JEE Knocout Test QP_Solution_Dec22 (1)

8/12/2019 JEE Knocout Test QP_Solution_Dec22 (1)

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JEE Knockout Test Solution

1. Time duration for this test is 180 minutes. This test consists of 90 questions. Themaximum marks are 360.

2. There are three parts in the question paper, namely, Part A: Mathematics, Part B:

Physics and Part C: Chemistry. Each question is awarded 4 (four) marks for correctresponse.

3. One-fourth (1/4) marks will be deducted for indicating incorrect response of eachquestion. No deduction from the total score will be made if no response is indicated foran item in the answer sheet.

4. This test contains 90 Multiple Choice Questions with single correct answer. EachQuestion has four choices (1), (2), (3) & (4), out of which Only One is Correct.

5. Filling up more than one response in any question will be treated as wrong response andmarks for wrong response will be deducted accordingly as per the instruction 3 above.

Part A

Mathematics

Question 1:If147 , then find .

3 4

iZ Z

i

(1) 27 (2) (2)7

(3) (27)i (4) (27) i

Chapter:Complex Numbers

Level of Difficulty:EasySolution:We have

7

3 4

iZ

i

Simplifying (i.e., rationalizing the denominator), we get

7 3 4 21 4 28 3

3 4 3 4 9 16

25 25

25

i i i i

i i

i

= 1i

Therefore,


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1414

2 7

2 7

7

7(1 )

3 4

[(1 ) ]

(1 2 )

( 2 )

ii

i

i

i i

i

Correct Option: (3).

Question 2:If 4 4 10,Z Z then the difference between the maximum and the minimum values of

Z is

(1) 2 (2) 3

(3) 41 5 (4) 0

Chapter:Complex Numbers

Level of Difficulty:Hard

Solution

4 4 10Z Z

Ps+Ps'= 2a

which implies that foci at 4 and4 and a= 5 as shown in the following figure.

Now,

b2= 25(1e

2)

= 25(5e)2

= 2516 = 9

b= 3

Zlies on the ellipse circumference Z denotes the distance from the origin. Therefore,

bi

-bi


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max

min

5

3

Z

Z

Thus, the difference between the maximum and the minimum values of Z is

max min 5 3 2Z Z


Question 3:If [x]27[x] + 10 < 0 and 4[y]216[y] + 7 < 0, then [x+y] cannot be ([] denotes greatest

integer function)

(1) 7 (2) 8

(3) 9 (4) Both (2) and (3)

Chapter:Quadratic Equations and Expressions

Level of Difficulty:Moderate

Solution: We have

2[ ] 7[ ] 10 0x x

([ ] 5) ([ ] 2) 0x x

2 [ ] 5x

[ ] 3or4x

[3, 5)x

and

24[ ] 16[ ] 7 0y y

(2[ ] 7)(2[ ] 1) 0y y

1 7[ ]

2 2y

[ ] 1 or 2 or 3y

[1, 4)y

Therefore,

[4,9)

[ ] {4,5,6,7,8}

x y

x y

Hence, [x+y] cannot be 9.

Correct Option:(3).


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Question 4:If a, band care the roots of the equationx3+ 2x2+ 1= 0, find .

a b c

b c a

c a b

(1) 8 (2)8

(3) 0 (4) 2

Chapter:Matrices, Determinants and System of Equations


Solution: As a, band care the roots ofx3+ 2x2+ 1 = 0, we have

a+ b+ c=2

ab+ bc+ ca= 0

abc=1

Now, for finding the value of

a b c

b c ac a b

, evaluating using first row, we get

a(bca2)b(b2ac) +c(abc2) = abca3b3+ abc+ abcc3

= 3abca3b3c3

=(a3+ b

3+ c

33abc)

=(a+ b+ c) (a2+ b2+ c2abbcca)

=(2) [(2)23(0)] = 8

Correct Option:(1).

Question 5:If1 2 cos sin

, ,0 1 sin cos

TA P Q P AP

, find 2014 .TPQ P

(1)20141 2

0 1

(2)1 4028

0 1

(3)2013 2014 2013( )TP A P (4) 2014TP A P

Chapter:Matrices, Determinants and System of Equations


Solution: If1 2 cos sin

, , ,0 1 sin cos

TA P Q P AP

we have

PQ20/4PT ( )( ) ( )

2014 times

T T T T P P AP P AP P AP P

( ) ( ) ( ) ( ) ( )T T T T T PP A PP A PP PP A PP


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Matrix multiplication is associative:

cos sin cos sin

sin cos sin cos

TPP

2

1 0

0 1I

Hence,PQ2014PT=A2014

21 2 1 2 1 2 1 4

0 1 0 1 0 1 0 1A A

3 1 4 1 2 1 6

0 1 0 1 0 1A

4 1 6 1 2 1 8

0 1 0 1 0 1A

20141 2 1 4028and =0 1 0 1

n nA A

Correct Option:(2).

Question 6:Find the value of3 3 3 3 3 3 3 3 3 3

2 6 12 20

1 1 2 1 2 3 1 2 3 4

up to 60 terms.

(1) 2 (2)1

2

(3) 4 (4)1

4

Chapter:Progressions, Sequences and Series


Solution

3 3 3 3 3 3 3 3 3 3

2 6 12 20...

1 1 2 1 2 3 1 2 3 4

3 3 3 3 3 3

1 2 2 3 3 4...

1 1 2 1 2 3

3 3 31

( 1)lim

1 2 ...

n

n

n n

n

21

( 1)lim

( 1)

2

n

n

n n

n n


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1

1lim 4

( 1)

n

n n n

1

1 14 lim

1

n

n n n

1 1 1 1 1 14 lim ...

1 2 2 3 1n n n

4 lim 41n

n

n


Question 7:In the expansion of (1 +x)2(1 +y)3(1 +z)4(1 + w)5, the sum of the coefficient of the terms

of degree 12 is

(1) 61 (2) 71

(3) 81 (4) 91

Chapter:Binomial Theorem


Solution

(1 +x)2(1 +y)

3(1 +z)

4(1 + w)

5

General term 2 3 4 5 a b d ea b d eC C C C x

2 3 4 5 14 14

12 1212

14 13

or 2a b d ea b d e C C C C C C

= 91

Correct Option:(4).

Question 8:Let Sbe the set of all functions from the set {1, 2, , 10} to itself. One function is selected

from S, the probability that the selected function is oneone onto is

(1)9

9!

10 (2)

1

10

(3)100

10! (4)

10

9!

