TS Inter 1st Year

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

TS Inter 1st Year Maths 1A Addition of Vectors Important Questions Very Short Answer Type

Question 1.
Find the unit vector in- the direction of the sum of the vectors ā = 2ī + 2j̄ – 5k̄ and b̄ = 2ī + j̄ + 3k̄. [May. 13]
Answer:
Given ā = 2ī + 2j̄ – 5k̄, b̄ = 2ī + j̄ + 3k̄
Now ā + b̄ = 2ī + 2j̄ – 5k̄ + 2ī + kj̄ + 3k̄ = 4ī + 3j̄ – 2k̄
|ā + b̄| = |4ī + 3j̄ – 2k̄| = \(\sqrt{(4)^2+(3)^2+(-2)^2}\) = \(\sqrt{16+9+4}\) = √29
The unit vector in the direction of ā + b̄ is \(\frac{\bar{a}+\bar{b}}{|\bar{a}+\bar{b}|}\)
= \(\frac{4 \overline{\mathrm{i}}+3 \overline{\mathrm{j}}-2 \overline{\mathrm{k}}}{\sqrt{29}}\) = \(\frac{1}{\sqrt{29}}\) (4ī + 3j̄ – 2k̄).

Let ā = ī + 2j̄ + 3k̄ and b̄ = 3ī + j̄. Find the unit vector in the direction of ā + b̄ [Mar. 16 (TS)]
Answer:
\(\frac{1}{\sqrt{34}}\)(4ī + 3j̄ + 3k̄).

Question 2.
Find the unit vector in the direction of vector ā = 2ī + 3j̄ + k̄ . [Mar. 14]
Answer:
Given ā = 2ī + 3j̄ + k̄
Now |ā| = \(\sqrt{(2)^2+(3)^2+(1)^2}\) = \(\sqrt{4+9+1}\) = √14
∴ The unit vector in the direction of ā vector

Question 3.
If the vectors – 3ī + 4j̄ + λk̄ and μī + 8j̄ + 6k̄ are collinear vectors, then find λ andμ. [Mar. 18 (AP); May 14, 12; 10; Mar. 14]
Answer:
If a₁ī + b₁j̄ + c₁k̄, a₂ī + b₂j̄ + c₂k̄ are collinear vectors, then \(\).
Let ā = – 3ī + 4j̄ + λk̄, b = μī + 8j̄ + 6k̄
Since ā, b̄ are collinear vectors, then \(\frac{-3}{\mu}=\frac{4}{8}=\frac{\lambda}{6}\) ⇒ \(\frac{-3}{\mu}=\frac{1}{2}=\frac{\lambda}{6}\) ⇒ \(\frac{-3}{\mu}=\frac{1}{2}, \frac{1}{2}=\frac{\lambda}{6}\)
μ = – 6, λ = 3
∴ λ = 3, μ = – 6

ā = 2ī + 5j̄ + k̄ and b̄ = 4ī + mj̄ + nk̄ are collinear vectors, then find m and n.
Answer:
10, 2.

Question 4.
Let ā, b̄ be non – collinear vectors. If ᾱ = (x + 4y) ā + (2x + y + 1) b̄ and β̄ = (y – 2x + 2) ā + (2x – 3y – 1) b̄ are such that 3ᾱ = 2β̄ , then find x and y.
Answer:
Given vectors are ᾱ = (x + 4y) ā + (2x + y + 1) b̄, β̄ = (y – 2x + 2) ā + (2x – 3y – 1) b̄
Given 3ᾱ = 2β̄ ⇒ 3[(x + 4y) ā + (2x + y + 1) b̄] = 2[(y – 2x + 2) ā + (2x – 3y -1) b̄]
⇒ 3(x + 4y) ā + 3 (2x + y +1) b̄ = 2(y – 2x + 2) ā + 2(2x – 3y -1) b̄
Since ā, b̄ are non-collinear then
3(x + 4y) = 2(y – 2x + 2)
3x + 12y = 2y – 4x + 4
7x + 10y – 4 = 0 ……………. (1)
3(2x + y + 1) = 2(2x – 3y – 1)
6x + 3y + 3 = 4x – 6y – 2
2x + 9y + 5 = 0 …………….. (2)
Solve (1) and (2)

Question 5.
Show that the points whose position vectors are – 2ā + 3b̄ + 5c̄, ā + 2b̄ + 3c̄, 7ā – c̄ are collinear when ā, b̄, c̄ are non-coplanar vectors. [May 05, 92; Mar. 02]
Answer:
Let A, B, C be the given points.
The position vectors of A, B, C with respect to the origin ‘O’ are

∴ A, B and C are collinear.

Question 6.
If the position vectors of the points A, B and C are – 2ī + j̄ – k̄, – 4ī + 2j̄ + 2k̄ and 6ī – 3j̄ – 13k̄ respectively and \(\overline{\mathbf{A B}}\) = λ.\(\overline{\mathbf{A C}}\), then find the value of A.
Answer:
The position vectors of the points A, B and C with respect to origin ‘O’ are

Question 7.
If \(\overrightarrow{\mathbf{O A}}\) = ī + j̄ + k̄, \(\overline{\mathrm{AB}}\) = 3ī – 2j̄ + k̄, \(\overline{\mathrm{BC}}\) = ī + 2j̄ – 2k̄ and \(\overline{\mathrm{CD}}\) = 2ī + j̄ + 3k̄, then find the vector OD. [Mar. 19 (TS); Mar. 13; May 96]
Answer:

Question 8.
Let ā = 2ī + 4j̄ – 5k̄, b̄ = ī + j̄ + k̄ and c̄ = j̄ + 2k̄. Find the unit vector in the opposite direction of ā + b̄ + c̄. [Mar.19 (TS); May 15 (AP); Mar.19, 15 (AP); Mar.12, 10, 09, 04, B.P.]
Answer:
Given vectors are ā = 2ī + 4j̄ – 5k̄, b̄ = ī + j̄ + k̄ and c̄ = j̄ + 2k̄
Now, ā + b̄ + c̄ = 2ī + 4j̄ – 5k̄ + ī + j̄ + k̄ + j̄ + 2k̄ = 3ī + 6j̄ – 2k̄
|a + b + c| ⇒ \(\sqrt{3^2+6^2+(-2)^2}\) = \(\sqrt{9+36+4}\) = √49 = 7
∴ The unit vector in the opposite direction of
ā + b̄ + c̄ = \(\frac{-(\bar{a}+\bar{b}+\bar{c})}{|\bar{a}+\bar{b}+\bar{c}|}\) = \(\frac{-(3 \bar{i}+6 \bar{j}-2 \bar{k})}{7}\)

Question 9.
Find the vector equation of the line passing through the point 2ī + 3j̄ + k̄ and parallel to the vector 4ī – 2j̄ + 3k̄. [Mar. 17 (TS), 15 (AP); May 10, 07, 01; Mar. 07; 1. 92]
Answer:
Let ā = 2ī + 3j̄ + k̄, b̄ = 4ī – 2j̄ + 3k̄
The vector equation of the line passing through the point 2ī + 3j̄ + k̄ and parallel to the vector
4ī – 2j̄ + 3k̄ is r̄ = ā + tb̄, t ∈ R = 2ī + 3j̄ + k̄ + t(4ī – 2j̄ + 3k̄), t ∈ R
= 2ī + 3j̄ + k̄ + 4tī – 2tj̄ + 3tk̄
∴ r̄ = (2 + 4t)ī + (3 – 2t)j̄ +(1 + 3t)k̄, t ∈ R

Question 10.
OABC is a parallelogram. If \(\overline{\mathbf{O A}}\) = ā and \(\overline{\mathbf{O C}}\) = c̄, find the vector equation of the side \(\overline{\mathbf{B C}}\).
Answer:

OABC is a parallelogram.
\(\overline{\mathrm{OA}}=\overline{\mathrm{a}}, \overline{\mathrm{OC}}=\overline{\mathrm{c}}=\overline{\mathrm{AB}}\)
\(\overline{\mathrm{OB}}=\overline{\mathrm{OA}}+\overline{\mathrm{AB}}\)
\(\overline{\mathrm{OB}}\) = ā + c̄
The vector equation of a side BC
i.e., the vector equation of a line passing through C(c̄) and B(ā + c̄) is r̄ = (1 – t)ā + tb̄, t ∈ R
r̄ = (1 – t)c̄ + t(ā + c̄) = c̄ – tc̄ + tā + tc̄
⇒ r̄ – c̄ + tā, t ∈ R

Question 11.
If ā, b̄, c̄ are the position vectors of the vertices A, B and C respectively of ∆ABC, then find the vector equation of the median through the vertex A. [Mar. 13]
Answer:
The position vectors of the vertices A, B and C with respect to the origin are
\(\overline{\mathrm{OA}}\) = ā, \(\overline{\mathrm{OB}}\) = b̄, \(\overline{\mathrm{OC}}\) = c̄
Since D is the midpoint of BC then the position vector of D is \(\overline{\mathrm{OD}}\) = \(\frac{\overline{\mathrm{OB}}+\overline{\mathrm{OC}}}{2}\) = \(\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{2}\). The vector equation of the median through the vertex ‘A’ i.e., The vector equation of a line passing through A(ā) and D\(\left(\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{2}\right)\) is r̄ = (1 – t) ā + tb̄, t ∈ R
r̄ = (1 – t)ā + t\(\left(\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{2}\right)\), t ∈ R
⇒ r̄ = (1 – t)ā + (b̄ + c̄), t ∈ R

Question 12.
Find the vector equation of the line joining the points 2ī + j̄ + 3k̄ and – 4ī + 3j̄ – k̄. [Mar. 18, 16 (TS); Mar. 16 (AP); 11, 04, 95; May 09, 08, 95]
Answer:
Let ā = 2ī + j̄ + 3k̄, b̄ = – 4ī + 3j̄ – k̄
The vector equation of the line passing through the points 2ī + j̄ + 3k̄ and – 4ī + 3j̄ – k̄
r̄ = (1 – t)ā + tb̄, t ∈ R
r̄ = (1 – t)(2ī + j̄ + 3k̄) + t(- 4ī + 3j̄ – k̄)
= 2ī + j̄ + 3k̄ – 2tī – tj̄ – 3tk̄ – 4tī + 3tj̄ – tk̄
= 2ī + j̄ + 3k̄ – 6tī + 2tj̄ – 4tk̄
∴ r̄ = (2 – 6t)ī + (1 + 2t)j̄ + (3 – 4t)k̄, t ∈ R

Question 13.
Find the vector equation of the plane passing through the points ī – 2j̄ + 5k̄, – 5j̄ – k̄ and – 3ī + 5j̄. [Mar. 19, 17 (AP), May 15 (AP); May 14, 13, 11, 93; Mar. 06]
Answer:
Let ā = ī – 2j̄ + 5k̄, b̄ = – 5j̄ – k̄, c = – 3ī + 5j̄
Vector equation of the plane passing through the points ā, b̄, c̄ is
r̄ = (1 – s – t) ā + sb̄ + tc̄ where s, t ∈ R
= (1 – s – t) (ī – 2j̄ + 5k̄) + s (- 5j̄ – k̄) + t (- 3ī + 5j̄), s, t ∈ R
= (ī – 2j̄ + 5k̄) + s(- 5j̄ – k̄ – ī + 2j̄ – 5k̄) + t(- 3ī + 5j̄ – ī + 2j̄ – 5k̄)
r̄ = ī – 2j̄ + 5k̄ + s(- ī – 3j̄ – 6k̄) + t(- 4ī + 7j̄ – 5k̄)

Find the vector equation of the plane passing through the points (0, 0, 0), (0, 5, 0) and (2, 0. 1). [Mar. 18 (AP)]
Answer:
r̄ = (5t) j̄ + 5(2ī + k̄); t, s ∈ R.

Question 14.
If α β γ are the angles made by the vector 3ī – 6j̄ + 2k̄ with the positive directions of the co-ordinate axes, then find cos α, cos β, cos γ.
Answer:
Unit vectors along the coordinate axes are respectively ī, j̄, k̄.
Let p̄ = 3ī – 6j̄ + 2k̄
Given (p̄, ī) = α, (p̄, j̄) = β, and (p̄, k̄) = γ

Telangana TSBIE TS Inter 1st Year English Study Material Grammar Spelling: Missing Letters Exercise Questions and Answers.

TS Inter 1st Year English Grammar Spelling: Missing Letters

Q.No. 15 (8 × 1/2- 4 Marks)

English spelling poses problems to learners. This is because of a) no one letter-one sound relationship b) silent letters and c) words borrowed from other languages and so on.

Though there are some rules and generalizations that MAY HELP improve one’s spelling, it is always good to develop a keen eye to the spelling of the words one comes across and to practise writing words as many times as possible, that too correctly.

• In the Intermediate Public Examination, usually the spelling of the words used in the selected lessons is tested.
• The words that have vowel clusters (as in rec£jve, entertain etc.) or the words that have doubled consonants (as in possible, success etc.) are generally set in the question paper.
• Careful observation of the spelling of commonly used but typically spelt words (like height, tuition etc.) will go a long way in helping the student improve spelling to a great extent.

Examples of some commonly misspelled words :

1. For adjectives that end in a consonant +y, ‘-y’ changes to ‘-ily’.
easy-easily
happy-happily
angry-angrily
Exceptions
shy-shyly
sly-slyly
coy-coyly

2. For adjectives that end in a consonant +le the ‘-le’ changes to ‘-ly’ after the consonant,
probable-probably
sensible-sensibly
idle-idly
noble-nobly
For adjectives that end in – ic, we add -ally.
tragic-tragically
automatic-automatically
Exception
public-publicly
The doubling of final consonants.
When a one syllable word ends with one vowel and consonant we double the final consonant before adding a suffix that begins with a vowel.
bat-batting
dub-dubbing
sad-sadder
dig-digger
travel-travelled
confer-conferred

3. Spelling of regular verbs.
The past tense and past participle are the same in regular verbs.
i) We simply add – ed to the base form of the verb.
clear-cleared
walk-walked
visit-visited

ii) If the word ends in -e, we add -d.
like-liked
smile-smiled
hope-hoped

iii) For verbs ending in y preceded by two consonants, we change y to i before adding the suffix – ed.
hurry-hurried
reply-replied
try-tried

iv) For the verbs that end in a consonant, the final consonant is doubled when a suffix with an initial vowel letter is added.
rob-robbed
mop-mopped
drop-dropped

4. For ‘ie’ and ‘ei’
The general rule is that ‘i’ comes before ‘e’.
chief
grief
relieve
But after ‘c’ ‘e’ comes before ‘i’.
receive
conceit
deceive
Note the exceptions
neighbour
weight
eight
weigh
seize
height

Here is a list of commonly misspelled words.

