12}{2}\)
= \(\frac{22}{2}\) = 11
The deviations of the respective observations from the median i.e., x_i – M are
6 – 11, 7 – 11, 10 – 11, 12 – 11, 13 – 11, 4 – 11, 12 – 11, 16 – 11 = – 5, – 4,- 1, 1, 2, – 7, 1, 5
The absolute values of the deviations are 5, 4, 1, 1, 2, 7, 1, 5.
∴ The required mean deviation about median is
M.D = \(\sum_{i=1}^n \frac{\left|x_i-M\right|}{n}=\sum_{i=1}^8 \frac{\left|x_i-M\right|}{n}\)
= \(\frac{5+4+1+1+2+7+1+5}{8}\)
= \(\frac{26}{8}=\frac{13}{4}\) = 3.25.

Question 7.
Find the mean deviation about mean for the following data. [AP – Mar. 2019]

Solution:
We can form the following table for the given data:

Here, Σf_i = 40
Σf_ix_i = 320
∴ Mean of the given data is \(\overline{\mathrm{x}}=\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\mathrm{N}}=\frac{320}{40}\) = 8
From the table, Σf_i |x_i – \(\overline{\mathrm{x}}\)| = 140
Mean deviation about mean = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\widetilde{\mathrm{x}}\right|}{\mathrm{N}}=\frac{140}{40}\) = 3.5.

Question 8.
Find the mean deviation about mean for the following data.

Solution:
We can form the following table for the given data:

Here, N = Σf_i = 45
Σf_ix_i = 534
∴ Mean of the given data is \(\overline{\mathrm{x}}=\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\mathrm{N}}=\frac{534}{45}\) = 11.8
From the table, Σf_i |x_i – x| = 33.0
Mean deviation about mean = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\mathrm{N}}\)
= \(\frac{33}{45}\) = 0.73.

Question 9.
Find the mean deviation about mean for the following data.

Solution:

Here, N = Σf_i = 80
Σf_ix_i = 4000
∴ Mean of the given data is \(\bar{x}=\frac{\Sigma f_i x_i}{N}=\frac{4000}{80}\) = 50
From the table, Σf_i |x_i – \(\bar{x}\)| = 1280
Mean deviation about mean \(\frac{\Sigma \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\mathrm{N}}=\frac{1280}{80}\) = 16.

Question 10.
Find the mean deviation about median for the following data. [May ‘14]

Solution:
We can form the following table for the given data:

Median = \(\frac{N^{\text {th }}}{2}\) observation
= \(\frac{30}{2}\) = 15th observation = 13.
Mean deviation about median, M.D = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right|}{\mathrm{N}}=\frac{149}{30}\) = 4.966 = 4.97.

Question 11.
Find the mean deviation about median for the following data.

Solution:
We can form the following table for the given data:

Median = \(\frac{N^{\text {th }}}{2}\) observation
= \(\frac{26}{2}\) = 13th observation = 7.
Mean deviation about median, M.D = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right|}{\mathrm{N}}=\frac{84}{26}\) = 3.23.

Question 12.
Find the variance and standard deviation for the discrete data: 5, 12, 3, 18, 6, 8, 2, 10. [AP – Mar. 2015; TS – Mar. 2019]
Solution:
Given data is 5, 12, 3, 18. 6, 8, 2, 10
The mean of the given data is \(\overline{\mathrm{x}}=\frac{\sum \mathrm{x}_{\mathrm{i}}}{\mathrm{n}}\)
= \(\frac{5+12+3+18+6+8+2+10}{8}=\frac{64}{8}\) = 8
To find the variance we construct the following table:

Here, Σ(x_i – \(\bar{x}\)) = 194
∴ Variance, \(\sigma^2=\frac{1}{n} \sum_{i=1}^8\left(x_i-\bar{x}\right)^2=\frac{194}{8}\) = 24.25
Standard deviation, a = \(\sqrt{\text { variance }}\)
= \(\sqrt{24.25}\)= 4.92 (approxi.)

Question 13.
Find the variance for the discrete data : 6, 7, 10, 12, 13, 4, 8, 12.
Solution:
Given data is 6, 7, 10, 12, 13, 4, 8, 12
The mean of the given data is \(\overline{\mathrm{x}}=\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{\mathrm{n}}\)
= \(\frac{6+7+10+12+13+4+8+12}{8}=\frac{72}{8}\) = 9
To find the variance we construct the following table:

Here, Σ(x_i – \(\bar{x}\))² = 74
∴ variance, \(\sigma^2=\frac{1}{n} \sum_{\mathrm{i}=1}^8\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2=\frac{74}{8}\) = 9.25.