10

Chapter: Probability


Solution:Letfbe function from {1, 2,,10} to itself total functions possible is 1010

. The number of

oneone onto functions possible is 10!

Hence, the probability of selected function to be oneone onto is


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10 9

10! 9!

10 10

Correct Option:(1).

Question 9: Two friends visit a restaurant randomly during 5 pm to 6 pm. Among the two, whoever

comes first waits for 15 min and then leaves. The probability that they meet is

(1)1

4 (2)

1

16

(3)7

16 (4)

9

16

Chapter: Probability


Solution: Let the friends come to the restaurant at 5 h x min and 5 h y min, respectively, where, [0,60]x y .

Hence, the sample space consists of all points (x, y) lying in 60 60 square as shown above and for

favourable cases, 15x y , that is, 15 xy 15 which is shown by shaded region in the graph

shown below.

Hence, the probability that they will meet is given by

145 45

2160 60

23 7

14 16

Correct Option:(3).


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Question 10: If ( ) ( ) ( )r a m n b n l c l m and [ ] 4l mn , find .( )

a b c

r l m n

(1)1

4 (2)

1

2

(3) 1 (4) 2

Chapter: Vectors


Solution

( ) ( ) ( )r a m n b n l c l m

where

[ ] 4l mn

4r l a

4r m b

4r n c

which imply that

1

4( )

a b c

r l m n

Correct Option

The correct option is (1).

Question 11: The volume of tetrahedron, for which three co-terminus edges are , anda b c , is k units.

Then, the volume of a parallelepiped formed by , 2a b b c and 3a c is

(1) 6k (2) 7k

(3) 30k (4) 42k

Chapter: Vectors


Solution

The volume tetrahedron is given by

1[ ] [ ] 6

6k a b c a b c k

The volume of parallelepiped is given by

[ 2 3 ]a b b c a c

[ 2 3 ] [ 2 3 ]a b c ac b b c a c


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[ 3 ] [ 2 3 ] [ 3 ] [ 2 3 ]a b a c a c a c b b a c b c a c

[ ] [ 2 3 ]a b c b c a

[ ] 6[ ]a b c a b c

7[ ]a b c

Volume is 42k.

Correct Option:(4).

Question 12: Statement 1: Negation of the statement if price increases, then demand falls is price

increases and demand does not fall

Statement 2:Negative ofp qis ~p q .

(1) Statement 1 is True; Statement 2 is True; Statement 2 is a correct explanation for Statement 1.

(2) Statement 1 is True; Statement 2 is true; Statement 2 is not a correct explanation for Statement 1.

(3) Statement 1 is True; Statement 2 is False.

(4) Statement 1 is False; Statement 2 is True.

Chapter: Mathematical Reasoning

Level of Difficulty:Easy

Solution: The negation ofp qis ~ .p q Hence, the negation of if price increases then demand falls

is price increases and demand does not falls.

Statement 1 is correct, but Statement 2 is incorrect.

Correct Option:(3).

Question 13: Simultaneously two dice are thrown five times. If getting a sum as multiple of 3 is

considered as success, then the variance of the distribution of number of successes is

(1)5

3 (2)

10

3

(3)10

9 (4)

140

121

Chapter: StatisticsLevel of Difficulty:Hard

Solution

The sum as multiple of 3, possibilities are 3, 6, 9 and 12, that is, (1, 2), (2, 1), (1, 5) (2, 4) (3, 3) (4, 2) (5,

1) (3, 6) (4, 5) (5, 4) (6, 3) and (6, 6). There are 12 possibilities and hence the probability of success is


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12 1

36 3

It is a binomial distribution with n= 5 and1

3p , which implies that the variance is

1 2 1053 3 9

npq

Correct Option:(3).

Question 14: If 270 < < 360, then find 2 2(1 cos ) .

(1) 2sin4

(2) 2sin

4

(3) 2sin

4

(4) 2cos

4

Chapter: Trigonometric Ratios and Identities


Solution: When, 270 < < 360, we have

2(1 cos ) 22 2cos2

which is non-negative. Now, the above equation can be written as

2(1 cos ) 2 cos

2

2cos2

cos 0 when135 180

2 2

Now, let us consider that

2 2(1 cos )

which is non-negative. That is,

2

2 2(1 cos ) 2 2cos2

2 1 cos2

2 2sin4

2 sin4


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1352sin sin 0when 90

4 4 2 4

Correct Option:(2).

Question 15: The number of solutions of sin1x+ sin1(1 +x) = cos 1xis/are

(1) 0 (2) 1

(3) 2 (4) infinite

Chapter: Inverse Trigonometric Functions


Solution

sin1(1 +x) is defined forx< 0 and1 1sin cos 1 1.

2x x x

The given equation is1 1 1sin sin (1 ) cosx x x

which can be written as

1 1 1cos cos (1 ) cos2 2

x x x

1 1cos (1 ) 2cosx x

1 1 2cos ( 1 ) 2 cos (2 1)x x

1 1 2

cos ( 1 ) cos (2 1) 2x x

1 1 2cos ( 1 ) cos (2 1)x x

21 2 1 1x x

0x

Which implies that the total number of solutions sin1x+ sin1(1 +x) = cos

1xis only one.

Correct Option:(2).

Question 16: IfP(A) is the power set of setA, which of the following statements is false?

(1) { : }A B x x A x B (2) A B B A

(3) N I Q R C (4) ( ) { : }P A X X A

Chapter: Sets


Solution


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1. :x x A x B means that all elements exist in setAalso exist in setB. Similarly, A B means

that all elements exist in set Aalso exist in set B. The Venn diagram can be drawn as shown in

any of the following two figures:

Therefore, option (1) is correct.

2. A B meansAis a subset ofBorB A means thatBis a super set ofA. Therefore, option (2) is

correct.

3. It is a well-known fact that N I or N I and hence both are correct. Therefore, option (3) is

correct.

4. The statement ( ) { : }P A X X A is wrong and the correct statement is ( ) { : }P A X X A , that

is, setX= setAis also included. Therefore, option (4) is incorrect or false.

Correct Option:(4).

Question 17: The value of2

lim ( , , , , {0})x

ax bx ca b c d e

dx e

depends on the sign of

(1) aonly (2) donly

(3) aand donly (4) a, band donly

Chapter: Limits


Solution

2 ( / )lim lim

( / )

lim

x x

x

ax bx c ax b c x

dx e d e x

a bx

d d

= +ifa

d

is positive.

=ifa

d

is negative.


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Correct Option:(3).