Exercise – A

Fill in the blanks with either “ei” or “ie”

1. n _ _ ther
2. br _ _ f
3. sh _ _ ld
4. cr _ _ d
5. tr _ _ d
6. fr _ _ nd
7. th _ _ f
8. gr _ _ f
9. l _ _ sure
10. c _ _ ling
11. s _ _ ze
Answer:
1. n e i ther
2. br i e f
3. sh o u ld
4. cr e e d / cr i e d
5. tr i e d
6. fr i e nd
7. th i e f
8. gr i e f
9. l e i sure
10. c e i ling
11. s e i ze

Exercise – B

Correct the following misspelled words.

1. foregn
2. lugage
3. liesure
4. knowlege
5. twelth
6. tomorow
7. gurantee
8. momonto
9. ilegal
10. restaurent
Answer:
1. foreign
2. luggage
3. leisure
4. knowledge
5. twelfth
6. tomorrow
7. guarantee
8. memento
9. illegal
10. restaurant

Exercise – C

1. Please ……………………….. (wait / weight) for me.
2. I went to market to ……………………….. (buy / bye) a toothbrush.
3. He may not ……………………….. (loose / lose) the match.
4. When Chief Justice came to our state, people from different walks of life went to ……………………….. (meat / meet) him.
5. Ahmad cut the paper into many a ……………………….. (piece / peace).
6. The letters on the board cannot be ……………………….. (scene / seen) clearly.
7. – My friend has a ……………………….. (stationery / stationary) shop.
8. It was a pretty ……………………….. (sight / cite).
Answer:
1. wait
2. buy
3. lose
4. meet
5. piece
6. seen
7. stationery
8. sight

Supply the missing letters in the following words.

Exercise – 1

i) sch _ _ l
ii) enc _ _ raging
iii) app _ _ rance
iv) exce _ _ ent
v) sp _ _ k
vi) a _ _ ention
vii) p _ _ pie
viii) kno _ _ edge
ix) di _ _ ipline
x) a _ _ ord
Answers
i) sch o o l
ii) enc o u raging
iii) app e q rance
iv) exe l l ent
v) sp e a k
vi) a t t ention
vii) p e o ple
viii) kno w l edge
ix) di s c ipline
x) a f f ord

Exercise – 2

i) tea _ _ er
ii) gl _ _ my
iii) le _ _ on
iv) re _ _ect
v) f _ _ thful
vi) infl _ _ nce
vii) le _ _ ers
viii) pl _ _ sant
ix) su _ _ est
x) si _ _ le
Answer:
i) tea c h er
ii) gl o o my
iii) le s s on
iv) re s p ect
v) f a i thful
vi) infl u e nce
vii) le a d ers
viii) pl e a sant
ix) su g g est
x) si m p le

Exercise – 3

i) childh _ _ d
ii) p _ _ ce
iii) frus _ _ ation
iv) ha _ _ en
v) gra _ _ ar
vi) col _ _ r
vii) ang _ _ sh
viii) li _ _ten
ix) obed _ _ nt
x) mu _ _ le
Answer:
i) child o o d
ii) p i e ce
iii) frs t r ation
iv) ha p p en
v) gra m m ar
vi) col o u r
vii) ang u i sh
viii) li g h ten
ix) obed i e nt
x) mu s c le

Exercise – 4

i) g _ _de
ii) ma _ _ er
iii) mar _ _ es
iv) fi _ _ ing
v) ye _ _ow
vi) h _ _ lithly
vii) sq _ _ re
viii) lau _ _ed
ix) su _ _ect
x) hi _ _ly
Answer:
i) g u i de
ii) ma n n er/ ma t t er/ ma d d er/ma p p er
iii) mar b l es / mar c h es
iv) fi b b ing
v) ye l l ow
vi) h e a lthy
vii) sq u a re
viii) lau g h ed
ix) su s p ect
x) hi g h ly

Exercise – 5

i) hi _ _ top
ii) ba _ _ an
iiii) r _ _ tine
iv) conc _ _ ve
v) m _ _ ntain
vi) mi _ _ ion
vii) in _ _ edible
viii) mons _ _ n
ix) ca _ _ y
x) reco _ _ ition
Answer:
i) hi l l top
ii) ba n y an
iii) r o u tine
iv) con e i ve
v) m o u ntain
vi) mi s s ion / mi l l ion
vii) in c r edible
viii) mons o o n
ix) ca r r y
x) reco g n ition

Exercise – 6

i) sa _ _ ling
ii) rup _ _ s
iii) hu _ _ and
iv) res _ _ rces
v) s _ _ rce
vi) su _ _ icient
vii) ma _ _ ive
viii) vi _ _ age
ix) init _ _ tive
x) a _ _ roval
Answer:
i) sa p p ling
ii) rup e e s
iii) hu s b and
iv) res o u rces
v) s o u rce
vi) su f f icient
viii) ma s s ive
ix) init i a tive
x) a p p roval

Exercise – 7

i) pers _ _ de
ii) flu _ _ er
iii) p _ _ nce
iv) ex _ _ ode
v) a _ _ empt
vi) br _ _th
vii) hu _ _ red
viii) pa _ _ ive
ix) co _ _ apse
x) thr _ _ ten
Answer:
i) per u a de
ii) flu t t er
iii) p o u nce
iv) ex p l ode
v) a t t empt
vi) br e a th
vii) hu n d red
viii) pa s s ive
ix) co l l apse
x) thr e a ten

Exercise – 8

i) thr _ _ gh
ii) sli _ _ tly
iii) gr _ _ nd
iv) wo _ _ y
v) sp _ _ d
vi) ang _ _ sh
vii) prev _ _ us
viii) mi _ _ t
ix) rec _ _ ve
x) p _ _ ce
Answer:
i) thr o u gh
ii) sli g h tly
iii) gr o u nd
iv) wo r r y
v) sp e e d
vi) ang u i sh
vii) prev i o us
viii) mi g h t
ix) rec e i ve
x) p i e ce / p e a ce

Exercise – 9

i) cro _ _ ed
ii) em _ _ atic
iii) tr _ _ ble
iv) consc _ _ us
v) pl _ _ sant
vi) dr _ _ dful
vii) chi _ _ ey
viii) pu _ _ le
ix) incr _ _ se
x) de _ _ ive / de _ _ ive
Answer:
i) cro p p ed / cro w n ed
ii) em p h atic
iii) tr o u ble
iv) cons i o us
v) pl e a sant
vi) dr e a dful
vii) chi m n ey
viii) pu z z le
ix) incr e a se
x) de p r ive / de c e ive

Exercise – 10

i) app _ _ r
ii) wo _ _ le
iii) tre _ _le
iv) con _ _ ary
v) cr _ _ m
vi) com _ _ ain
vii) f _ _ lt
viii) req _ _ st
ix) enc _ _ nter
x) acq _ _ int
Answer:
i) app e a r
ii) wo b b le
iii) tre m b le
iv) con t r ary
v) cr e a m
vi) com p l ain
vii) f a u lt
viii) req u e st
ix) enc o u nter
x) acq u a int

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

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

Very Short Answer Type Questions

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

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

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

Consider two gases distinguished by subscripts 1 and 2.

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

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

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

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

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

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

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

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

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

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

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

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

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

Short Answer Questions

Question 1.
Explain the kinetic interpretation of Temperature.
Answer:
According to kinetic theory of gases, the pressure P exerted by one mole of an ideal gas is given by

Thus, the square root of the absolute temperature of an ideal gas is directly proportional to root mean square velocity of its molecules.
Also, from eqn. (1)

But, \(\frac{1}{2}\)mc² is average translational K.E per molecule of a gas.

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

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

1) Monoatomic gas has three degrees of freedom.

But molar specific heat at constant volume C_v = \(\frac{dU}{dT}\)

2) A diatomic gas has 3 translational and two rotational degrees of freedom.
∴ Kinetic energy per molecule U₁ = 5.\(\frac{1}{2}\)K_B.T
For one gram mole total energy U = \(\frac{5}{2}\)K_B.T.N_A
Molar-specific heat at constant volume

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

Kinetic energy
= per gram mole of molecules
= U₁N_A = U = (3 + f)K_B.N_A.T = (3 + f)RT
Molar specific heat C_v = \(\frac{dU}{dT}\) = (3 + f) R
∴ C_P = (u + f)R
∴ For polyatomic gases C_v = (3 + f) R & C_P = (4 + f)R

Question 3.
Explain the concept of absolute zero of temperature on the basis of kinetic theory.
Answer:
According to kinetic theory of gases, the pressure ‘P* exerted by one mole of an ideal gas is,

Absolute zero:
When T = 0, from equation (1), C = 0.

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

It means, molecular motion ceases at absolute zero.

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

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

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

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

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

As the vessels are thermally insulated and no work is done, total energy remains conserved.
Therefore, \(\frac{3}{2}\) (P₁V₁ + P₂V₂) = \(\frac{3}{2}\) P(V₁ + V₂)
where P is resultant pressure

For mixture of two gases,
(µ₁ + µ₂)RT = P(V₁ + V₂)
where T is resultant temperature.

Substitute equation (1) and equation (2) in the above equation, we have

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

Question 6.
What is the ratio of r.m.s speed of Oxygen and Hydrogen molecules at the same temperature?
Answer:
The r.m.s. speed of oxygen molecules at temperature T is

The r.m.s. speed of Hydrogen molecules at same temperature (T) is
C_H = \(\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{H}}}}\) But molecular weight of hydrogen = 2
Ratio of r.m.s. speed of Oxygen and Hy-drogen molecules is
∴ C_H = \(\sqrt{\frac{3 \mathrm{RT}}{2}}\)
Ratio of r.m.s. speed of Oxygen and Hydrogen molecules is

∴ C_ghy : C_H = 1 : 4

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

The root mean square speed of molecules is

C_rms = 2.74 km/s

Question 8.
If a gas has ‘f degrees of freedom, find the ratio of C_p and C_v.
Answer:
Suppose a polyatomic gas molecule has ‘f degrees degrees of freedom (0 and ratio of Cp and of freedom.
∴ Internal energy of one gram mole of the gas is
U = f × \(\frac{1}{2}\)k_BT × N_A = \(\frac{f}{2}\)RT

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

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

Question 10.
When pressure increases by 2%, what is the percentage decrease in the volume of a gas, assuming Boyle’s law is obeyed?
Answer:
Let ‘P’ be the initial pressure and ‘V’ be the volume.
When pressure is increased by 2%, new pressure, P¹ = P + \(\frac{2}{100}\)P ⇒ P¹ = \(\frac{102}{100}\)P
According to Boyle’s law, PV = constant ⇒ PV = P’V’

% decrease in volume = \(\frac{2}{100}\) × 100 = 1.96%

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

Long Answer Questions

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

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

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

Pressure on Y – Z plane =

If behaviour of gas is isotropic that is equal in all directions then V_x = V_y = V_z
∴ Root mean square velocity of gas

∴ Pressure of gas P = \(\frac{1}{3}\)nm\(\overline{\mathrm{V^2}}\)
From Pascal’s law pressure is same throughout the container so pressure in any direction (say x, y or z) is same.

Kinetic interpretation of temperature :
From gas equation

For Ideal gas internal energy is purely kinetic energy.

But \(\frac{E}{N}=\frac{1}{2}\)mV² and \(\frac{E}{N}=\frac{3}{2}\) K_BT

∴ \(\frac{E}{N}\) ∝ T is the average kinetic energy of gas molecule is proportional to absolute temperature of gas. But it does not depend on volume of container.

Additional Problems

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

Question 2.
Molar volume is the volume occupied by 1 mole of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0°Q. Show that it is 22.4 litres.
Solution:
For one mole of an ideal gas, PV = RT
∴ V = \(\frac{RT}{P}\)
Put R = 8.31 .1 mole^-1 K^-1, T = 273 K,
P = 1 atmosphere = 1.013 × 10⁵ Nm^-2

= 0.0224 × 10⁶ cc = 22400 cc = 22.4 litre.

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

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

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

Telangana TSBIE TS Inter 1st Year English Study Material Grammar Silent Letters Exercise Questions and Answers.

TS Inter 1st Year English Grammar Silent Letters

Q.No. 16 (8 × 1/2 = 4 Marks)

Silent letters are a peculiar feature of English spelling. Letters are used in writing but the corresponding sounds are not uttered in pronunciation. This poses problems to learners both in spelling and pronunciation. A study says that 60% of the English words have silent letters !

Careful observation of both the spelling and pronunciation is the only way out to overcome this problem. There are, however, some generalizations about silent letters that will help learners understand this feature to some extent.

In the Intermediate Public Examination, normally, the words that appear in the prescribed selections with a silent consonant letter are set under this question.

Some Guidelines :

• ‘b’ before word final ‘t’ is silent – doubt debt.
• ‘b’ in final ‘mb’ clusters is also silent.
  comb
  womb
  tomb
• ‘g’ before word final ‘n’ is silent.
  sign
  resign
  foreign
  campaign
  reign
  sovereign
• ‘g’ in most ‘ough’, ‘augh’ combinations is silent.
  (But in some words with ‘augh’ it may sound as /f/ as in ‘laughter),
  though
  taught
  through
  benign
  thorough
• ‘h’ in word beginning position (not always) is silent.
  honest
  honour
• ‘k’ in word beginning followed by ‘n’ is silent.
  knowledge
  know
  knit
  knee
  knave
  knuckle
• ‘l’ before word final’m’ is silent.
  calm
  palm (but in realm and film – T is not silent)
• Word final ‘n’ preceded by ‘m’ is silent,
  condemn
  column
  autumn
• ‘p’ in word beginning position followed by ‘s’ is silent.
  ’psychology
  psalm
  pseudo
• ‘p’ in words like receipt is also silent.
• ‘r’ followed by a consonant letter is silent,
  world
  card
• ‘f is silent in clusters like ‘astle’, ‘isten’, ‘istle’ as in
  castle
  whistle
  listen
• ‘W’ is silent in words like
  Write
  wrap
  know
  fawn
  lawn

(These guiding principles are by no way exhaustive. They help the learner to a great extent in identifying the silent letters. About vowels – the problem is not this simple or unambiguous. So no generalisation is given here.)