Question 14.
Find the variance for the discrete data: 350, 361, 370, 373, 376, 379, 385, 387, 394, 395.
Solution:
Given data is 350, 361, 370, 373, 376, 379, 385, 387, 394, 395
The mean of the given data is
\(\overline{\mathrm{x}}=\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{\mathrm{n}}=\frac{350+361+370+373+376+379+385+387+394+395}{10}=\frac{3770}{10}\) = 377
To find the variance we construct the following table:

Here, Σ(x_i – x) = 1832
∴ Variance \(\frac{1}{n} \sum_{i=1}^{10}\left(x_i-\bar{x}\right)^2\)
= \(\frac{1832}{10}\) = 183.2.

Question 15.
The variance of 20 observations is 5. If each of the observations is multiplied by 2. Find the variance of the resulting observations. [Board Paper]
Solution:
Let the given observations be x₁, x₂, x₃, …………… x₂₀
Mean, \(\bar{x}=\sum_{i=1}^{20} x_i \times \frac{1}{n}\)
Given that, n = 20 and variance = 5
\(\frac{1}{n} \sum_{i=1}^{20}\left(x_i-\bar{x}\right)^2\) = 5
\(\frac{1}{20} \sum_{i=1}^{20}\left(x_i-\bar{x}\right)^2\) = 5
\(\sum_{i=1}^{20}\left(x_i-\bar{x}\right)^2\) = 100
If each observation is multiplied by 2 then the new observations are 2x₁, 2x₂, 2x₃, …………….., 2x₂₀
i.e., y₁, y₂, y₃, ……………, y₂₀
y_i = 2x₁
The mean of the new observations
\(\overline{\mathrm{y}}=\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{20} \mathrm{y}_{\mathrm{i}}\)
= \(\frac{1}{20} \sum_{\mathrm{i}=1}^{20} 2 \mathrm{x}_{\mathrm{i}}=2\left[\frac{1}{20} \sum_{\mathrm{i}=1}^{20} \mathrm{x}_{\mathrm{i}}\right]=2 \overline{\mathrm{x}}\)
⇒ \(\overline{\mathrm{x}}=\frac{\overline{\mathrm{y}}}{2}\)
Substituting the values of \(\bar{x}_{\mathbf{i}}\) and \(\overline{\mathbf{x}}\) in (1), we get
\(\sum_{\mathrm{i}=1}^{20}\left(\frac{\mathrm{y}_{\mathrm{i}}}{2}-\frac{\overline{\mathrm{y}}}{2}\right)^2\) = 100
\(\frac{1}{4} \sum_{i=1}^{20}\left(y_i-\bar{y}\right)^2\) = 100
\(\sum_{i=1}^{20}\left(y_i-\bar{y}\right)^2\) = 400
∴ The variance of the new observations = \(\frac{1}{n} \sum_{i=1}^{20}\left(y_i-\bar{y}\right)^2\)
= \(\frac{1}{20}\) (400) = 20 = 2² . 5

Note:
If each observation is multiplied by a constant ‘k’, the variance of the resulting observation becomes k² times the original variance.

Question 16.
If each of the observation x₁, x₂, ………………, x_n is increased by k, where k is a positive or negative number, then show that the variance remains unchanged.
Solution:
Let, \(\overline{\mathrm{x}}\) be the mean of x₁, x₂, …………….., x_n then
variance, \(\sigma^2=\frac{1}{n} \sum_{i=1}^n\left(x_i-\bar{x}\right)^2\)
\(\bar{x}=\frac{1}{n} \sum_{i=1}^n x_i\)
1f k is added to each observat ion then the new observations will be x₁ + k₁, x₂ + k, x₃ + k, ………………….. x_n + k i.e., y₁, y₂, y₃, …………., y_n.
Now, y_i = x_i+ k,
For i = 1, 2, 3 …………… n
Let \(\overline{\mathrm{y}}\) be the mean of the new observations then
\(\bar{y}=\frac{1}{n} \sum_{i=1}^n y_i=\frac{1}{n} \sum_{i=1}^n x_i+k\)
= \(\frac{1}{n} \sum_{i=1}^n x_i+k=\frac{1}{n} \sum_{i=1}^n x_i+\frac{1}{n} \sum_{i=1}^n k\)
= \(\overline{\mathrm{x}}+\frac{1}{\mathrm{n}} \mathrm{nk}=\overline{\mathrm{x}}+\mathrm{k}\).

Variance of the new observations
\(\sigma_y^2=\frac{1}{n} \sum_{i=1}^n\left(y_i-\bar{y}\right)^2\)
= \(\frac{1}{n} \sum_{i=1}^n\left(x_i+k-\bar{x}-k\right)^2\)
= \(\frac{1}{n} \sum_{i=1}^n\left(x_i-\vec{x}\right)^2\) = σ²
∴ The variance of new observations is same that of the original observations.