Alternate Solution

2 2

2

( / ) ( / )lim lim

( / ) ( / )x x

ax bx c a b x c x

dx e d x e x

Here,2

e d

xx . Therefore,

2

lim lim/

if 0 and 0if 0

if 0 and 00

if 0 and 0if 0

0 if 0 and 0

x x

ax bx c a

dx e d x

a dad

a d

a a dd

a d

Question 18: Ify= (x3)(x2)(x1)x(x+ 1)(x+ 2)(x+ 3), then2

2

d y

dxatx= 1 is

(1) 101 (2) 48

(3) 56 (4) 190

Chapter: Differentiation


Solution

We have

2 2 2( 9)( 4)( 1)y x x x x

6 4 2( 14 (49) 36)x x x x

7 5 314 49 36x x x x

Therefore,

6 4 27 70 147 36dy

x x xdx

Thus,

2 5 3

2 42 280 294d y x x x

dx

2

2

1

42 280 294

x

d y

dx

= 56

Correct Option:(3).


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Question 19: If sin cos( )2

dy xx

dx

, thenyis strictly increasing in

(1) (3, 4) (2)5 7

,2 2

(3) (2, 3) (4)1 3

,2 2

Chapter: Applications of Derivatives


Solution: Let us draw the graph of

( ) sin2

f x x

and

( ) cos( )g x x

on the samexy-plane as shown in the following figure.

From this graphical representation, it is clear that yis strictly increasing in

5 7,

2 2

f(x)

x

y


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Because for all values ofx,

5 7

2 2x

That is,

sin 02

x

and

cos ( ) 0x

which imply that

0dy

dx

which means thatyis strictly increasing.

Correct Option:(2).

Question 20: Given that

2 1 1

2 2

2 3

2 1(1 ) 2 1 1 1n n n

ndx x dx

nx n x x

. Find the value of

1

40 2.

1

dx

x

(You may or may not use reduction formula given.)

(1)11 5

48 64

(2)

11 5

48 32

(3)1 5

24 64

(4)

1 5

96 32

Chapter: Indefinite Integration


Solution


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3

11 1

4 2 4 10 0 2)20

11

3 2 2 2 20

01

1

2 200

1

5

62(4 1)(1 ) (11

1 5 5 30 .

6 6 46(2) 2(2)(1 ) (1 )

1 5 1 5 5 10

48 6 16 8 8 22(1)(1 ) 1

1 5 5 1 50 tan

48 6 16 8 4 16

dx x dx

x xx

x dx

x x

x dx

x x

x

1

0

7 5 50

6 16 8 4 16 4

22 5

6 16 64

11 5

48 64

Correct Option:(1).

Alternate Solution

1

40 21

dxI

x

Put tanx ; therefore, 2secdx d .

2

480

sec

sec

d

That is,

/4 6

0

2/4

0

/4 /4 /42 2

0 0 0

/4 /4 /4

0 0 0

/4 /4

0 0

(cos )

3cos cos3

4

9 1 3cos (cos3 ) cos cos3

16 16 8

4 cos 29 1 cos2 1 1 cos6 3

16 2 16 2 8 2

9 sin 2 1 sin 6

32 2 16 2 6

d

d

d d

cosd d

/4

0

3 sin 4 sin 2

8 2 4 2

9 1 1 1 3 10 0

32 4 2 16 2 4 6 8 2 2

5 11

64 48


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Question 21: If the length of the normal for each point on a curve is equal to the radius vector, then the

curve

(1) is a circle passing through origin

(2) is a circle having centre at origin and radius > 0

(3) is a circle having centre onx-axis and touchingy-axis

(4) is a circle having centre ony-axis and touchingx-axis

Chapter: Differential Equations


Solution: The length of the normal is

2

1 dy

ydx

The length of radius vector of a point (x,y) on the curve is xi yj , that is, 2 2 .x y It is given that

2 2 21 ( ')x y y y

Squaring on both sides of this equation, we get

2 2 2 2

22 2 2 2

2

2

[1 ( ') ]

or

x y y y

dyx y y y

dx

dyx y dx

dy dyy x y x

dx dx

Now,

dyy x

dx

ydy xdx

Integrating on both sides, we get

2 2

2 2 2 2

2 2

2 or Constant

y x c

x y c x y

This answer does not exist in the given options. So, consider the other alternative:

y dy x dx

Integrating on both sides, we get


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2 2

2 2

y xc

2 2 Constantx y

and this constant is > 0 in practical sense.

Correct Option:(2).

Question 22: Find the value of/4

4

0

(sin ) .x dx

(1)3

16

(2)

3 1

32 4

(3)3 3

32 4

(4)

3 7

16 8

Chapter: Definite Integration


Solution

We have

/44

0

(sin )I x dx

(1)

We know that

2 1 cos2sin2

xx

Therefore,

24 4

2

1 cos2sin (sin )

2

1[1 2cos2 (cos2 ) ]

4

1 1 cos 41 2cos2

4 2

1 3 cos 42cos2

4 2 2

xx x

x x

xx

xx

Substituting this value of 4sin x in Eq. (1), we get


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/4

0

/4/4 /4

0 00

3 1 1cos 2 cos 4

8 2 8

3 1 1[sin 2 ] [sin 4 ]

8 4 32

3 1 1. (1 0) (0 0)8 4 4 32

3 1

32 4

I x x dx

x x x

Correct Option:(2).

Alternate Solution: We have

/44

0

(sin )I x dx

which can be written as

2 2

2 2 2

2

(sin ) (1 cos )

1sin 4sin cos

4

1 cos 2 1(sin2 )

2 4

1 1 1 1 cos 4sin 2

2 4 4 2

sin 2 1 1sin 4

2 4 8 32

3 sin 2 sin 4

8 4 32

J x x dx

xdx x x dx

xdx x dx

xx x dx

x xx x c

x x

x c

Using the given limits, the above equation becomes

/4 /4 /4/4

00 0 0

3 sin 2 sin 4[ ]

8 4 32

3 1

32 4

x xI J x

Question 23: The area of the loop formed by 2 3(1 )y x x dx is

(1)1 4

0x x dx

(2)

14

0

2 x x dx


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(3)1 4

1x x dx

(4)

1/24

0

4 x x dx

Chapter: Area Under the Curve


Solution: We have

2 3(1 )y x x (1)

For 1x , 2y is negative. Since the square of a real number cannot be negative, ydoes not exist at x =

0 or atx= 1;y= 0. Let1

2x . Therefore, from Eq. (1), we get

2 1 1 712 8 16

7

4

y

y

Also, forx< 0, 2y is negative. Therefore, the required area is

1 13

0 0

14

0

2 2 ( ) 1

2

y dx x x dx

x x dx


Quick Tip: Whenx< 0, no curve exists; whenx< 1, no curve exists. Therefore,y2cannot be negative ify

is real.