TEXTUAL EXAMPLES

Silent letter – Example words

b – lamb, bomb, tomb, climb, dumb, subtle, plumber, womb, crumb, thumb
c – muscle, blackguard, yacht, indict, scene
d – Wednesday, handkerchief, handbag, handsome, adjourn, adjective, judge
g – gnaw, gnome, phlegm, foreign, resign, campaign, align, sovereign
h – honour, heir, ghost, night, rhyme, rhythm, when, where, hour
k – know, knee, knock, knot, kneel, knowledge
t – talk, folk, salmon, colonel, calf, calm, half, walk, baulk
m – mnemonic . . .
n – hymn, solemn, damn, autumn, column
p – recepit, psychic, pneumonia, psyche
q(u) – lacquer
r – girl, bird, card, teacher, leader, curd, world
s – isle, aisle, viscount, island
t – thistle, fasten, mortgage, soften, watch, tsunami
w – whole, sword, two, who, wrist, wrong
y – prayer, mayor
z – rendezvous
gh – sigh, high, daughter, naughty, caught, brought, bought, night, haughty

Exercise -1

I. Underline the silent letters in the following words.

1) knell
2) consign
3) thorough
4) yellow
5) often
6) delight
7) benign
8) almond
9) yolk
10) limb
11) pseudonym
12) cupboard
13) Indict
14) climb
15) half
16) bouquet
17) wreath
18) dumb
19) depot
20) answer
21) aisle
22) exhibition
23) night
24) condemn
25) pneumonia
26) design
27) christmas
28) reign
29) palm
30) debut

31) precis
32) whistle
33) poignant
34) knead
35) castle
36) subtle
37) feign
38) debt
39) knight
40) knack
41) debris
42) comb
43) succumb
44) fight
45) deign
46) chalk
47) folk
48) bustle
49) align
50) malign
51) honest
52) lodge
53) pawn
54) doubt
55) bridge
56) coup
57) rapport
58) ghost
59) through
60) listen
Answer:
1) knell
2) consign
3) thorough
4) yellow
5) often
6) delight
7) benign
8) almond
9) yolk
10) limb

11) pseudonym
12) cupboard
13) Indict
14) climb
15) half
16) bouquet
17) wreath
18) dumb
19) depot
20) answer
21) aisle
22) exhibition
23) night
24) condemn
25) pneumonia
26) design
27) christmas
28) reign
29) palm
30) debut
31) precis
32) whistle
33) poignant
34) knead
35) castle
36) subtle
37) feign
38) debt
39) knight
40) knack
41) debris
42) comb
43) succumb
44) fight
45) deign
46) chalk
47) folk
48) bustle
49) align

50) malign
51) honest
52) lodge
53) pawn
54) doubt
55) bridge
56) coup
57) rapport
58) ghost
59) through
60) listen

II. Identify the silent consonant letters in the following words.

Exercise – 1

i) bright
ii) scene
iii) hour
iv) neighbour
v) wrong
vi) knell
vii) wreath
viii) palm
ix) limb
x) design
Answer:
i) bright – gh
ii) scene – c
iii) hour – h – r
iv) neighbour – g, h, r
v) wrong – w
vi) knell – k
vii) wreath – w
viii) palm – l
ix) limb – b
x) design – g

Exercise – 2

i) chalk
ii) knock
iii) depot
iv) teacher
v) often
vi) thought
vii) honest
viii) almond
ix) know
x) talk
Answer:
i) chalk – 1
ii) knock – k
iii) depot – t
iv) teacher – r
v) often – t
vi) thought – gh
vii) honest – h
viii) almond – l
ix) know – k, w
x) talk – l

Exercise – 3

i) lodge
ii) castle
iii) feign
iv) laugh
v) debut
vi) malign
vii) talk
viii) psyche
ix) lighten
x) muscle
Answer:
i) lodge – d
ii) castle – l
iii) feign – g
iv) laugh – …
v) debut – t
vi) malign – g
vii) talk – l
viii) psyche – p
ix) lighten – gh
x) muscle – c

Exercise – 4

i) yolk
ii) would
iii) pneumonia
iv) consign
v) drawing
vi) what
vii) knead
viii) doubt
ix) island
x) aisle
Answer:
i) yolk – l
ii) would – l
iii) pneumonia – p
iv) consign – g
v) drawing – w
vi) what – h
vii) knead – k
viii) doubt – b
ix) island – s
x) aisle – s

Exercise – 5

i) thorough
ii) who
iii) benign
iv) receipt
v) rhythm
vi) diversity
vii) nursery
viii) column
ix) curd
x) kneel
Answer:
i) thorough – gh
ii) who – w
iii) benign – g
iv) receipt – p
v) rhythm – h
vi) diversity – r
vii) nursery – r
viii) column – n
ix) curd – r
x) kneel – k

Exercise – 6

i) bustle
ii) although
iii) parliament
iv) fight
v) knee
vi) brought
vii) bomb
viii) could
ix) hymn
x) which
Answers
i) bustle – t
ii) although – gh
iii) parliament – r, i
iv) fight – gh
v) knee – k
vi) brought – gh
vii) bomb – b
viii) could – l
ix) hymn – n
x) which – h

Exercise – 7

i) align
ii) ghost
iii) leader
iv) straight
v) calf
vi) plumber
vii) wrap
viii) thistle
ix) attempt
x) burden
Answer:
i) align – g
ii) ghost – h
iii) leader – r
iv) straight – gh
v) calf – l
vi) plumber – b, r
vii) wrap – w
viii) thistle – t
ix) attempt – t
x) burden – r

Exercise – 8

i) through
ii) sovereign
iii) slightly
iv) tsunami
v) watch
vi) tomb
vii) caught
viii) naughty
ix) half
x) leopard
Answer:
i) through – gh
ii) sovereign – g
iii) slightly – gh
iv) tsunami – t
v) watch – t
vi) tomb – b
vii) caught – gh
viii) naughty – gh
ix) half – l
x) leopard – o, r

Exercise – 9

i) wrist
ii) daughter
iii) receipt
iv) solemn
v) hatter
vi) mnemonic
vii) 4umb
viii) damn
ix) should
x) folk
Answer:
i) wrist – w
ii) daughter – gh
iii) receipt – p
iv) solemn – n
v) hatter – r
vi) mnemonic – m
vii) dumb – b
viii) damn – n
ix) should – l
x) folk – l

Exercise – 10

i) knock
ii) autumn
iii) cupboard
iv) tight
v) walk
vi) sword
vii) subtle
viii) psalm
ix) handsome
x) gnaw
Answer:
i) knock – k
ii) autumn – n
iii) cupboard – p
iv) tight – gh
v) walk – l
vi) sword – w, r
vii) subtle – b
viii) psalm – p, l
ix) handsome – d
x) gnaw – g, w

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

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

Very Short Answer Type Questions

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

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

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

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

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

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

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

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

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

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

Question 7.
Indicate the graph showing the variation of specific heat of water with temperature. What does it signify?
Answer:
The variation of specific heat of water with the temperature is as shown in the graph.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Hence adiabatic compression increases the pressure more than isothermal compression.

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

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

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

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

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

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

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

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

Short Answer Questions

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

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

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

Increase in internal energy dU = nC_vdT

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

Limitations of 1st law of thermody namics:

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

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

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

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

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

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

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

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

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

Hence, C_p > C_v.

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

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

Let C_v be the amount of heat energy supplied. It is utilised only to raise the temperature of the gas by 1°C.
∴ dU = C_vdT

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

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

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

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

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

∴ dW = PdV = RdT OR (C_P – C_v) dT = RdT
But change of temperature = dT = 1°C (from definition of specific heat)
∴ C_p – C_v = R

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

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

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

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

Question 5.
Obtain an expression for the work done by an ideal gas during adiabatic change and explain.
Answer:
Work done during Adiabatic process :
Consider n mole of perfect gas contained in a cylinder having insulating walls. When piston moves through a small distance dx, then small work (dW) will be done.

∴ dW = (Pa) dx = PdV,
where a is area of cross-section of the piston. Therefore, when the system goes from initial state A (P₁, V₁) to the final state B ( P₂, V₂) the amount of work done,
W = \(\int_{v_1}^{v_2} P d V\) ………….. (1)
For an adiabatic change,
PV^γ = K (a constant ) or P = KV^-γ

Substituting for P in equation (1), the work done in an adiabatic process

∴ Work done in adiabatic process,

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

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

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

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

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

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

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

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

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

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

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

Question 8.
Write a short note on Quasistatic process.
Answer:
Quasi-static process:
A Quasistatic process can be defined as an infinitesimally show process in which at each and every intermediate stage the system remains in thermal and mechanical (thermodynamic) equilibrium with the surroundings through out the entire process.

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

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

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

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

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

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

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

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

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

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

Efficiency of heat engine,

where
Q₁ = heat energy supplied by source
Q₂ = heat energy delivered to sink
Block diagram of heat engine is as shown.

Long Answer Questions

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

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

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

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

Examples:

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

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

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

Step 1 :
In Carnot cycle, the 1st step consists of isothermal expansion of gases. So temperature T is constant, P, V changes are

Step 2 :
In this stage gases will expand adiabatically. So energy to the system Q is constant.

Step 3 :
In this stage gases will be compressed isothermally. So P₁V change are

Step 4 :
In the fourth stage the gas suffers adiabatic compression and returns to original stage.

The total work done W = Q₁ – Q₂ i.e., the difference to heat energy absorbed from source and heat energy given to sink
Efficiency of Carnot engine

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

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

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

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

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

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

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

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

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

Efficiency of heat engine,

where
Q₁ = heat energy supplied by source
Q₂ = heat energy delivered to sink
Block diagram ot heat engine is as shown.

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

It extracts heat energy Q₂ from sink i.e., from low temperature body with the help of external work and delivers heat energy Q₁ to high temperature body called source.
Work done W = Q₁ – Q₂.
In refrigerators external work is done on working substance.
A block diagram of refrigerator is as shown.

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

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

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

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

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

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

Step 1 :
In Carnot cycle, the 1st step consists of isothermal expansion of gases. So temperature T is constant, P, V changes are

Work done in isothermal process

Step 2 :
In this stage gases will expand adiabatically. So energy to the system Q is constant.

Step 3 :
In this stage gases will be compressed isothermally. So PjV changes are

Step 4 :
In the fourth stage the gas suffers adiabatic compression and returns to original stage.

Total work done in Carnot Cycle
W = W_1,2 + W_{2, 3} + W_{3, 4} + W_{4, 1}

The total work done W = Q₁ – Q₂ i.e., the difference to heat energy absorbed from source and heat energy given to sink Efficiency of Carnot engine

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

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

Problems

Question 1.
If a monoatomic ideal gas of volume 1 litre at N.T.P. is compressed (I) adiabatically to half of its volume, find the work done on the gas. Also find (ii) the work done if the compression is isothermal. (γ = 5/3)
Solution:
i) During an adiabatic process T₁V₁^γ-1 = T₂V₂^γ-1

ii) Work done during isothermal compression is
W = 2.3026 nRT log₁₀ \(\frac{V_2}{V_1}\)
n = number of moles = \(\frac{1}{22.4}\)
T = 273 K; R = 8.314 J mol^-1K^-1

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

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

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

Question 1.
Without expanding the determinant show that \(\left|\begin{array}{lll}
\mathbf{b}+\mathbf{c} & \mathbf{c}+\mathbf{a} & \mathbf{a}+\mathbf{b} \\
\mathbf{c}+\mathbf{a} & \mathbf{a}+\mathbf{b} & \mathbf{b}+\mathbf{c} \\
\mathbf{a}+\mathbf{b} & \mathbf{b}+\mathbf{c} & \mathbf{c}+\mathbf{a}
\end{array}\right|\) = 2\(\left|\begin{array}{lll}
\mathbf{a} & \mathbf{b} & \mathbf{c} \\
\mathbf{b} & \mathbf{c} & \mathbf{a} \\
\mathbf{c} & \mathbf{a} & \mathbf{b}
\end{array}\right|\) [Mar. 15 (AP); May 98, 96, 91]
Answer:

Question 2.
Show that \(\left|\begin{array}{ccc}
1 & a^2 & a^3 \\
1 & b^2 & b^3 \\
1 & c^2 & c^3
\end{array}\right|\) = (a – b) (b – c) (c – a) (ab + bc + ca). [Mar. 17(AP), 09: May 15 (AP); 02]
Answer:

= (a – b) (b – c) (c – a) [0 (c³ – c² (a + b + c) – (a + b) (0 – a – b – c) + (a² + ab + b²) (0 – 1)]
= (a – b) (b – c) (c – a) [0 + a² + ab + ac + ab + b² + bc – a² – ab – b²]
= (a – b) (b – c) (c – a) (ab + bc + ca) = RHS.

Question 3.
Show that \(\left|\begin{array}{ccc}
\mathbf{a}-\mathbf{b}-\mathbf{c} & \mathbf{2 a} & \mathbf{2 a} \\
\mathbf{2 b} & \mathbf{b}-\mathbf{c}-\mathbf{a} & \mathbf{2 b} \\
\mathbf{2 c} & \mathbf{2 c} & \mathbf{c}-\mathbf{a}-\mathbf{b}
\end{array}\right|\) = (a + b + c)³. [Mar. 11; May 11]
Answer:

Question 4.
Find the value of x if \(\left|\begin{array}{ccc}
x-2 & 2 x-3 & 3 x-4 \\
x-4 & 2 x-9 & 3 x-16 \\
x-8 & 2 x-27 & 3 x-64
\end{array}\right|\) = 0 [Mar 15 (TS); Mar. 06]
Answer:

⇒ (x – 2) (30 – 24) – (2x – 3) (10 – 6) + (3x – 4) (4 – 3) = 0
⇒ (x – 2)6 – (2x – 3)4 + (3x – 4)(1)
⇒ 6x – 12 – 8x + 12 + 3x – 4 = 0
⇒ x – 4 = 0
⇒ x = 4.