Note :
If each observation is added by a number ‘a’ then the variance remains unchanged.

Question 17.
The coefficient of variation of two distributions are 60 and 70 and their standard deviations are 21 and 16 respectively. Find their arithmetic means.
Solution:
I distribution:
Let \(\overline{\mathbf{x}}_1\) and σ₁ be the mean and standard deviation of I distribution.
Given that, Coefficient of variation, C.V = 60
Standard deviation, σ₁ = 21
∴ CV = \(\frac{\sigma_1}{x_1}\) × 100
60 = \(\frac{21}{\overline{\mathrm{x}}_1}\) × 100
\(\overline{\mathrm{x}}_1=\frac{21}{60} \times 100\)
\(\overline{\mathrm{x}}_1\) = 35

II distribution:
Let \(\overline{\mathrm{x}}_2\) and σ₂ be the mean and standard deviation of II distribution.
Given that, C.V = 70
Standard deviation, σ₂ = 16
∴ CV = \(\frac{\sigma_2}{\overline{\mathrm{x}}_2}\) × 100
70 = \(\frac{16}{\bar{x}_2}\) × 100
\(\overline{\mathrm{x}}_2\) = \(\frac{16}{70}\) × 100
= \(\frac{160}{7}\) = 22.85.

Question 18.
The mean of 5 observations is 4.4. Their variance is 8.24. If three of the observations are 1, 2 and 6, find the other two observations.
Solution:
Let, the remaining 2 observations be x, y.
∴ The series is 1, 2, 6, x, y.
Given that, mean, \(\overline{\mathbf{x}}\) = 4.4
\(\frac{1+2+6+\mathrm{x}+\mathrm{y}}{5}\) = 4.4
9 + x + y = 22
x + y = 13 ……………….(1)
Given that, σ² = 8.24
\(\frac{\sum_{i=1}^n x_i^2}{n}-(\bar{x})^2\) = 8.24
\(\frac{1+4+36+x^2+y^2}{5}\) = (4.4)² = 8.24
\(\frac{41+x^2+y^2}{5}\) = 8.24 + 19,36
\(\frac{41+x^2+y^2}{5}\) = 27.6
41 + x² + y² = 138
x² + y² = 138 – 41
x² + y² = 97 ……………..(2)
From(1), y = 13 – x
From (2), x² + (13 – x)² = 97
x² + 169 + x² – 26x – 97 = 0
2x² – 26x + 72 = 0
x² – 13x + 36 = 0
x² – 9x – 4x + 36 0
x(x – 9) – 4 (x – 9) = 0
(x – 9) (x – 4) = 0
x = 9 or x = 4
If x = 4 then y = 13 – 4 = 9
If x = 9 then y = 13 – 9 = 4
∴ The remaining two observations are 4 and 9.

Question 19.
The arithmetic mean and standard deviation of a set of 9 items are 43 and 5 respectively. If an item of value 63 is added to that set, find the new mean and standard deviation of 10 item set given.
Solution:
Given that, number of items, n = 9
Mean, \(\overline{\mathrm{x}}\) = 43
\(\frac{1}{n} \sum_{i=1}^9 x_i\)= 43
\(\frac{1}{9} \sum_{i=1}^9 x_i\) = 43
\(\sum_{i=1}^9 x_i\) = 387
Now, given 10th item = 63
∴ Sum of the 10 observations \(\sum_{i=1}^9 x_i\) + 63 = 387 + 63 = 450
The mean of the new observations \(\frac{1}{n} \sum_{i=1}^{10} x_i\) = \(\frac{1}{10}\) (450) =
Given that, standard deviation, σ = 5
σ² = 25
\(\frac{1}{n} \sum_{i=1}^9 x_i^2-(\bar{x})^2\) = 25
\(\frac{1}{9} \sum_{i=1}^9 x_i^2-(43)^2\) = 25
\(\frac{1}{9} \sum_{i=1}^9 x_i^2\) – x = 25 + (43)²
= 25 + 1849 = 1874
\(\frac{1}{9} \sum_{i=1}^9 x_i^2\) = 1874 × 9 = 16866
Given that, 10th item = 63
∴ \(\frac{1}{9} \sum_{i=1}^10 x_i^2\) = \(\frac{1}{9} \sum_{i=1}^9 x_i^2\) + (63)²
= 16866 + 3969 = 20835
The variance of new observations = \(\frac{1}{n} \sum_{i=1}^{10} x_i^2-\bar{x}^2\)
= \(\frac{1}{10}\) (20835) – (45)²
= 2083.5 – 2025 = 58.5
∴ Standard deviation = √58.5 = 7.64.