1


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Question 24: The equation of a plane passing through the line of intersection of the planes

2 10 0x y z and 3 5x y z and passing through the origin is

(1) 5 3 0x z (2) 5 3 0x z

(3) 5 4 3 0x y z (4) 5 4 3 0x y z

Chapter: Three-Dimensional Geometry


Solution

We know that the equation of the plane passing through the line of intersection of planes 1 0p and

2 0p is

1 2 0p p

That is,

( 2 10) (3 5) 0x y z x y z (1)

Since, this plane passes through the origin, (0,0,0) satisfies this equation. This implies that

( 10) ( 5) 0

2

Substituting the value of in Eq. (1), we get

( 2 10) 2(3 5) 0x y z x y z

That is,

5 3 05 3 0x zx z

Correct Option:(2).

Question 25: Find the locus of a point whose distance from x-axis is twice the distance from the point

(1, 1,2) .

(1) 2 2 2 4 6 0y x y z (2) 2 2 2 4 6 0x x y z

(3) 2 2 2 4 6 0x x y z (4) 2 2 2 4 6 0z x y z

Chapter: Three-Dimensional Geometry

Level of Difficulty: Hard

Solution: Let the point ( , , )PP P

P x y z be the required point.

The distance of the point fromx-axis is 2 2 .P Py z

The distance from the point (1, 1, 2) is


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22 2( 1) 1 ( 2)P P Px y z

2 2 2 2 2( 1) ( 1) ( 2)P P P P Py z x y z

2 2 2 4 6 0P P P Px x y z

Therefore, the locus of pointPis

2 2 2 4 6 0x x y z

Correct Option:(3).

Question 26: If the axes are rotated through 60 in the anticlockwise sense, find the transformed form of

the equation 2 2 2x y a .

(1) 2 2 23 3 2X Y XY a (2) 2 2 2X Y a

(3) 2 2 22 3 2Y X XY a (4) 2 2 22 3 2X Y XY a

Chapter: Rectangular Coordinate System

Level of Difficulty: Moderate

Solution: Let ( , )x y and ( , )X Y be the old and the new coordinates, respectively. Since, the axes are

rotated in the anticlockwise direction, 60 . Therefore,

cos60 sin60

sin60 cos60

x X

y Y

1 32 2

3 1

2 2

x X

y Y

3

2 2

3

2 2

XY

x

y YX

3 3and

2 2 2 2

x Yx Y y X

2 2

23 3

2 2 2 2

X yY X a

2 2 2 2( 3 2 3 ) (3 2 3 )X Y XY X Y XY yc

2 2 22 2 4 3 4X Y XY a


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2 2 22 3 2Y X XY a

which is the required equation

Correct Option:(3).

Question 27:The straight lines 3 4 0, 3 4 0x y x y and 0x y form a triangle which is

(1) equilateral (2) right-angled

(3) acute-angled and isosceles (4) obtuse-angled and isosceles

Chapter: Straight Lines and Pair of Lines


Solution: The following figure depicts the condition. By observation from the figure, ABC is clearly an

obtuse angled and isosceles triangle.

Correct Option:(4).

Alternate Solution: The following figure depicts the condition.


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From the figure, we get

A: 3 4 and 2; 2x y y x x y

B: (1,1) by solving the equations.

C: 3 4 0 and 2; 2x y y x x y

Thus,

2 2

1 9 10

4 4 4 2

10 10 16(2)cos 0

2( 10)( 10)

AB BC

AC

B

Therefore, the given triangle is isosceles and obtuse angled triangle.


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Question 28: Find the equation of the circle whose diameter is the common chord of the circles

2 2 2( )x a y a and 2 2 2( )x y b b

(1) 2 2 2 2( )( ) 2 ( ) 0x y a b ab bx ay (2) 2 2 2 2( )( ) 2 ( ) 0x y a b ab bx ay

(3)2 2 2 2

( )( ) 2 ( ) 0x y a b ab bx ay (4) 2 2 2 2

( )( ) 2 ( ) 0x y a b ab ax by

Chapter: Circle

Level of Difficulty: Easy

Solution: Since 2 2 2( )x a y a , it is concluded that this circle has centre on x-axis and the circle passes

through origin.

Since 2 2 2( )x y b b , it is concluded that this circle has centre on y-axis and the circle passes through

origin.

Therefore,

2 2 2 0 (1)x y ax

2 2 2 0 (2)x y by

Subtracting Eq. (2) from Eq. (1), we get the common chord of the two circles:

(3)ax by

Solving Eqs. (1) and (3); we get the point of intersection of the two given circles:

22 2 0

ax x ax

b

2

20 or 2 0

ax x x a

b

2

2 2

20or

abx x

a b

2

2 2

20or

a by y

a b

Therefore, the equation of circle having (0, 0) and2 2

2 2 2 2

2 2,

ab a b

a b a b

as end points of one of its diameter

is

2 2

2 2 2 2

2 2( 0) ( 0) 0

ab a bx x y y

a b a b

2 22 2

2 2 2 2

2 20

ab a bx x y y

a b a b


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2 2 2 2( )( ) 2 ( ) 0x y a b ab bx ay

Correct Option:(1).

Question 29:Tangents are drawn from the point (4, 2) to the curvex2

+ 9y2

= 9. The angle between these

tangents is

(1)3 3

5 17 (2)

43

10

(3)43

5 (4)

3

17

Chapter: Ellipse


Solution

The combined equation of pair of tangents drawn from a point (x1,y1) to the ellipse2 2

2 2 1 0

x yS

a b is

S12= SS11. Therefore,

2 2 22 21 1 1 1

2 2 2 2 2 2

2 2 22 2

2 2

2

1 1 1

4 42 1 1 2 1

9 9 9

3 7 16 8 36 52 0

2tan

xx yy x yx y

a b a b a b

x xy y

x y xy x y

h ab

a b

where, a= 3, b= 7 and h=8. Therefore,

2 64 21 43tan

10 5

Note:is acute angle between the pair of tangents. Therefore,

2 2 2 2( ) 2 2 2a b c a b c ab ac bc

Correct Option:(2).