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

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

Question 7.
Show that \(\left|\begin{array}{ccc}
a^2+2 a & 2 a+1 & 1 \\
2 a+1 & a+2 & 1 \\
3 & 3 & 1
\end{array}\right|\) = (a – 1)³ [Mar. 13, 07]
Answer:

= (a – 1)² [(a + 1) (1 – 0) – 1 (2 – 0) + 0 (6 – 3)]
= (a – 1)² [a + 1 – 2] = (a – 1)² (a – 1) = (a – 1)³ = R.H.S.

Question 8.
Show that \(\left|\begin{array}{ccc}
\mathbf{a} & \mathbf{b} & \mathbf{c} \\
\mathbf{a}^2 & \mathbf{b}^2 & \mathbf{c}^2 \\
\mathbf{a}^3 & \mathbf{b}^3 & \mathbf{c}^3
\end{array}\right|\) = abc (a – b) (b – c) (c – a). [May. 06]
Answer:

= abc(a – b)(b – c) [0(c² – bc – c²) – 0(c² – ac – bc) + 1(b + c – a – b)]
= abc (a – b) (b – c) (c – a) = R.H.S

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

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

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

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

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

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

Question 12.
Solve the equations 3x + 4y + 5z = 18, 2x – y + 8z = 13 and 5x – 2y + 7z = 20 by Gauss-Jordan method. [May 15(TS): May 06, 01: Mar. 01]
Answer:
Given system of Linear equations are 3x + 4y + 5z = 18, 2x – y + 8z = 13 and 5x – 2y + 7z = 20

In this case, system of equations have unique solution.
i.e., x = 3, y = 1, z = 1.
∴ The solution of the given system of equations is x = 3, y = 1, z = 1.

Question 13.
Solve the equation 2x – y + 3z = 9, x + y + z = 6, x – y + z = 2 by Gauss-Jordan method. [Mar. 18(AP): Mar. 11, 10; May 11]
Answer:
The given system of linear equations are 2x – y + 3z = 9, x + y + z = 6, x – y + z = 2

In this case, system of equations has unique solution i.e., x = 1; y = 2; z = 3.
∴ The solution of given system of equations is x = 1; y = 2; z = 3.

Question 14.
Solve the following system of equations by Gauss – Jordan method : x + y + z = 9, 2x + 5y + 7z = 52, 2x + y – z = 0. [May 10, 07; Mar. 09, 1, 99]
Answer:
Given system of linear equations are x + y + z = 9, 2x + 5y + 7z = 52; 2x + y – z = 0. The matrix form is AX = D

In this case, the system of equations has unique solution i.e., x = 1; y = 3; z 5.
∴ The solution of given system of equations is x = 1; y = 3; z = 5.

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

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

Question 15.
Examine whether the system of equations x + y + z = 9, 2x + 5y + 7z = 52, 2x + y – z = 0 are consistent or Inconsistent and If consistent, find the complete solution. [May 15(TS); May 11]
Answer:
The given system of linear equations are x + y + z = 9, 2x + 5y + 7z = 52, 2x + y – z = 0.

∴ Rank [A] = 3
Now, Rank [AD] = 3
Since, the 3 × 3 sub matrix is \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) whose det is 1 ≠ 0.
∴ Rank [A] = Rank [AD] = 3.
In this case, system of equations has unique solution. i.e., x = 1, y = 3, z = 5.
Hence, the given system is consistent.

Question 16.
Examine whether the system of equations x + y + z = 6, x – y + z = 2, 2x – y + 3z = 9 are consistent or inconsistent and if consistent, find the complete solution. [Mar.11, 05]
Answer:
Given system of linear equations are x + y + z = 6, x – y + z = 2, 2x – y + 3z = 9
The given system of equations can be written as AX = D

∴ Rank of AD = 3.
Rank [A] = Rank [AD] = 3.
In this case, system of equations has unique solution. i.e., x = 1, y = 2, z = 3.
Hence, given system is consistent.

Question 17.
Examine whether the system of equations x + y + z = 1, 2x + y + z = 2, x + 2y + 2z = 1 are consistent or Inconsistent and if consistent, find the complete solution. [Mar. 15 (TS); May 05]
Answer:
Given system of linear equations are x + y + z = 1; 2x + y + z = 2; x + 2y + 2z = 1
The system of equations can be written as AX = D

∴ Rank [A] = Rank [AD]
In this case, system of equations has infinitely many solutions. x = 1; y + z = 0
∴ The solution of the given system of equations is x = 1, y + z = 0.
Hence, the given system is consistent.

Question 18.
Examine whether the following system of equations x + y + z = 6, x + 2y + 3z = 10, x + 2y + 4z = 1 are consistent or Inconsistent and if consistent, find the complete solution. [May. 02]
Answer:
The given system of linear equations are x + y + z = 6; x – 2y + 3z = 10; x + 2y – 4z = 1
The system of equations can be written as AX = D

∴ Rank [AD] = 3
∴ Rank [A] = Rank [AD] = 3
∴ In this case, system of equations has unique solution.
i.e., x = – 7; y = 22; z = – 9
Hence, the given system is consistent.

Question 19.
If A = \(\left[\begin{array}{ccc}
3 & 2 & -1 \\
2 & -2 & 0 \\
1 & 3 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
-3 & -1 & 0 \\
2 & 1 & 3 \\
4 & -1 & 2
\end{array}\right]\) and X = A + B then find X.
Answer:

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

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

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

Question 23.
If A = \(\left[\begin{array}{ccc}
0 & 1 & 2 \\
2 & 3 & 4 \\
4 & 5 & -6
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
-1 & 2 & 3 \\
0 & 1 & 0 \\
0 & 0 & -1
\end{array}\right]\) find B – A and 4A – 5B.
Answer:

Question 24.
If A = \(\left[\begin{array}{lll}
0 & 1 & 2 \\
2 & 3 & 4 \\
4 & 5 & 6
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
1 & -2 & 0 \\
0 & 1 & -1 \\
-1 & 0 & 3
\end{array}\right]\) find A – B and 4B – 3A.
Answer:

Question 25.
If A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
-4 & 2 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 5 \\
2 & 1
\end{array}\right]\) do AB and BA exist? If they exist find them. Do A and B commute with respect to multiplication?
Answer:
Given A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
-4 & 2 & 5
\end{array}\right]\), B = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 5 \\
2 & 1
\end{array}\right]\)
The order of matrix A is 2 × 3
The order of matrix B is 3 × 2
The no.of columns in A The no.of rows in B.
∴ AB is defined

The no.of columns in B = The no.of rows in A.
∴ BA is defined.

∴ AB ≠ BA
∴ A and B is not commute with respect to multiplication.

Question 26.
Find A² where A = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]\).
Answer:

Question 27.
If A = \(\left[\begin{array}{ll}
\mathrm{i} & 0 \\
0 & \mathrm{i}
\end{array}\right]\) find A².
Answer:

Question 28.
If A = \(\left[\begin{array}{ccc}
1 & 1 & 3 \\
5 & 2 & 6 \\
-2 & -1 & -3
\end{array}\right]\) then find A³.
Answer:

Question 29.
If A = \(\left[\begin{array}{cc}
-1 & 2 \\
0 & 1
\end{array}\right]\) then find AA’. Do A and A’ commute with respect to multiplication of matrices?
Answer:

∴ A and A’ do not commute with respect to multiplication of matrices.

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

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

Question 32.
Find the adjoint and the inverse of the matrix A = \(\).
Answer:
Given A = \(\left[\begin{array}{cc}
1 & 2 \\
3 & -5
\end{array}\right]\)
Cofactor of 1 is A₁ = + (- 5) = – 5
Cofactor of 2 is B₁ = – (3) = – 3
Cofactor of 3 is A₂ = – (2) = – 2
Cofactor of – 5 is B₂ = +(1) = 1
∴ The cofactor matrix of A is

Now det A = ad – bc = 1(- 5) – 2(3)
= – 5 – 6 = – 11 ≠ 0
Hence A is invertiable.

Question 33.
If A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\) then show that A² – 4A – 5I = 0. [Mar. 16(AP)]
Answer:
Given A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\)

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

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

Question 36.
Solve the system of equations
2x – y + 3z = 9, x + y + z = 6, x – y + z = 2 by using Cramer’s rule.
Answer:

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

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

∴ The solution of given system of equations is x = 7, y = – 10, z = 4.

Question 39.
Solve the equations 2x – y + 3z = 8, – x + 2y + z = 4, 3x + y – 4z = 0 by Gauss – Jordan method.
Answer:
Given system of linear equations are 2x – y + 3z = 8; – x + 2y + z = 4; 3x + y – 4z = 0.
Matrix equation form is AX = D

In this case, the system of equations has unique solution. i.e., x = y = z = 2
∴ The solution of given system of equations is x = 2; y = 2; z = 2.

Question 40.
Solve the equations 2x – y + 8z = 13, 3x + 4y + 5z = 18, 5x – 2y + 7z = 20 by Gauss – Jordan method.
Answer:
Given system of linear equations are 2x – y + 8z = 13, 3x + 4y + 5z = 18, 5x – 2y + 7z = 20.
Matrix equation form is AX = D

In this case, the system of equations has unique solution. i.e., x = 3, y = 1, z = 1.
∴ The solution of given system of equations is x = 3; y = 1; z = 1.

Some More Maths 1A Matrices Important Questions

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

∴ Order of X is 2 × 3.

Question 2.
Construct a 3 × 2 matrix whose elements are defined by a_ij = \(\frac{1}{2}\) |i – 3j|.
Answer:
In general a 3 × 2 matrix is given by

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

Question 4.
A certain book shop has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are Rs. 80, Rs. 60 and Rs. 40 each respectively. Using matrix algebra, find the total value of the books in the shop.
Answer:
Number of 3 types of books is expressed by the row matrix A

= [120 96 120]
Selling price of 3 types of books is expressed by the column matrix B

Total value of the books in the shop is given by AB
AB = [120 96 120] \(\left[\begin{array}{l}
80 \\
60 \\
40
\end{array}\right]\)
= [120 × 80 + 96 × 60 + 120 × 40]
= [9600 + 5760 + 4800]
= [20160]
∴ Total value of the books = Rs. 20160

Question 5.
If A = \(\left[\begin{array}{ll}
2 & 1 \\
1 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
3 & 2 & 0 \\
1 & 0 & 4
\end{array}\right]\), find AB and BA, if it exists.
Answer:
Given A = \(\left[\begin{array}{ll}
2 & 1 \\
1 & 3
\end{array}\right]\), B = \(\left[\begin{array}{lll}
3 & 2 & 0 \\
1 & 0 & 4
\end{array}\right]\)
The order of matrix A is 2 × 2
The order of matrix 13 is 2 × 3
The no.of columns in A = The no.of rows in B
∴ AB is defined

The no.of columns in B ≠ The no.of rows in A
∴ BA is not defined.

Question 6.
Give examples of two square matrices A and B of the same order for which AB = 0. But BA ≠ 0.
Answer:

Question 7.
If A = \(\left[\begin{array}{rr}
7 & -2 \\
-1 & 2 \\
5 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
-2 & -1 \\
4 & 2 \\
-1 & 0
\end{array}\right]\) then find AB’ and BA’. [Mar. 18 (AP)]
Answer:

Question 8.
Find the minors of – 1 and 3 in the matrix \(\left[\begin{array}{ccc}
2 & -1 & 4 \\
0 & -2 & 5 \\
-3 & 1 & 3
\end{array}\right]\).
Answer:

Question 9.
Find the cofactors of the elements 2, – 5 in the matrix \(\left[\begin{array}{ccc}
-1 & 0 & 5 \\
1 & 2 & -2 \\
-4 & -5 & 3
\end{array}\right]\).
Answer:

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

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

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

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

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

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

Question 14.
Solve the following system of equations by using Cramer’s rule. [Mar. 15 (TS)]
x – y + 3z = 5, 4x + 2y – z = 0, – x + 3y + z = 5
Answer:
Given system of equations can be written as:

Question 15.
If A = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\) and B = \(\frac{1}{2}\left[\begin{array}{lll}
\mathbf{b}+\mathbf{c} & \mathbf{c}-\mathbf{a} & \mathbf{b}-\mathbf{a} \\
\mathbf{c}-\mathbf{b} & \mathbf{c}+\mathbf{a} & \mathbf{a}-\mathbf{b} \\
\mathbf{b}-\mathbf{c} & \mathbf{a}-\mathbf{c} & \mathbf{a}+\mathbf{b}
\end{array}\right]\) then show that ABA^-1 is a diagonal matrxi.
Answer:
Given A = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\),
B = \(\frac{1}{2}\left[\begin{array}{lll}
\mathbf{b}+\mathbf{c} & \mathbf{c}-\mathbf{a} & \mathbf{b}-\mathbf{a} \\
\mathbf{c}-\mathbf{b} & \mathbf{c}+\mathbf{a} & \mathbf{a}-\mathbf{b} \\
\mathbf{b}-\mathbf{c} & \mathbf{a}-\mathbf{c} & \mathbf{a}+\mathbf{b}
\end{array}\right]\)
Cot actor of 0 is A₁ = + (0 – 1) = – 1
Cofactor of ‘1’ is B₁ = – (0 – 1) = 1
Cofactor of 1 is C₁ = + (1 – 0) = 1
Cofactor of 1 is A₂ = – (0 – 1) = 1
Cofactor of 0 is B₂ = + (0 – 1) = – 1
Cofactor of 1 is C₂ = – (0 – 1) = 1
Cofactor of 1 is A₃ = + (1 – 0) = 1
Cofactor of 1 is B₃ = – (0 – 1) = 1
Cofactor of 0 is C₃ = +(0 – 1) = – 1
∴ Cofactor matrix of

det A = 0(0 – 1) – 1 (0 – 1) + 1 (1 – 0)
= 0 + 1 + 1 = 2 ≠ 0
∴ A is invertiable.

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

Cofactor of 3 is A₁ =+(- 3 + 4) = 1
Cofactor of – 3 is B₁ = – (2 – 0) = – 2
Cofactor of 4 is C₁ = (- 2 + 0) = – 2
Cofactor of 2 is A₂ = – (- 3 + 4) = – 1
Cofactor of – 3 is B₂ = + (3 – 0) = 3
Cofactor of 4 is C₂ = – (- 3 + 0) =3
Cofactor of 0 is A₃ = + (- 12 + 12) = 0
Cofactor of – 1 is B₃ = – (12 – 8) = – 4
Cofactor of 1 is C₃ = + (- 9 + 6) = – 3

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

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

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

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

Question 21.
Solve the following system of equations by using Matrix inversion method.
2x – y + 3z = 9, x + y + z = 6, x – y + z = 2. [Mar. 16 (TS)]
Answer:

Question 22.
Solve x + y + z = 9, 2x + 5y + 7z = 52 and 2x + y – z = 0 by using matrix inversion method. [Mar. 17 (AP)]
Answer:

Question 23.
Solve the following system of equations by Cramer’s rule: 2x – y + k = 8, – x + 2y + z = 4, 3x + y – 4z = 0 [Mar. 18 (TS)]
Answer:
Given equations are
2x – y + 3z = 8,
– x + 2y + z = 4,
3x + y – 4z = 0

Telangana TSBIE TS Inter 1st Year English Study Material Grammar Phonetic Transcription Exercise Questions and Answers.