Quick Tip

S12= SS11

22

tan h ab

a b

, that is,


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1 2

1 2

21 2 1 2

1 2

tan1

( ) 4

1

m m

m m

m m m m

m m

Alternate Solution: Any line passing through the point (4, 2) is given by

2 ( 4)y m x

4 2y mx m

For this line to be tangent to the given ellipse, put this yinto the equation of the ellipse and make

D= 0

That is,

2

2

2 2 2 2

( 4 2) 19

(1 9 ) (36 72 ) 16(9) 16(9) 27 0

x mx m

m x x m m m m

Now,

2

2 2 2 2

2 2 2 2

2 2 2 4 2

2

0 4 0

(36 72 ) 4(1 9 )(16.9 16.9 27) 0

(36 ) (1 2 ) 36(1 9 )(16 16 3) 0

(1 4 4 ) 36(16 16 3 9.16 9.16 27 ) 0

7 16 3 0

D B AC

m m m m m

m m m m m

m m m m m m m m

m m

Now,

21 2 1 21 2

1 2 1 2

2

2

( ) 4tan

1 1

16 34.

7 2tan

31

7

7 16 4.3.7

10 7

14(43)

10

43

5

m m m mm m

m m m m

where is the acute angle between the tangents.


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Question 30:The chord of contact of a point A AA( , )x y of2 4y x passes through (3, 1) and point A lies

on 2 2 25x y . Then

(1)2

A A5 24 11 0x x (2)2

A A13 8 21 0x x

(3) 2A A5 24 61 0x x (4)2A A13 21 31 0x x

Chapter: Parabola


Solution: The following figure depicts the condition. Chord of contact of a point A AA( , )x y with respect

to 2 4y x is A A2( ).y y x x Since this chord passes through the point (3, 1), we have

A A2( 3)y x

AB and AC are tangents to the parabola.

BC is chord of contact of point A with respect to the parabola 2 4 .y ax

Given that point A lies on 2 2 25x y , we have


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2 2A A

2 2A A

2 2A A A

2A A

2A A

25

4( 3) 25

4( 9 6 ) 25

5 24 36 25 0

5 24 11 0

x y

x x

x x x

x x

x x

Quick Tip: The chord of contact of point A AA( , )x y with respect to2 4 0y ax is

A A2( )( ) 0y y a x x

Correct Option:(1).


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Part B Physics

Question1. The figure below shows two different arrangements of materials 1, 2 and 3 to form a wall. All

the three slabs are of equal thickness. The thermal conductivities are 1 2 3k k k . The left side of the wall

is 20 o C higher than the right side. Which of the following statements is correct for steady state heat flow

across the wall?

(1) The temperature difference across 1 will be the greatest in both the arrangements.

(2) The rate of heat transfer across the wall will be the same for both the arrangements.

(3) The temperature difference across 1 will be greater in the first arrangement than in the second

arrangement.

(4) The rate of heat transfer across the wall will be greater in the first arrangement than in the second

arrangement.

Chapter:Conduction


Solution

In both the arrangements, the three materials are in series. Since, the temperature difference across the

wall is the same for both the arrangements, the heat current will also be the same for both. Hence, the rate

of heat transfer across the wall will be the same for both the arrangements.

Correct Option:(2)

Question2.A bubble from the tank of a scuba diver contains43.5 10 mol of gas. The bubble expands

as it rises to the surface from a freshwater depth of 10.3 m. Assuming that the gas is an ideal gas and the

temperature remains constant at 291 K, the amount of heat that flows into the bubble is

(1)1

0.423 ln2

(2)1

0.846 ln2


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(3) 0.846 ln(2) (4) 0.423 ln(2)

Chapter:Thermodynamics


Solution

The process of expansion of the gas is an isothermal process. Therefore, 0U andQ W .

2

1

ln V

Q W nRT V

Since the process is isothermal,

2 1

1 2

V P

V P ,

where1Pis the pressure inside the bubble at the bottom and 2P is the pressure at the top of the surface.

2 1 atmP .

1 2 Pressure of 10.3 m columnP P of water

= 1atm + 1 atm [since pressure of 10.3 m column of water = 1

atm]

2 atm

Hence,

2 1

1 2

2V P

V P

So, the amount of heat that flows into the bubble will be

2

1

4

ln

3.5 10 8.31 291 ln 2

0.846 ln 2

Q W

VnRT

V

Correct Option:(3)

Question3. An ideal gas is taken along the cycle ABCA as shown in the figure given below with

1 40PaP , 3

1 1.0mV , 2 10PaP , and3

2 4.0mV . The amount of heat transferred into the system

during the entire cyclic process is


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(1)45 J (2) 45 J

(3) 55 J (4) 55 J

Chapter:Thermodynamics


Solution

For the cyclic process, change in internal energy 0U . So, using first law of thermodynamics we get

Q W

Now, the work done will be

W = the area enclosed inside the cycle

1 2 2 1

3 3

1( )( )

21

(40Pa 10Pa)(4.0m 1.0m )2

45 J

P P V V

Thus, heat is transferred into the system during the entire cyclic process will be 45J.Q

Correct Option:(2)

Question4.A ball suspended by a light inextensible string from a ceiling is shifted by an angle 0 (as

shown in the figure below) so as to strike the vertical wall normally with the coefficient of restitution e.The maximum angle made by the string with vertical after the first collision is

0

O


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(1) 1 2 0cos 1 (1 cos )e (2) 1 2 0sin 1 (1 sin )e

(3) 0 (4) 1 2 0cos (1 cos )e

Chapter:Collisions


Solution

0

O

h1

l

Let the mass of ball be mand length of the string suspending the ball be l. Then the height h1with respect

to the position at collision is given by

01 1 cosh l

Let v1be the speed of the ball before the first collision. Conserving mechanical energy up to first collision

i i f f K U K U

21 11

0 02

mgh mv

02

1

11 cos

2gl v

01 2 1 cosv gl

If v2is the speed of the ball just after the collision, then

2 1 02 1 cosv ev e gl

If ball rises to height h2after the collision then the angle made by string with the vertical is related to it

as

2 1 cosh l

Conserving mechanical energy after the first collision up to rise to height h2

i i f f K U K U

22 21

0 02

mv mgh

021 2 1 cos 1 cos

2e gl gl


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02 1 cos 1 cose

2 0cos 1 1 cose

1 02cos 1 1 cose

Correct Option:(1)

Quick Tip: By substituting e= 1, you should get 0and by putting e= 0, you should get 0. That you get

only with 1 2 0cos 1 (1 cos )e and 1 2 0sin 1 (1 sin )e and no other option. By realizing the

relation of height with angular deviation, that is, 1 cosh l , you can zero in the right option of

1 2 0cos 1 (1 cos )e .

Question5.Planet A has mass

24

3.00 10 kgM and radius 7

2.00 10 mR , and it completes a fullrotation in time 35.0hT . What is the free-fall accelerationgon its equator? (Take G=

11 2 26.67 10 Nm / kg )

(1) 20.500ms (2) 20.050ms

(3) 20.450ms (4) 20.550ms

Chapter:Gravitation


Solution

The gravitational acceleration on the planets surface is given by

g 2

11 24

7 2

2

(6.67 10 )(3.00 10 )

(2.00 10 )

0.500ms

GMa

R

The centripetal acceleration of any mass on the planets surface at the equator is

2

c

2

2 7

2

2

2

4 (2.00 10 )

(35 3600)

0.0497 ms

a R

RT


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In rotational frame any mass will experience a centrifugal force (= mac) which is directed away from the

center. Hence, achas to be subtracted from ag. Free-fall acceleration at the equator of the planet will be

g c

2 2

2

0.500ms 0.0497ms

0.450ms

g a a

Correct Option:(3)

Question6.The position of a particle moving along thex-axis is given by 37.8 9.2 2.1x t t , with x in

meters and tin seconds. What is the velocity at 3st ?

(1) 121 ms (2) 135 ms

(3) 147.5 ms (4) 154ms

Chapter:Motion along a Straight LineLevel of Difficulty:Easy

Solution

Position of a particle moving along thex-axis is

37.8 9.2 2.1x t t

The velocity of the particle will be

3

2

7.8 9.2 2.1

9.2 6.3

dxv

dt

d

t tdt

t

At 3st , the velocity of the particle is

2 19.2 6.3(3 ) 47.5 msv

Correct Option:(3)

Question7.A particle is launched into projectile motion over level ground. At point A, 3.0m above

ground level, the particle has velocity 1 1

(5.0 ms ) ( 6.0 ms )v i j

. What was its angle of launch?

(1) 1 6.25

tan5

(2) 1 3.65

tan5

(3) 1 8.59

tan5

(4) 1 9.74

tan5

Chapter:Motion in Two Dimensions and Projectile Motion


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Solution

At h=3.0m above ground level,

1 1 (5.0 ms ) ( 6.0 ms )v i j

Let the initial velocity be x yu u i u j .

Then 15.0 msxu because for the projectile, the horizontal orx-component of velocity remains constant.

Using third kinematics relation along the vertical direction, we get

2 2

1

( 6) 2(9.8) (3.0)

9.74 ms

y

y

u

u

The required angle of launch will be

1 1 9.74tan tan 5

y

x

u

u

Correct Option:(4)

Question8.A flat body of mass 2.00 kg is sliding on a frictionless surface, as two constant forces act on

it: 1(10N)F j and 2 ( 4N)F j . At 0t , the velocity of the body is

1

0(4.00 ms )v i . What is the

angle of the bodys velocity (relative to the positivex-direction) at 3.00st ?

(1) 1 3

tan2

(2) 1

3tan

4

(3) 1 9

tan4

(4) 1 5

tan8

Chapter:Newtons Laws of Motion


Solution

The acceleration of the body will be

1 2

2

10 4

2.00

(3.00 )ms

F Fa

m

j j

j

So, to find out the velocity at 3.00st , we can write


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0

4.00 (3.00 )(3.00)

4.00 9.00

v v a t

i j

i j

Thus the components of velocity are14.00 ms

xv and 19.00 ms .yv

If be the angle made by this velocity with the positivex-direction, then

1

tan

9.00

4.00

9tan

4

y

x

v

v

Correct Option:(3)

Question9. A body is rotating around a fixed axis with angular velocity 3 rad s1with constant angular

acceleration of 1 rad s2at some time. Find the magnitude of acceleration of a particle 5 m away from the

axis after the body has turned by 90.

(1) 2 25 1 9 ms (2) 25 9 ms

(3)25 ms (4) 25 82 ms

Chapter:Rotational KinematicsLevel of Difficulty:Moderate

Solution

Initial angular velocity 0 3 rad s1.

Angular acceleration= 1 rad s2.

If angular velocity after turning by = 90 is , then we have,

2 2

0 2

2

2

9 2 1 2

9

Acceleration will have two components, tangential and centripetal. The centripetal component of

acceleration is

2c29 5 msra

and the tangential component is


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2

t 5 1 msa r

The net acceleration will be

2 2

c t

2 2

2 2

5 9 1

5 1 9 ms

a a a

Correct Option:(1)

Question10. Two boxes, A and B, are connected to each other as shown in the figure given below.

The system is released from rest and the box B falls through a distance of1.00m . The surface of the table

is frictionless. What is the kinetic energy of box B just before it reaches the floor?

(1) 2.45J (2) 4.90J

(3) 9.80J (4) 29.4J

Chapter:Work and Energy


Solution

Let Tbe the tension in the string as shown in the figure given below.


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We can write the force equation for both the masses as:

B B

A

(1)

(2)

m g T m a

T m a

Adding Eq. (1) and Eq. (2), we get

B B A

B

A B( )

m g m a m a

ma g

m m

Substituting value of ain Eq. (2), we get

A B

A B( )

3.00 1.00

3.00 1.00

3.00

4.00

m mT g

m m

g

g

Since the system is released from rest, initial kinetic energy i 0J.K

Final kinetic energy of box B is equal to the net work done on it by the tension force and the gravitational

force.

f

B

B

2

( ) (1.00)

3.001.00 (1.00) [where 9.8ms ]

4.00

9.8

4

2.45 J

K W

Fd

m ad

m g T

g g g

Correct Option:(1)

Question11. Two uniform steel bars are freely hanging from a ceiling. The length of the second bar is

double that of the first. The ratio of elongation of first bar to that of second is(1) 1 : 4 (2) 4 : 1

(3) 2 : 1 (4) 1 : 2

Chapter:Elasticity


Solution


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Let the first bar be of massM, lengthL, cross sectionA, mass per unit length mand Youngs modulus Y.

Therefore,

M =L

Elongation of segment dx, at a distancexfrom the free bottom is given by

dxdx Tx

AY

dxmxg

AY

Mx dxg

L AY

Lx dxg

L AY

dxxg

AY

Elongation of a freely hanging body

0

0

2

0

2

2

2

L

L

L

x dx

dxxg

AY

gxdx

AY

g x

AY

gL

AY

Therefore,

2x L

Hence, if length is doubled, elongation will quadruple.

Correct Option:(1)

Question12.The height of liquid in a capillary tube depends upon

(1) the outer radius of the tube. (2) the inner radius of the tube.