TS Inter 1st Year English Grammar Phonetic Transcription

Q.No. 17 (4 × 1 = 4 Marks)

Meaningful speech sounds are the basic raw material for any language. These sounds are represented by symbols in writing. We refer to these symbols as ‘alphabet’.

Most of the Indian languages have a fixed relationship between the sounds and their symbols. In other words, one symbol always stands for one sound and vice versa. Therefore we find absolutely no problem while reading or writing Indian languages once we learn the alphabet of that language.

The same is not the case with English. One letter may stand for many sounds. Example Q as / k / in car, as / s / in century as /t/ in change. One sound is also represented by various letters. Ex : / f / is represented by ‘ph’ in philosophy, by ‘f in fan, by ‘gh’ in rough. This results in a lot of problems for the learners, particularly for foreign learners in writing the spelling and pronouncing the written words.

A way out of this problem is a set of forty four symbols, called phonetic symbols. Each of these symbols stands for one sourid only. Learning these symbols arms us with the necessary weapons to war against the problems in pronounciation and spelling.

Study the following forty four phonetic symbols carefully and learn to identify and use them. Phonetic symbols are always placed between two Slant lines.

Vowels

Consonants

Writing the symbols that represent the sounds in a word is called phonetic transcript. In many examinations phonetic transcription is given and the examinee is asked to write the spelling of those words.

Exercises

Exercise – A

Read the words and fill in the spaces with the appropriate vowel symbol. The first one is done for you. If necessary, don’t hesitate to use a dictionary.

Exercise – B

Read the words according to the vowel symbols mentioned.

Exercise – C

Go through the words and identify the sounds the end with. The first one is done for you.

Exercise-D

Read the following words. You will notice that in some words the letters ‘th’ are pronounced as /θ/ and in some others, as /ð/ and in some others, as 161. Write the sound you noticed. The first one is done for you.
the /ð/ this /ð/ through /θ/ then /ð/
thus /ð/ thought /θ/ thick /θ/ mother /ð/

Exercise – E

We get confused with the sounds /w/ and /v/. The sound /w/ is pronounced with rounded lips. The sound /v/ is pronounced with the articulation of front upper teeth and lower lip and with more force.
Now, pronounce the words aloud.
wheel   ventilator
worst    verse
wet      veto

Exercise – F

Read the following transcriptions and write the words in ordinary spelling. The first one is done for you.

Exercise – G

Pronounce the following words and transcribe them in the column my transcription Later, consult a dictionary and make necessary corrections.

Exercise – H

Words with short vowels are entirely different from their long counterparts. Understanding the difference is vital for pronunciation. Read the words in the following table and write a few more words from your text.

Write the following transcriptions using ordinary English spelling.

Exercise – 1

Answer:
i) purpose
ii) accomplish
iii) beautiful
iv) question
v) faith
vi) miserable

Exercise – 2

Answer:
i) speak
ii) constantly
iii) attention
iv) unfortunate
v) want
vi) individual

Exercise – 3

Answer:
i) transgression
ii) nervy
iii) harbinger
iv) recognize
v) strive
vi) pesticide

Exercise – 4

Answers
i) provide
ii) literate
iii) frustration
iv) imagination
v) fear
vi) adamant

Exercise – 5

Answer:
i) stretch
ii) incredible
iii) plant
iv) condition
v) hospital
vi) entire

Exercise – 6

Answer:
i) education
ii) husband
iii) pension
iv) recently
v) mountain
vi) close

Exercise – 7

Answer:
i) desperate
ii) lull
iii) impelled
iv) resistance
v) pride
vi) faint

Exercise – 8

Answer:
i) success
ii) effort
iii) excitement
iv) worry
v) previous
vi) athletic

Exercise – 9

Answer:
i) acquaint
ii) attic
iii) horizontal
iv) gridiron
v) curb
vi) vengeance

Exercise – 10

Answer:
i) emphatic
ii) appearance
iii) mention
iv) gentleman
v) tremble
vi) sleep

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

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

Question 1.
If A = \(\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\), , then show that for all the positive integers n, Aⁿ = \(\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\) [May 98, 91]
Answer:

∴ S(k + 1) is true.
∴ By the principle of mathematical induction, S(n) is true for all n ∈ N.
∴ Aⁿ = \(\left[\begin{array}{cc}
\cos n \theta & \sin n \theta \\
-\sin n \theta & \cos n \theta
\end{array}\right]\), ∀ n ∈ N.

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

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

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

Question 5.
If A = \(\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right]\), then show that Aⁿ = \(\left[\begin{array}{cc}
1+2 n & -4 n \\
n & 1-2 n
\end{array}\right]\) for any integer n ≥ 1, by using mathematical induction. [May 08, 02]
Answer:

∴ S(k + 1) is true.
∴ By using the principle of mathematical Induction, S(n) is true for all n ∈ N.
∴ Aⁿ = \(\left[\begin{array}{cc}
1+2 n & -4 n \\
n & 1-2 n
\end{array}\right]\) ∀ n ∈ N.

Question 6.
For any n × n matrix A, prove that A can be uniquely expressed as a sum of a symmetric matrix and a skew symmetric matrix. (Mar. ‘03)
Answer:
Let A be a square matrix.

∴ A can be expressed as a sum of a symmetric matrix and a skew symmetric matrix.

Question 7.
Show that \(\left|\begin{array}{ccc}
\mathbf{1} & \mathbf{a} & \mathbf{a}^2 \\
\mathbf{1} & \mathbf{b} & \mathbf{b}^2 \\
\mathbf{1} & \mathbf{c} & \mathbf{c}^2
\end{array}\right|\) = (a – b) (b – c) (c – a). [Mar. 17 (TS). 05]
Answer:

= (a – b) (b – c) (c – a) [0(c² – c) – 1(0 – 1) + (a + b) (0 – 0)]
= (a – b) (b – c) (c – a) (1) = (a – b) (b – c) (c – a) = RHS

Question 8.
Show that \(\left|\begin{array}{lll}
b c & b+c & 1 \\
c a & c+a & 1 \\
a b & a+b & 1
\end{array}\right|\) = (a – b) (b – c) (c – a). [Board Paper]
Answer:

= (a – b) (c – a) [b (1 – 0) – 1 (c – 0) + 0 (ac + bc – ab)]
= (a – b) (c – a) [b – c] = (a – b) (b – c) (c – a) = R.H.S.

Question 9.
Show that \(\left|\begin{array}{ccc}
b+c & c+a & a+b \\
a+b & b+c & c+a \\
a & b & c
\end{array}\right|\) = a³ + b³ + c³ – 3abc. [May 13, 07; Mar. 08]
Answer:

= (a + b + c) [c² – bc – ac + ab + a² – 2ab + b²]
= (a + b + c) [a² + b² + c² – ab – bc – ca]
= a³ + b³ + c³ – 3abc

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

Question 11.
Show that \(\left|\begin{array}{lll}
a-b & b-c & c-a \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\) = 0. [May. 08]
Answer:

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

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

Now, det A = 1(16 – 9) – 3(4 – 3) + 3 (3 – 4)
= 1(7) – 3(1) + 3(- 1) = 7 – 3 – 3 = 1
Hence A is invertiable.
A^-1

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

Find the adjoint and the inverse of the matrix \(\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 1 & 0 \\
3 & 2 & 1
\end{array}\right]\). [Mar. 05; May 98]
Answer:

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

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

Cofactor of – 1 is A₁ = + (1 – 4) = – 3
Cofactor of – 2 is B₁ = – (2 + 4) = – 6
Cofactor of – 2 is C₁ = + (- 4 – 2) = – 6
Cofactor of 2 is A₂ = – (- 2 – 4) = 6
Cofactor of 1 is B₂ = + (- 1 + 4) = 3
Cofactor of – 2 is C₂ = – (2 + 4) = – 6
Cofactor of 2 is A₃ = + (4 + 2) = 6
Cofactor of – 2 is B₃ = – (2 + 4) = – 6
Cofactor of 1 is C₃ = + (- 1 + 4) = 3
∴ Cofactor matrix of

∴ Adj A = 3A’
det A = – 1 (1 – 4) + 2 (2 + 4) – 2 (- 4 – 2)
= – 1 (- 3) + 2(6) – 2 (- 6) = + 3 + 12 + 12 = 27 ≠ 0
∴ A is invertiable.
A^-1 = \(\frac{{adj} A}{{det} A}=\frac{1}{27}\left[\begin{array}{ccc}
-3 & 6 & 6 \\
-6 & 3 & -6 \\
-6 & -6 & 3
\end{array}\right]\)

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

Question 17.
If 3A = \(\left[\begin{array}{rrr}
1 & 2 & 2 \\
2 & 1 & -2 \\
-2 & 2 & -1
\end{array}\right]\), then show that A^-1 = A’. [Mar. 14, 09; May. 12]
Answer:

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

Telangana TSBIE TS Inter 1st Year English Study Material Grammar Syllables Exercise Questions and Answers.

TS Inter 1st Year English Grammar Syllables

Q.No. 19 (4 × 1 = 4 Marks)

A syllable is the next higher unit to a speech sound and forms a word or part of a word. It contains one (and only one) vowel sound (not letter). The number of consonant sounds in a syllable may be ‘Zero to Seven’.

A word may have one syllable or more.

• Words with one syllable each are called monosyllabic words.
• Words with two syllables each are called disyllabic words.
• Words with three syllables each are called trisyllabic words.
• Words with more than three syllables each are called polysyllabic words.

The, number of vowel sounds in a word gives us the number of syllables in that word. By noticing the vowel symbols in the phonetic transcript of a given word, we can arrive at the number of syllables in that word. Look at the following examples :

pen / pen only one vowel sound – one syllable – monosyllabic paper / peips (r) / two vowel sounds – two syllables – disyllabic gravity / graeviti / three vowel sounds – three syllables – trisyllabic.

discovery / dɪˈskʌvɚɹi / four vowel sounds – four syllables – poly (tetra) syllabic, organization / ˌɔːrɡənəˈzeɪʃən / five vowel sounds – five syllables – poly (penta) syllabic.

There are, however, certain words in which the number of vowel sounds is not equal to the number of syllables. Look at the following examples :
brittle / ˈbɹɪtl̩ / only ong vowel sound – but two syllables
prism / pnzm / only one vowel sound – but two syllables
mutton / mAtn / only one vowel sound – but two syllables

The reason for this variation is that the consonant sounds / l /, / m / and / n / help form a syllable. These sounds in such words are, therefore, called syllabic consonants.
Examine some more examples of this kind :

Careful observation of phonetic transcription or correct pronunciation of words will help students find out the number of syllables in a given word.

Exercise – A

In the following table four categories of words are given. Read them aloud paying attention to the syllabic division.

Exercise – B

Read the words in the table and write the number of syllables in the columns. Look up the words in a dictionary to check your answers. The first one has been done for you.

Mention the number of syllables in the following words.

Exercise – 1

i) misery
ii) direction
iii) remember
iv) information
v) encourage
vi) excellent
Answer:
i) 3 trisyllabic
ii) 3 – trisyllabic
iii) 3 – trisyllabic
iv) 4 – polysyllabic
v) 3 – trisyllabic
vi) 3 – trisyllabic

Exercise – 2

i) person
ii) weakness
iii) dark
iv) thought
v) fact
vi) discipline
Answer:
i) 2 – disyllabic
ii) 2 – disyllabic
iii) 1 – monosyllabic
iv) 1 – monosyllabic
v) 1 – monosyllabic
vi) 3 – trisyllabic

Exercise – 3

i) lawyer
ii) literacy
iii) square
iv) harbinger
v) adamant
vi) muse
Answer:
i) 2 – disyllabic
ii) 4 – polysyllabic
iii) 1 – monosyllabic
iv) 3 – trisyllabic
v) 3 – trisyllabic
vi) 1 – monosyllabic

Exercise – 4

i) before
ii) doctor
iii) mother
iv) imagination
v) essence
vi) quarter
Answer:
i) 2 – disyllabic
ii) 2 – disyllabic
iii) 2 – disyllabic
iv) 5 – polysyllabic
v) 2 – disyllabic
vi) 2 – disyllabic

Exercise – 5

i) glance
ii) propel
iii) silence
iv) realize
v) excitement
vi) climax
Answers:
i) 1 – monosyllabic
ii) 2 – dissyllabic
iii) 2 – disyllabic
iv) 2 – disyllabic / 3 – tnsyllabic
v) 3 – trisyllabic
vi) 2 – disyllabic

Exercise – 6

i) understand
ii) decision
iii) shout
iv) supremely
v) encouragement
vi) flashlight
Answer:
i) 3 – trisyllabic
ii) 3 – trisyllabic
iii) 1 – monosyllabic
iv) 3 – trisyllabic
v) 4 – polysyllabic
vi) 2 – disyllabic

Exercise – 7

i) pension
ii) source
iii) confer
iv) captivate
v) modest
vi) contribution
Answer:
i) 2 – disyllabic
ii) 1 – monosyllabic
iii) 2 – disyllabic
iv) 3 – trisyllabic
v) 2 – disyllabic
vi) 4 – polysyllabic

Exercise – 8

1) popular
ii) today
iii) side
iv) plant
v) rainwater
vi) condition
Answer:
i) 3 – trisyllabic
ii) 2 – disyllabic
iii) 1 – monosyllabic
iv) 1 – monosyllabic
v) 3 – trisyllabic
vi) 3- trisyllabic

Exercise – 9

i) punctual
ii) increase
iii) room
iv) mantelpiece
v) breakfast
vi) gracious
Answer:
i) 2 – disyllabic
ii) 2 – disyllabic
iii) 1 – monosyllabic
iv) 3 – trisyllabic
v) 2 – disyllabic
vi) 2 – disyllabic

Exercise – 10

i) particular
ii) handful
iii) apearance
iv) often
v) apartment
vi) idea
Answer:
i) 4 – polysyllabic
ii) 2 – disyllabic
iii) 3 – trisyllabic
iv) 2 – disyllabic
v) 3 – trisyllabic
vi) 2 – disyllabic

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

TS Inter 1st Year Maths 1A Product of Vectors Important Questions Short Answer Type

Question 1.
Prove that angle in a semi-circle is a right angle by using vector method. [MAR ’13, ’08, ’99]
Answer:
Let AB be a diameter of a circle with centre O.
Let OA = a, then OB = -a
Let P be a point on the circle and OP = r
OA = OB = OP


∠APB = 90°
∴ Angle in a semicircle is 90°.