(3) the temperature of the tube. (4) the length of the tube.

Chapter:Surface Tension and Viscosity


Solution

Rise of liquid in a capillary tube is given by the relation


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2 cosTh

r g

where r= inner radius of tube, T= surface tension and= density of liquid. Thus, height of liquid

depends on the inner radius of the capillary tube not the outer radius.

Correct Option:(2)

Question13.A particle executes SHM alongx-axis about the center atx=awith frequencyf. Initially

the particle is at rest at origin. Its equation of motion is

(1) (cos 2 1)x a ft (2) (1 cos 2 )x a ft

(3) 2 (cos 2 1)x a ft (4) 2 (1 cos 2 )x a ft

Chapter:Simple Harmonic Motion


Solution

Aboutx = 0, SHM equation is

sin 2 fx a t

For particle initially at rest at maximum positive displacement,2

.

sin 2 cos 22

x a t a tf f

(1)

Shifting axis of SHM tox = a, Eq. (1) would be

cos 2

cos(2 )

os2 1c

x

fx a t

tf

a a t

a

f

a

Correct Option:(1)

Question14. A spherical source of radius 1 m emits sound equally in all directions in a non-dispersive

medium. The ratio of the magnitude of the amplitudes of the waves at two points at a distance of 25 m

and 36 m from the source is

(1) 36:25 (2) 5:6

(3) 6:5 (4) 216:125

Chapter:Sound Waves



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Solution

Power transfer across a spherical surface with the same center as the source would be same across all such

surfaces by conservation of energy. Intensity at any point at a distance of rfrom the center, if power

generated by the source isP, would be

24

PI

r

Since 2I A when other independent parameters remain same, so we can write

2

2

1A

r

1A

r

Hence, ratio would be 36:25 because amplitude is inversely proportional to distance.

Correct Option:(1)

Question15.Two 3.25 Fcapacitors are connected in series across the terminals of an 50.0Hz

alternating current (AC) generator that has a peak voltage of 244 V . What is the rms current in the

circuit?

(1) 0.124A (2) 0.088A

(3) 0.308A (4) 0.248A

Chapter:Alternating Current


Solution

Since the capacitors are in series, the effective capacitance of the circuit will be

(3.25)(3.25)1.625F

(3.25 3.25)C

Hence, capacitive reactance will be

6

3

1

21

2 (50.0)(1.625 10 )

1.96 10

CX

fC

Peak current will be,


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00

3

244

1.96 10

0.124 A

C

VI

X

Therefore, the rms current will be

0rms

2

0.124

2

0.088 A

II

Correct Option:(2)

Question16.A stream of electrons is going round a circle in the counter clockwise direction as shown in

the figure below.

At the center of the circle, they produce a magnetic field that is directed

(1) into the page. (2) out of the page.

(3) to the left. (4) to the right.

Chapter:Magnetic Field due to a Current


Solution

The direction of the conventional current is opposite to the direction of the flow of the electrons. So,applying the right hand rule, we conclude that the magnetic field at the center will be directed into the

page.

Correct Option:(1)


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Question17.In Youngs double slit experiment, blue light ( 440 nm) gives a second order bright

fringe at a certain location on a flat screen. What wavelength of visible light would produce a dark fringe

at the same location? Assume that the range of visible wavelengths extends from 380 nm to 750 nm .

(1) 467 nm (2) 587 nm

(3) 671 nm (4) 736 nm

Chapter:Light Waves


Solution

Since it is a second-order bright fringe, we have

2

2 2(440) 880D D Dy

d d d

where dis separation between the slits andDis distance of screen from slits.

For getting a dark fringe at the same position,

( 1 / 2) 880

1880 (1)

2

m D D

d d

m

where 0, 1, 2,...m

For 0m , using Eq. (1) we get

10 880

2

1760 nm

For 1m , using Eq. (1) we get

11 880

2

587 nm

,

and for 2m , using Eq. (1) we get

12 880

2352 nm

So, the only value that falls within the given range of 380 nm to 750 nm is 587 nm.

Correct Option:(2)


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Question18.A diverging lens of focal length 14.0 cm is 30.0 cm to the left of a converging lens of focal

length12.0 cm . An object is placed 16.0 cm to the left of the diverging lens. How far is the final image

from the converging lens (after refraction through the converging lens)?

(1) 7.5 cm (2) 15.7 cm

(3) 17.6 cm (4) 27.4 cm

Chapter:Refraction


Solution

Applying lens formula for the diverging lens,

1 1 1

1 1 1

1 1

16.0 14.0

15

112

7.47 cm

v u f

v u f

v

This image becomes the virtual object for the second (converging) lens. Thus, the object distance for the

converging lens = 7.47 30.0 37.47 cm.

We can now apply the lens formula to the converging lens.


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1 1 1

1 1 1

1 1

37.47 12.0

37.47 12.0

25.47

17.6 cm

v u f

v u f

v

Thus, the final image will be formed 17.6 cm to the right of the converging lens.

Correct Option:(3)

Question19. If(n+ 1) divisions on the vernier scale (VSD) of a vernier calipers coincide with ndivisions

on the main scale (MSD) and each division on the main scale is of cunits, then the least count of

instrument is

(1)1

c

n (2)

2

c

n

(3)c

n (4)

1

c

n

Chapter: Experimental Skills


SolutionLeast count (LC) = 1 MSD1 VSD

In the given question

(n) MSD (n+ 1) VSD

1VSD MSD1

n

n

So, the least count will be

LC = I MSD MSD1

n

n

1MSD

1

1MSD

1

n n

n

n

Since each division on the main scale is of cunits


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LC1

c

n

Correct Option:(1)

Question20.A wire of cross-sectional radius ais bent to form a circle of radiusR(R>> a). A charge Qis

uniformly distributed on the ring. An infinite line of positive charges of linear charge density passes

through center of the ring, perpendicular to the plane of the ring. If Youngs modulus of the material of

the wire is Y, then the change in length of the wire after placing the infinite line of charge at the center of

charged ring is

(1)2 2

02

Q

a Y

(2)

2 2

0

Q

a Y

(3)2 2

0

2 Q

a Y

(4)

2 2

0

3

2

Q

a Y

Chapter:Electric Field


Solution

Consider an infinitesimal element of the circle, as shown in the figure below.

Charge on the infinitesimal element =2

Qdx

R

Electric field due to infinite line of charges at distanceRis given by

02E

R

0

2 sin2 2

QT dx

R R

0

(2 )2 2

QT dx

R R


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Since is small

02 2

dx QT dx

R R R

204

QT

R

2/

/

T aY

L L

2

2T RL

a Y

2 2

02

QL

a Y

Correct Option:(1)

Question21. Read the two given statements S1 and S2 carefully and mark the correct options

S1:The potential difference between the shells of two concentric charged spherical shells depends on

charge of inner shell.