Question 2.
If P, Q, R and S are points whose position vectors are i̅ – k̅, -i̅ + 2j̅, 2i̅ – 3k̅ and 3i̅ – 2j̅ – k̅ respectively, then find the component of RS on PQ. [Mar. ’98]
Answer:
The position vectors of the points P, Q, R and S with respect to the origin ‘O’ are
\(\overline{\mathrm{OP}}\) = i̅ – k̅,
\(\overline{\mathrm{OQ}}\) = -i̅ + 2j̅,
\(\overline{\mathrm{OR}}\) = 2i̅ – 3k̅,
\(\overline{\mathrm{OS}}\) = 3i̅ – 2j̅ – k̅
Now \(\overline{\mathrm{PQ}}=\overline{\mathrm{OQ}}-\overline{\mathrm{OP}}\) = -i̅ + 2j̅ -i̅ + k̅ = -2i̅ + 2j̅ + k̅
\(\overline{\mathrm{RS}}=\overline{\mathrm{OS}}-\overline{\mathrm{OR}}\) = 3i̅ – 2j̅ – k̅ – 2i̅ + 3k̅ = i̅ – 2j̅ + 2k̅

Question 3.
Prove that the angle ‘θ’ between any two diagonals of cube is given by cos θ = \(\frac{1}{3}\). [Mar ’12, ’11, ’10; Mar. ’10]
Answer:
Let \(\overline{\mathrm{OA}}\) = i̅, \(\overline{\mathrm{Ob}}\) = j̅ ,\(\overline{\mathrm{Oc}}\) = k̅
Let OA = OB = OC = 1 unit
In a cube, diagonals are OF, CD, BG, AE

Question 4.
Show that the points (5, – 1, 1), (7, – 4, 7), (1, – 6, 10) and (- 1, – 3, 4) are the vertices of a rhombus by vectors. [Mar. ’13]
Answer:
Let A(5, – 1, 1), B(7, – 4, 7), C(1, – 6, 10) and D(- 1, – 3, 4) are the given points.

∴ AB = BC = CD = DA = 7 units and AC ≠ BD.
∴ A, B, C, D are the points which are the vertices of a rhombus.

Question 5.
Find the area of the triangle whose vertices are A(1, 2, 3), B(2,3, 1) and C(3,1, 2). [Mar. ’14, ’06]
Answer:
Let the position vectors of A, B, C with respect to the origin are
\(\overline{\mathrm{OA}}\) = i̅ + 2j̅ + 3k̅, \(\overline{\mathrm{OB}}\) = 2i̅ + 3j̅ + k̅, \(\overline{\mathrm{OC}}\) = 3i̅ + j̅ + 2k̅
\(\overline{\mathrm{AB}}=\overline{\mathrm{OB}}-\overline{\mathrm{OA}}\) = 2i̅ + 3j̅ + k̅ – i̅ – 2j̅ – 3k̅ = i̅ + j̅ – 2k̅
\(\overline{\mathrm{AC}}=\overline{\mathrm{OC}}-\overline{\mathrm{OA}}\) = 3i̅ + j̅ + 2k̅ – i̅ – 2j̅ – 3k̅ = 2i̅ – j̅ – k̅

\(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}}\) = \(\left|\begin{array}{ccc}
\overline{\mathbf{i}} & \overline{\mathrm{j}} & \overline{\mathrm{k}} \\
1 & 1 & -2 \\
2 & -1 & -1
\end{array}\right|\)
= i̅(-1 -2) – j̅(-1 + 4) + k̅(-1-2) = -3i̅ -3j̅ – 3k̅

∴ The area of triangle whose vertices are A, B, C is \(\frac{1}{2}|\overline{\mathrm{AB}} \times \overline{\mathrm{AC}}|\) = \(\frac{3 \sqrt{3}}{2}\)

Question 6.
If a̅ + b̅ + c̅ = 0, then prove that a̅ × b̅ = b̅ × c̅ = c̅ × a̅. [Mar. ’03; May ’98]
Answer:
Given a̅ + b̅ + c̅ = 0
⇒ a̅ = -b̅ – c̅
⇒ a̅ × b̅ = (-b̅ – c̅) × b̅
= -(b̅ × b̅) – (c̅ × c̅)
= -0 + b̅ × c̅

= a̅ × b̅ = b̅ × c̅ ………………(1)
⇒ a̅ + b̅ + c̅ = 0
b̅ = -a̅ – c̅

⇒ b̅ × c̅ = (-a̅ – c̅) × c̅ =-(a̅ × c̅) – (c̅ × b̅) = c̅ × a̅ – 0
b̅ × c̅ = c̅ × a̅ ……………………..(2)
From (1) & (2) ⇒ a̅ × b̅ = b̅ × c̅ = c̅ × a̅

Question 7.
Find the unit vector perpendicular to the plane passing through the points (1, 2, 3), (2,-1,1) and (1,2,-4). [Mar. ’17(AP) ’05; May ’10]
Answer:
Let the position vectors of the points A, B, C with respect to the origin ‘O’ are
\(\overline{\mathrm{OA}}\) = i̅ + 2 j̅ + 3k̅; \(\overline{\mathrm{OB}}\) = 2 i̅ – j̅ + k̅; \(\overline{\mathrm{OC}}\) = i̅ + 2 j̅ – 4k̅
\(\overline{\mathrm{AB}}=\overline{\mathrm{OB}}-\overline{\mathrm{OA}}\) = 2i̅ – j̅ + k̅ – i̅ – 2j̅ – 3k̅ = i̅ – 3 j̅ – 2k̅
\(\overline{\mathrm{AB}}=\overline{\mathrm{OB}}-\overline{\mathrm{OA}}\) = i̅ + 2j̅ – 4k̅ – i̅ – 2j̅ – 3k̅

The unit vector perpendicular to the plane passing through the points A, B and C is
\(\pm \frac{(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})}{|\overline{\mathrm{AB}} \times \overline{\mathrm{AC}}|}=\pm \frac{(21 \overline{\mathrm{i}}+7 \overline{\mathrm{j}})}{7 \sqrt{10}}=\pm \frac{(3 \overline{\mathrm{i}}+\overline{\mathrm{j}})}{\sqrt{10}}\)

Find a unit vector perpendicular to the plane determined by the points P(1, – 1, 2), Q(2, 0, – 1) and R (0, 2, 1).
Answer:
±\(\frac{1}{\sqrt{6}}\)(2i̅ + j̅ + k̅)

Question 8.
If a̅ = 2i̅ + 3j̅ + 4k̅, b̅ = i̅ + j̅ – k̅ and c̅ = i̅ – j̅ + k̅, then compute a̅ × (b̅ × c̅) and verify that it is perpendicular to a̅. [Mar. ’19(TS); May ’06; May ’03]
Answer:
Given vectors are a̅ = 2i̅ + 3j̅ + 4k̅, b̅ = i̅ + j̅ – k̅, c̅ = i̅ – j̅ + k̅
b̅ × c̅ = \(\left|\begin{array}{ccc}
\overline{\mathrm{i}} & \overline{\mathrm{j}} & \overline{\mathrm{k}} \\
1 & 1 & -1 \\
1 & -1 & 1
\end{array}\right|\)
= i̅(1 – 1) -j̅(1 + 1) + k̅(-1- 1) = 2i̅ – 2k̅

a̅ × (b̅ × c̅) = \(\left|\begin{array}{ccc}
\overline{\mathrm{i}} & \overline{\mathrm{j}} & \overline{\mathrm{k}} \\
2 & 3 & 4 \\
0 & -2 & -2
\end{array}\right|\)
= i̅(-6 + 8) – j̅(-4 – 0) + k̅(-4-0)
= 2i̅ + 4j̅ – 4k̅

Now [a̅ × (b̅ × c̅)].a̅ = (2i̅ + 4j̅ – 4k̅).(2i̅ + 3j̅ + 4k̅) = 4 + 12 – 16 = 0
a̅ × (b̅ × c̅) is perpendicular to a̅.

Question 9.
For any four vectors a, b, c and d show that (a̅ × b̅) . (c̅ × d̅) = \(\left|\begin{array}{cc}
\bar{a} \cdot \bar{c} & \bar{a} \cdot \bar{d} \\
\bar{b} \cdot \bar{c} & \bar{b} \cdot \bar{d}
\end{array}\right|\) and in particular \((\overline{\mathrm{a}} \times \overline{\mathrm{b}})^2=\overline{\mathrm{a}^2} \overline{\mathrm{b}^2}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}})^2\). [Mar. ’02, ’00]
Answer:
LHS = (a̅ × b̅). (c̅ × d̅) = a̅. (b̅ × (c̅ × d̅)) = a̅. [(b̅. d̅) c̅ – (b̅. c̅) d̅]
= (a̅ . c̅)(b̅ . d̅) – (a̅ . d̅)(b̅ . c̅) = \(\left|\begin{array}{cc}
\bar{a} \cdot \bar{c} & \bar{a} \cdot \bar{d} \\
\bar{b} \cdot \bar{c} & \bar{b} \cdot \bar{d}
\end{array}\right|\)

In the above formula if c̅ = a̅ and d̅ = b̅, then
(a̅ × b̅) . (c̅ × d̅) = \(\left|\begin{array}{cc}
\bar{a} \cdot \bar{c} & \bar{a} \cdot \bar{d} \\
\bar{b} \cdot \bar{c} & \bar{b} \cdot \bar{d}
\end{array}\right|\) = (a̅.a̅)(b̅.b̅) – (a̅.b̅)(a̅.b̅) = \(\overline{\mathrm{a}^2} \overline{\mathrm{b}^2}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}})^2\)

Question 10.
Let a̅, b̅ and c̅ be unit vectors such that b is not parallel to c and a̅ × (b̅ × c) = \(\frac{1}{2}\) b̅. Find the angles made by a̅ with each of b̅ and c̅. [May ’01]
Answer:
Since a̅, b̅ and c̅ be unit vectors then |a̅| = 1, |b̅| = 1, |c̅| = 1
Given a̅ × (b̅ × c̅) = -b̅
⇒ (a̅.c̅)b̅ – (a̅.b̅)c = \(\frac{1}{2}\)b̅
Since b̅ and c̅ are non-collinear vectors equating corresponding coefficients on both sides.
a̅.c̅ = \(\frac{1}{2}\)

|a||c| cos (a, c) = \(\frac{1}{2}\), -(a̅.b̅) = 0
1.1.cos(a̅, c̅) = \(\frac{1}{2}\), a̅.b̅ = 0
cos(a̅,c̅) = \(\frac{1}{2}\), a̅ ⊥ b̅
(a̅, c̅) = 60°, (a̅, b̅) = 90°

Question 11.
Let a̅ = i̅ + j̅ + k̅, b̅ = 2i̅ – j̅ + 3k̅, c = i̅ – j̅ and d̅ = 6i̅ + 2j̅ + 3k̅ . Express d̅ in terms of b̅ × c̅, c̅ × a̅ and a̅ × b̅. [May ’12]
Answer:
Given a̅ = i̅ + j̅ + k̅, b̅ = 2i̅ – j̅ + 3k̅, c̅ = i̅ – j̅, d̅ = 6i̅ + 2j̅ + 3k̅
= 1 (0 + 3) – 1 (0 – 3) + 1(-2 + 1) = 3 + 3 – 1 = 5
Now, d̅. a̅ = (6 i̅ + 2 j̅ + 3k̅). (i̅ + j̅ + k̅) = 11
d̅.b̅ = (6i̅ + 2j̅ + 3k̅).(2i̅ – j̅ + 3k̅) =19
d̅.c̅ = (6i̅ + 2j̅ + 3k̅).(i̅ – j̅) = 4
Take d̅ = x(b̅ x c̅) + y(c̅ x a̅) + z(a̅ x b̅), then we have x = \(\frac{\overline{\mathrm{d}} \cdot \overline{\mathrm{a}}}{\left[\begin{array}{lll}
\overline{\mathrm{a}} & \overline{\mathrm{b}} & \overline{\mathrm{c}}
\end{array}\right]}\), y = \(\frac{\overline{\mathrm{d}} \cdot \overline{\mathrm{b}}}{\left[\begin{array}{lll}
\overline{\mathrm{a}} & \overline{\mathrm{b}} & \overline{\mathrm{c}}
\end{array}\right]}\), z = \(\frac{\overline{\mathrm{d}} \cdot \overline{\mathrm{c}}}{\left[\begin{array}{lll}
\overline{\mathrm{a}} & \overline{\mathrm{b}} & \overline{\mathrm{c}}
\end{array}\right]}\)
∴ x = \(\frac{11}{5}\), y = \(\frac{19}{5}\), z = \(\frac{4}{5}\)
d̅ = \(\frac{11}{5}\) (3i̅ + 3j̅ – k̅) + \(\frac{19}{5}\)(-i̅ – j̅ + 2k̅) + \(\frac{4}{5}\)(4i̅ – j̅ – 3k̅)

Question 12.
For any four vectors a̅, b̅, c̅ and d̅, show that
(i) (a̅ × b̅) × (c̅ × d̅) = [a̅ c̅ d̅] b̅ – [b̅ c̅ d̅] a̅ and
(ii) (a̅ × b̅) × (c̅ × d̅) = [a̅ b̅ d̅]c̅ – [a̅ b̅ c̅]d̅. [Mar. ’18(AP); May ’99]
Answer:
(i) (a̅ × b̅) × (c̅ × d̅) = [(c̅ × d̅).a̅]b̅ – [(c̅ × d̅).b̅]a̅ = [a̅.(c̅ × d̅)]b̅ – [b̅.(c̅ × d̅)]a̅ = [a̅ c̅ d̅] b̅ – [b̅ c̅ d̅] a̅
(ii) (a̅ × b̅) × (c̅ × d̅) = [(a̅ × b̅). d̅]c̅ – [(a̅ × b̅). c̅]d̅ = [a̅ b̅ d̅]c̅ – [a̅ b̅ c̅]d̅

Question 13.
a, b, c are non-zero vectors and a is perpendicular to both b̅ and c̅. If |a̅| = 2, |b̅| = 3, |c̅| = 4 and (b̅, c̅) = \(\frac{2 \pi}{3}\), then find |[a̅ b̅ c̅]|. [May ’08]
Answer:
Given |a̅| = 2, |b̅| = 3, |c̅| = 4 and (b̅, c̅) = \(\frac{2 \pi}{3}\)
a is perpendicular to both b̅ and c̅.
Now b̅ × c̅ is a vector perpendicular to both b̅ & c̅.
a̅ is parallel to b̅ × c̅ (a̅, b̅ × c̅) = 0° or 180°

Now [a̅ b̅ c̅] = a̅. (b̅ × c̅) = |a̅| |b̅ × c̅| cos (a̅, b̅ × c̅) = |a̅| |b̅| |c̅| sin (b̅, c̅). cos (a̅, b̅ × c̅)
= 2.3.4.sin \(\frac{2 \pi}{3}\) . cos(0° or 180°) = 24. \(\frac{\sqrt{3}}{2}\)(±1) = 12√3 = |[a b c]| = 12√3

Question 14.
If [b̅ c̅ d̅] + [c̅ a̅ d̅] + [a̅ b̅ d̅] = [a̅ b̅ c̅], then show that the points with position vectors a̅, b̅, c̅ and d̅ are coplanar. [Mar ’14; Mar. ’00]
Answer:
Let the position vectors of the points A, B, C and D with respect to the origin O’ are \(\overline{\mathrm{OA}}\) = a, \(\overline{\mathrm{OB}}\) = b, \(\overline{\mathrm{OC}}\) = c, \(\overline{\mathrm{OD}}\) = d.
Given [b̅ c̅ d̅] + [c̅ a̅ d̅] + [a̅ b̅ d̅] = [a̅ b̅ c̅] ……………………(1)
Now \([\overline{\mathrm{AB}}  \overline{\mathrm{AC}}  \overline{\mathrm{AD}}]=[\overline{\mathrm{OB}}-\overline{\mathrm{OA}}  \overline{\mathrm{OC}}-\overline{\mathrm{OA}}  \overline{\mathrm{OD}}-\overline{\mathrm{OA}}]\)
= [b̅ – a̅ c̅ – a̅ d̅ – a̅]
= (b̅ – a̅). [(c̅ – a̅) × (d̅ – a̅)] (b̅ – a̅).[c̅ × d̅ – c̅ × a̅ – a̅ × d̅ + a̅ × a̅]
= (b̅ – a̅).[c̅ × d̅ + a̅ × c̅ + d̅ × a̅ + 0]
= b̅. (c̅ × d̅) + b̅. (a̅ × c̅) + b̅. (d̅ × a̅) – a̅ . (c̅ × d̅) – a̅. (a̅ × c̅) – a̅. (d̅ × a̅)
= [b̅ c̅ d̅] + [b̅ a̅ c̅] + [b̅ d̅ a̅] – [a̅ c̅ d̅] – [a̅ a̅ c̅] – [a̅ d̅ a̅]
= [b̅ c̅ d̅] – [a̅ b̅ c̅] – [b̅ a̅ d̅] + [c̅ a̅ d̅] – 0 – 0
= [b̅ c̅ d̅] – [a̅ b̅ c̅] + [a̅ b̅ d̅] + [c̅ a̅ d̅]
= [b̅ c̅ d̅] + [c̅ a̅ d̅] + [a̅ b̅ d̅] – [a̅ b̅ c̅]
= [a̅ b̅ c̅] – [a̅ b̅ c̅] = 0 (from (1))
∴ The vectors AB, AC, AD are coplanar.
∴ The four points A, B, C and D are coplanar.

Question 15.
Find the volume of the tetrahedron whose vertices are (1, 2, 1), (3, 2, 5), (2, – 1, 0) and (- 1, 0, 1). [Mar. ’15(TS); May ’07; Mar. ’04]
Answer:
Let the position vectors of the points A, B, C and D with respect to the origin ‘O’ are
\(\overline{\mathrm{OA}}\)= i̅ + 2j̅ + k̅, \(\overline{\mathrm{OB}}\) = 3i̅ +2j̅ +5k̅, \(\overline{\mathrm{OC}}\) = 2i̅ – j̅, \(\overline{\mathrm{OD}}\) = -i̅ + k̅
Now \(\overline{\mathrm{AB}}=\overline{\mathrm{OB}}-\overline{\mathrm{OA}}\)A
= 3i̅ + 2j̅ + 5k̅ – i̅ – 2j̅ – k̅
= 2i̅ + 4k̅

\(\overline{\mathrm{AC}}=\overline{\mathrm{OC}}-\overline{\mathrm{OA}}\)
=2i̅ – j̅ – i̅ – 2j̅ – k̅
= i̅ – 3j̅ – k̅

\(\overline{\mathrm{AD}}=\overline{\mathrm{OD}}-\overline{\mathrm{OA}}\)
= -i̅ + k̅ – i̅ – 2j̅ – k̅
= -2i̅ – 2j̅

The volume of the tetrahedron whose vertices are A, B, C and D is \(\frac{1}{6}\left[\begin{array}{lll}
\overline{\mathrm{AB}} & \overline{\mathrm{AC}} & \overline{\mathrm{AD}}
\end{array}\right]\)
\(\frac{1}{6}\left|\begin{array}{ccc}
2 & 0 & 4 \\
1 & -3 & -1 \\
-2 & -2 & 0
\end{array}\right|\)
= \(\frac{1}{6}\)|2(0 – 2) – 0(0 – 2) + 4(-2 – 6)|
= \(\frac{1}{6}\)|-4-32|
= \(\frac{36}{6}\) = 6
∴ Volume = 6 cubic units.

Question 16.
Prove that the four points 4i̅ + 5j̅ + k̅,-(j̅ + k̅), 3i̅ + 9j̅ + 4k̅ and -4i̅ + 4j̅ + 4k̅ are coplanar. [Mar. ’99]
Answer:
Let the position vectors of the points A, B, C and D with respect to the origin ‘O’ are \(\overline{\mathrm{OA}}\) = 4i̅ + 5j̅ + k̅,\(\overline{\mathrm{OB}}\) = -(j̅ + k̅), \(\overline{\mathrm{OC}}\) = 3i̅ + 9j̅ + 4k̅,\(\overline{\mathrm{OD}}\) = -4i̅ + 4j̅ + 4k̅
\(\overline{\mathrm{AB}}=\overline{\mathrm{OB}}-\overline{\mathrm{OA}}\) = -j̅ – k̅ – 4 i̅ – 5j̅ – k̅ = -4i̅ – 6j̅ – 2k̅
\(\overline{\mathrm{AC}}=\overline{\mathrm{OC}}-\overline{\mathrm{OA}}\) = 3i̅ + 9j̅ + 4k̅ – 4i̅ – 5j̅ – k̅ = -i̅ + 4j̅ + 3k̅
\(\overline{\mathrm{AD}}=\overline{\mathrm{OD}}-\overline{\mathrm{OA}}\) = – 4i̅ + 4j̅ + 4k̅ – 4i̅ – 5j̅ – k̅ = -8i̅ – j̅ + 3k̅

Now \([\overline{\mathrm{AB}} \overline{\mathrm{AC}} \overline{\mathrm{AD}}]=\left|\begin{array}{ccc}
-4 & -6 & -2 \\
-1 & 4 & 3 \\
-8 & -1 & 3
\end{array}\right|\)
= -4(15) + 6(21) – 2(33)
= -60 + 126 – 66
= 126 – 126
= 0
∴ The vectors AB, AC, AD are coplanar.
∴ The four points A, B, C and D are coplanar.

Question 17.
If a̅ = 2i̅ + j̅ – k̅, b̅ = – i̅ + 2j̅ – 4k̅ and c̅ = i̅ + j̅ + k̅, then find (a̅ × b̅).(b̅ × c̅). [Mar. ’19(AP); Mar. ’17(TS)]
Answer:
Given a̅ = 2 i̅ + j̅ – k̅; b̅ = – i̅ + 2j̅ – 4k̅; c̅ = i̅ + j̅ + k̅
Now a̅ × b̅ = \(\left|\begin{array}{ccc}
\overline{\mathrm{i}} & \overline{\mathrm{j}} & \overline{\mathrm{k}} \\
2 & 1 & -1 \\
-1 & 2 & -4
\end{array}\right|\)
= i̅ (-4 + 2) – j̅ (-8 – 1) + k̅ (4 + 1)
= -2i̅ + 9j̅ + 5k̅

b̅ × c̅ = \(\left|\begin{array}{ccc}
\overline{\mathrm{i}} & \overline{\mathrm{j}} & \overline{\mathrm{k}} \\
-1 & 2 & -4 \\
1 & 1 & 1
\end{array}\right|\)
= i̅ (2 + 4) – j̅ (-1 + 4) + k̅ (-1 – 2)
= 6i̅ – 3j̅ – 3k̅

(a̅ × b̅).(b̅× c̅) = (-2i̅ + 9j̅ + 5k̅ ).(6i̅ – 3j̅ – 3k̅) = -12 – 27 – 15 = -54

Telangana TSBIE TS Inter 1st Year English Study Material Grammar Odd Sound Out Exercise Questions and Answers.

TS Inter 1st Year English Grammar Odd Sound Out

Q.No. 18 (4 × 1 = 4 Marks)

A set of three words are given. One or two letters which are common in all the three words are underlined. The underlined letter stands for one sound in two of the given words and for a different sound in the other word. The word with a different sound is to be written as the response.

The pronunciation of English words is quite tricky and confusing. Some vowels and consonants are pronounced differently in different places. Learning all these varieties is necessary to master pronunciation. It is possible only with practice.

Look at the following words. Circle the word that sounds different with regards to the sound of the bold letters.

Circle the words that sound different with regard to the sound of the bold letters.

Exercise -1

Exercise – 2

Exercise – 3

Exercise – 4

Exercise – 5

Exercise – 6

Exercise – 7

Exercise – 8

Exercise – 9

Exercise – 10

Telangana TSBIE TS Inter 1st Year English Study Material Grammar Information Transfer Exercise Questions and Answers.

TS Inter 1st Year English Grammar Information Transfer

Q.No. 20 (1 × 4 = 4 Marks)

Information can be expressed through verbal (description) and non verbal (diagrams) modes. Some of the non-verbal modes are :

1. Pie-charts,
2. Bar graphs,
3. Tree diagrams,
4. Flow charts and
5. Tables.

The process of changing a text from Verbal to Non-verbal mode or vice versa is called Information Transfer. This is a very useful and important skill for students. Acquiring this skill enables the students make notes quickly, understand various texts effectively and present ideas clearly and briefly.

Non-verbal expressions are remarkable for their brevity, clarity, simplicity, accessibility and provision for comparative, contrastive and analytical studies.

1. PIE-CHARTS

A pie-chart is a circle divided into parts. Each part represents a particular thing. And eah part is in proportion to the ratio of that thing to its total. Studying the given pie chart slowly helps one understand the information given there. Then the information can be presented in verbal mode. Once the mode of representing the given information in the form of a pie-chart is understood, verbal text can be transferred into a pie-chart.

In a pie charl; the information is presented in the form of a circle. The circle is divided into sections called sectors. The contribution of each unit in the chart is represented in percentages.

Example 1 :
The following pie chart depicts the results of a survey regarding distribution of different Blood Groups in a college.
Blood Groups in a College

From the figure we can see that 35% of the students of a college have 0 Group of Blood and these students form the largest group. The next largest group comprises students with B Group of Blood. 30% of students belong to this category. 25% of students have AB Group of Blood. Finally, we see that only 10% of students have A Group of Blood. Thus, from the piechart we can conclude that while many students have O Group of Blood. Very few have A Group.

Example 2 :
The following piechart depicts the favourite subject of students in a class. We can see from the figure that five subjects have been taken into consideration – Economics, Civics, Commerce, English and 2^nd Language. Students who like Economics form the largest group. A quarter of the students of the class i.e 25% expressed preference for this subject. 20% of the students like English and the same percentage i.e 20% of the students like Commerce. Next in popularity is Civics, liked by 18% of the class. Finally, trailing closely behind Civics, comes 2^nd Language, which is the favourite subject of 17% of the students.
Favourite Subjects of Students

Exercises and Activities

Question 1.
The following paragraph gives the information about the most widely spoken languages in India. Convert the passage into a pie chart.

Hindi is the most widely spoken language in India. The fact that 44% of Indians speak Hindi across India justifies its title as our National Language. 9% of Indians speak Bengali followed by Marathi which is spoken by 8%. Telugu comes next in the list with 7%, Tamil and Gujarati account for 6% and 5% respectively. All other languages together share the remaining percentage.
Answer:
Pie chart showing languages spoken in India

Hindi – 44%
Bengali – 9%
Marathi – 8%
Telugu – 7%
Tamil – 6%
Gujarathi – 5%
All other
languages – 21%

Question 2.
Read the following paragraph and convert the information into a pie chart.

There are seven continents in the world. Asia is the largest continent with an area of 30% followed by Antarctica with 28%. North America occupies 17% of the land on the earth. South America stands fourth in the list with 12% of land. Africa and Australia are the fifth and sixth largest ones with their respective shares of 6% and 5%. Europe is the last in the list which occupies 2% of the land only.
Answer:
Areas of Continents

Continents
Asia – 30%
Antarctica – 28%
North America – 17%
South America – 12%
Africa – 6%
Australia – 5%
Europe – 2%

Question 3.
Observe the pie chart given below. It contains information about the mode of transport used by students of a certain junior college. Write a small paragraph.
Mode of Transport of Students

Answer:
Mode of Transport of Students
The given pie chart presents the mode of transport used by students of a particular junior college. A major part of them 40% – use the public transport, i.e. bus. A half of the share of bus, that is 20% of them travel by autorickshaws. Two wheelers and cars carry 15% each of the students. Just 10% of them use the cleanest and the healthiest mode – walking.

Question 4.
The pie chart given below shows how people spend their time on smart phones. Convert the information into a paragraph.
Time spent on Smart. Phones

Answer:
Time spent on Smart Phones
Time spent on smart phones is presented in the given pie chart. The lion’s share, i.e. 35% of the time goes to games. Social networking follows games with its share of 29% of the time. Utilities Consume 20% time. The share of music and videos is 8%. Others take 5% time. News comes last with just 3% time.

2. BAR BRAPHS

A bar graph is a diagram in which values of variables are shown by the length of rectangular columns with equal width. It is another visual representation of data. It helps to compare the values presented in a group. The bars can be plotted vertically or horizontally. A vertical bar chart is sometimes called a column bar chart.

Example 1 :
Given below is the iar graph that shows the cost of certain vegetables over a period of 4 months. Let us now make a detailed analysis.

The bar graph given below shows the cost of carrots and potatoes over a period of four months – January, February, March and April. Carrots were more costly than potatoes during all the months. In January carrots cost Rs. 35 a kilo, while potatoes cost a little less, at Rs. 30 a kilo. The cost of carrots increased to Rs. 40 in February, while there was a sharp fall in the cost of potatoes.

There was a sharp rise in the cost of both the vegetables after that and in March the cost of carrots was Rs. 50 per kilo while that of potatoes was Rs. 40. In April once again there was a steep increase in the cost of carrots but the cost of potatoes remained the same as in March. Thus we observe that the cost of carrots kept increasing over the months but that of potatoes fluctuating.
COST OF VEGETABLES (in Rs per kg)

Example 2 :
The following bar graph represents the favourite sports of various group of students studying in a college. Students of four sections HEC, CEC, BPC and MPC were asked about their preferences in sports. The number of students in each section varied. Three sports were considered – football, cricket and kabaddi. HEC students expressed great interest in cricket. 50 out of 85 students, i.e. more than half liked cricket. Very few in that section, just 5, were fond , of football. 30 liked kabaddi.

In the CEC section, consisting of 100 students, an equal number of students, i.e. 40 liked kabaddi and cricket. 20 liked football. With regard to the science sections, cricket was more popular among BPC students. An equal number in both the sections, 30, were fond of football. The figures for kabaddi too were more or less the same. The BPC section consisted of 88 students while MPC students were 75 in number. On the whole, one can conclude that cricket is the most popular sport in the college, followed by kabaddi.
FAVOURITE SPORTS OF STUDENTS

Exercises And Activities

Question 1.
The passage below represents the data of improvement of English language skills due to Internet usage. Present it in a bar graph.

Internet plays an important role in improving Reading skills. 94% participants in this study agreed that they improved their Reading skills by using Internet while 91% opined that they improved Translation skills. Internet usage helped 87% of participants in enhancing their vocabulary skills. 80% of participants unanimously agreed that they improved their Writing skills, Speaking skills and Grammar.
Answer:
Bar Graph Showing Skills due to Internet Usage

Question 2.
The following passage shows the favourite sports of the students of a school. Represent the data in a bar graph.

Cricket is the most favourite sport of the students which is liked by 80 students. Tennis falls behind Cricket with a slight difference. It is the favourite of 75 students. Swimming and Football are liked by 40 and 45 students respectively while Badminton is the favourite of 30 students. Hockey is the least favouring sport of the students which is liked by 20 students only.
Answer:
Bar Graph Showing Favourite Sport of Students

Question 3.
Analyse the bar graph given below and write about it in a paragraph.
MARKS OF STUDENTS

Answer:
The bar chart presents marks of three students in three subjects. Meena scored 70 in Telugu, 65 in Maths and in English just 50. Mala scored 65 in Maths, 50 in Telugu and only 40 in English. Megha secured 70 each in English and Maths but scored 60 in Telugu.

Question 4.
The given below bar graph shows how much dietary fibre is found in certain fruits. Convert the information into a paragraph.

Answer:
Fibre Content in Fruits
The given bar graph presents the details of fibre content in various fruits. The guava stands tall with six (6) grams of dietary fibre per a serving of one cup. Next comes the pear with five (5) grams per unit. The third in the order is the apple with four (4) grams per a cup. The banana and the orange have almost the same quantity of dietary fibre – three (3) grams per cup.

3. TREE DIAGRAMS

A tree diagram is another way of representing information. It has a branching tree-like structure. It shows how its components are related to one another. It helps us understand the relevant information in a short time.

Example 1 :
There are three types of muscle in the human body. They are smooth, cardiac and skeletal muscles. Smooth muscles are controlled by involuntary responses. Examples of smooth muscles are muscles in the digestive tract and blood vessels. The second type of muscle is cardiac muscle. It is also an involuntary muscle. Muscles that cover the heart are examples of cardiac muscles. The third type of muscle is the skeletal muscle. It is controlled by voluntary response. All the muscles attached to the bones such as biceps, deltoid are examples of skeletal muscles.

The above paragraph can be depicted in the form of a tree diagram as follows.

Example 2 :
A man who managed a popular hotel was asked the secret of his success. He said that only when customers were happy with the dining experience would they keep returning to the hotel. Dining would be a pleasant experience only if the food served was of a high standard. Good service too was equally important. He elaborated that food should be tasty and fresh. Service should be prompt and courteous.
Given below is a tree diagram representing the man’s views.

Exercises And Activities

Question 1.
Read the following paragraph and transfer the information into a tree diagram.

The oldest musical instrument in the world is the drum, made initially in one of the three ways. First, frame drums were made by stretching the skin over bowl-shaped frames. Next, rattle drums were made by filling gourds or skins with dried grains, shells, or rocks. Finally, tubular drums were made from hollowed logs or bones covered with skins. Both frame and tubular drums were struck with the hand or with beaters to produce sounds. In contrast, rattle drums were shaken or scraped to make rhythmic sounds. For thousands of years, drums have been used to transmit messages to call soldiers to battle and make music.
Answer:
Tree Diagram showing Types of Drums

Question 2.
Read the following paragraph and transfer the information into a tree diagram.

There are so many species of animals that we find living on the earth. Scientists grouped these animals into different classes based on certain similarities they share. Animals are divided into vertebrates, ones with backbones and invertebrates, those without backbones. The vertebrates are basically divided into five classes. They are commonly known as mammals, birds, fish, reptiles and amphibians. Arachnids and insects are the two commonly known classes in the invertebrates group.
Answer:
Tree Diagram showing Species of Animals

Question 3.
The following tree diagram depicts the classification of Vitamins. Present the information in a paragraph.

Answer:
Classification of Vitamins
The given tree diagram presents the classification of vitamins. Vitamins are broadly of two types. They are : 1) Soluble vitamins in water and 2) Soluble in fats. Vitamin B and Vitamin C fall in the category of ‘Soluble in water’. Vitamins A, D, E and K (four) belong to the group of vitamins soluble in fat and Vitamin B is sub-divided into Bl, B2, B3, B6 and B12 (five) types.

Question 4.
Study the following tree diagram and write it in a paragraph.

Answer:
Types of Oils
The given tree diagram explains the types of oils. Based on the source, oils are of three categories. They are : 1) Oils from nuts, 2) Oils from vegetation (plants / flowers) and 3) Oils from minerals. Examples are 1) groundnut oil, 2) oils from flowers and 3) oils from the crust of the earth. Groundnut oil is used in cooking. Oils from flowers go into the making of soap, medicines and perfumes (scents). Mineral oil fuels machines and automobiles.

4. FLOW CHARTS

We draw flow charts when we present information in the form of a process. For instance, we construct flow charts to put the information of the industrial production from raw product to finished product in a logical order in successive steps. Flow charts are simple to construct and easy to understand. Each step in the sequence is written in a diagram shape. These successive stages or steps are linked by connecting directional arrows. They guide readers to understand flow charts logically and follow the process from beginning to end. In these flow charts we find elongated circles, rectangles and diamond shaped diagrams.

Example 1 :
Describe how the following passage is presented in a flow chart. The passage shows the time table for children in a boarding school. You are supposed to wake up at 5 am every day and lights – out time is 9.30 pm. Siesta time is between 1 and 2 in the afternoon. Assembly begins at 8 am sharp in the school hall. You have to report to your House Prefect by 7.30 am on all school days. You may play any game between 4 and 6 pm. You must not be late for study time which is between 6 and 8 in the evening. School timings are from 8.30 am to 3.30 pm with an hour-long lunch break at 1 pm. These details are shown in a flow chart.

Time table of children in a boarding school

Example 2 :
Read the following paragraph and transfer the information into a flow chart.

Rayon is a man-made fiber. It is a reconstituted natural fiber – cellulose. Rayon is made by dissolving cellulose in a solution of sodium hydroxide or caustic soda. The cellulose is obtained from shredded wood pulp. The dissolved cellulose is formed into threads by forcing it through a spinneret in a dilute sulphuric acid setting bath. The threads are drawn from the setting bath, wound on a reel, washed, dried on a heated roller, and finally wound onto a bobbin.

Process of Making Rayon

Exercises And Activities

Question 1.
The following paragraph describes how clothes are washed.

Draw a flow chart based on the information given. First, fill a bucket half full with water. Then, add a spoonful of washing powder. Stir vigorously till the power mixes with water and forms foam. Put the unwashed clothes into it. Wait for fifteen minutes. Take out clothes and scrub with a brush to remove stains. Now, rinse the clothes with clean water.

Wring out the clothes gently by twisting and compressing them. This removes excess water from the clothes. This saves the time of drying. Now dry the washed clothes by putting them on the clothes line. Collect the washed and dried clothes later.
Answer:
How to wash clothes ?

Question 2.
Convert the following paragraph into a flow chart.

Silver occurs in the ores of several metals. The frothing process of extracting silver accounts for about 75% of all silver recovered. Here the ore is ground to a powder, placed in large vats containing a water suspension of frothing agents, and thoroughly agitated by air jets. Depending on the agents used, either the silver-bearing ore or the gangue adhering to the bubbles of the foam is skimmed off and washed. The final refining is done using electrolysis.
Answer:
Flow Chart depicting Frothing Process of Extracting Silver

Question 3.
The following flow chart describes how paper is manufactured in a paper mill. Write the details in a paragraph.
Manufacture of paper

Answer:
The given flow chart describes the process of manufacturing paper. First, the raw materials like wood, grass, bamboo and rags are procured. Secondly they are cut into pieces, immersed in water and made into pulp. Then the pulp is mixed with lime for whitening. Later, the pulp is boiled and passed through wire meshes. At this stage, we obtain wet paper. Finally, it is passed over heated rollers. Then we get the end product, in the form of thin sheets of paper.

Question 4.
Draw a flow chart based on the information given below.

The following process is the description of how a post office transfers a letter from a sender to a receiver. First, the sender posts the letter in a post box. Next, the box is opened. Then the contents in it are sorted out. Then they are kept in a bag and the bag is tied. The destination is written on the bag. The bags are sent to the district post office. The district post office sends the bags to the destination village / town post offices. The destination post office receives the letters. The received letters are arranged and sorted out. The post man delivers the letters to the addressees.
Answer:
Flow Chart depicting the Process of Delivering Letters

5. TABLES

We can also represent information in the form of a table.
Example 1 :
Given below are the marks secured by Aravind, Akash and Ramesh in their half-yearly examinations of class X.

After reading the information given in the table we can write a paragraph like this.

In this table, the marks secured by 3 students are compared. While all the three students scored equal marks in Hindi, there is a slight variation of marks in Mathematics and Social Studies. However, there is a great variation of marks in English. From the table it can be concluded that Aravind needs to concentrate more on English and Science, whereas Akash needs to focus on Telugu and English. Ramesh, who scored the highest marks among the three, needs to focus on Telugu.

Example 2 :
The following table shows the number of gold medals won by 8 participating countries in the XII South Asian games 2016. First read the data given in the table.

Now read the paragraph given below.

The above table gives the information of the number of gold medals won by 8 participating countries in the XII South Asian games 2016. India secured the first rank with 188 gold metals. It was far ahead of the other countries. Sri Lanka was ranked 2, securing only 25 gold medals. Pakistan got only 12 gold medals and was ranked 3. With 7 golds, Afghanistan is in the 4th place. Bangladesh won 4 golds while Nepal secured just 3 golds. Maldives and Bhutan which stood at the bottom of the table got no gold medals. This table shows the commendable performance of India in the XII South Asian Games.

Exercises And Activities

Question 1.
Read the following paragraph and transfer the information into a table.

A reading test assesses reading comprehension by employing multiple testing techniques, represented by eight main types of questions. Question types, such as Multiple-Choice, Matching, Diagram Labelling, Summary Completion, Sentence Completion, Short Answer Questions with percentage, i.e., 37.50%, 18.13%, 16.25%, 10%, 9.36%, and 8.76%, take place respectively. The number of questions for each of these questions types is variable. Basic English grammar, cloze summary, percentages are although with lower portions and are also considered in the reading test.
Answer:
Table Showing Types of Questions in Reading & Tests

Question 2.
Convert the following paragraph into a table.

There are many elements in the earth’s crust. Oxygen occupies 46%; Silicon 28%; Aluminum 8%; Iron 5%; Calcium 3.6%; Sodium 2.8%; Potassium 2.6%; Magnesium 2%; certain other elements occupy 2% of the earth’s crust. This is what we mean by the abundance of elements in the earth’s crust.
Answer:
Table Showing Elements in Earth’s Crust

Question 3.
Study the table below showing a few Asian countries with their capitals and currencies. Write a paragraph containing all the information in the table.

Answer:
The table presents the capitals and their currencies of 5 Asian countries. Kabul is the capital of Afghanistan and their currency is Afgani. China’s capital is Beijing and their currency is Yuan. With Yen as their currency Japan administers the country from Tokyo, the capital city. Saudi Arabia’s capital is Riyadh and their currency is Riyal. Finally Singapore has as its capital Singapore city and their currency is Singapore dollar.

Question 4.
Look at the following table. It gives information about nutrients (in gms) present in 100 ml. of milk. Present the information in the form of a paragraph.

Note : k stands for thousand; g stands grammes.
Answer:
The given table provides us information about tire nutrition value of milk. 100 ml of milk gives us 78 kcals of energy. Fat is 5.0 gms. Carbohydrates are 4.4 gms. Sugar Nil. Proteins 2.3 gms. Calcium 8.9 mg. and Minerals 0.8 grams.