S2:Potential due to charge of outer shell remains same at every point inside the sphere.

(1) S1 and S2 are True and S2 is the correct explanation for S1.

(2) S1 is True and S2 is True but S2 is NOT a correct explanation for S1.

(3) S1 is True and S2 is False.

(4) S1 is False and S2 is True.

Chapter:Electric Potential


Solution

Consider the two concentric shells A and B shown in the figure below.

Potential due to shell A


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1 2A

0 2

1

4

Q QV

R

Potential due to shell B

1 2B

0 1 2

1

4

Q QV R R

The potential difference will be

B A 1

0 1 2

1 1 1

4V V V Q

R R

Hencethe potential difference between the shells of two concentric charged spherical shells depends on

charge of inner shell because potential due to charge of outer shell remains the same at every point inside

the sphere.

Correct Option:(1)

Question22.A conducting wire of length 1 m, being used as a potentiometer wire, has radius linearly

changing from r to 2racross its ends. The distance of the null point from the end with radius r,if the

resistances in the gaps are in the ratio 2 : 3, is

(1) 25 cm (2) 40 cm

(3) 20 cm (4) 33.33 cm

Chapter:Electric Current in Conductors


Solution

Since the radius of the wire increases linearly with distancexfrom left end, we have the radiusRat a

distancexas

R= r+xr

l

The resistance of an element dxat a distancexis

22dx dxrR xr

rl


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If the null point occurs at a distance x from the left end, we have the ratio of the two parts (left and

right) in the ratio 2 : 3 . This implies

2

0

2

23

x

x

dx

xrr

ldx

xrr

l

32

0( )

xdx

l x

2 2( )l

x

dx

l x

3

01 1

2

x

x ll x l x

3 1 1 1 122l l x l x l

Substituting l= 100 cm and simplifying above expression, we get

1 1 1 13 2

100 100 100 200x x

x= 25 cm

Correct Option:(1)

Question23.When a shunt of resistance 2is connected in parallel to galvanometer, its range to measure

current becomes 11 times. Then the resistance required to be connected in series to convert it into

voltmeter of range 20 times is

(1) 380 (2) 280

(3) 180 (4) 480

Chapter:Electric Current in Conductors


Solution

Now the shunt Scan be expressed as

g

g

1

RS

I

I


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g2

11 1

R

g 20R

The required resistance can be calculated using

g

g

1V

R RV

20 (20 1)

380

Correct Option:(1)

Question24.The graph given below, plots photocurrent versus anode potential for a cathode with 4eVwork function. The energy of the incident photon is

(1) 6 eV (2) 4 eV

(3) 2 eV (4) 8 eV

Chapter:Photoelectric effect


Solution

Given = 4eV

From the graph, stopping potential V0= 2V. This implies that

Kmax= eV0= 2 eV

But from Einsteins photoelectric equation

h= Kmax+

h= (4+ 2)eV= 6eV

Correct Option:(1)


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Question25.An inductanceLand a resistanceRare connected in series to an ideal battery. A switch in

the circuit is closed at time 0, at which time the current is zero. The rate of increase of the energy stored in

the inductor is maximum

(1)just after the switch is closed.

(2) at time /t L R after the switch is closed.

(3) at time 2 /t L R after the switch is closed.

(4) at time ( / ) ln 2t L R after the switch is closed.

Chapter:Inductor


Solution

The transient current in anLRcircuit is given by

1 exp /

V t

I R L R

Therefore,

exp/

dI V t

dt L L R

Energy stored in the inductor is given by

21

2E LI

Rate of change of energy,

2

exp 1 exp/ /

dE di V t t L I

dt dt R L R L R

2 2

2 exp 1 2exp

/ /

d E V t t

dt L L R L R

Now, fordE

dtto be maximum,

2

2 0

d E

dt

Thus,

2

exp 1 2exp 0/ /

V t t

L L R L R

1exp

/ 2

ln 2/

( / ) ln 2

t

L R

t

L R

t L R


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Correct Option:(4)

Question26.For a logic gate, the voltage waveforms A and B are shown as input in the figure given

below and C as output. The logic circuit gate is

(1) OR gate

(2) AND gate

(3)NAND gate(4)NOR gate

Chapter:Digital Electronics


Solution

The truth table corresponding to waveforms is as shown here. It indicates that the given logic circuit gate

is AND gate.

A B C

0 0 0

1 0 0

0 1 0

1 1 1


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Hence, the correct option is (2).

Correct Option:(2)

Question27.The atomic mass of 10B is 10.811 u. The binding energy of 10B nucleus is (given that the

mass of electron is 0.0005498 u, the mass of proton is mp= 1.007276 u, and the mass of neutron is mn=

1.008665 u)

(1)681.201 MeV (2) 678.932 MeV

(3) 378.932 MeV (4) None of these

Chapter:Atomic Nucleus


Solution

Mass of 10B nucleus isM= 10.811u

Mass of proton mp= 1.007276 u

Mass of neutron mn= 1.008665u

The number of protons and neutrons in10

B nucleus is 5 each.

The mass defect will be

1 007276 5 1 008665 10 811

5 03638 5 043325 10 811

10 079705 10 811

0 731295u

5 5

5

p nm

. . .

. . .

.

m

.

m

.

M

The binding energy will be

2

931.5 MeV

0.731295 931.5 MeV

681.201MeV

E mc

m

Correct Option:(1)


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Question28.In case ofp-njunction diode at high value of reverse bias, the current rises sharply. The

value of reverse bias is known as

(1) cut-off voltage.

(2) Zener voltage.

(3) inverse voltage.

(4) critical voltage.


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Chapter:Semiconductors


Solution

In case ofp-njunction diode at high value of reverse bias, the current rises sharply. The value of reverse

bias is known as Zener voltage.

The Zener diode or breakdown diode, as it is sometimes called, is basically the same as the

standard p-n junction diode but is specially designed to have a low predetermined reverse breakdown

voltage that takes advantage of this high reverse voltage. TheZener diodeis the simplest types of voltage

regulator and the point at which a Zener diode breaks down or conducts is called the Zener voltage.

Correct Option:(2)

Question29.A T.V. transmission tower has a height of 180 m. The broadcast is available up to a distance

of (Radius of Earth 6.4 106m)

(1) 48 km (2) 24 km

(3) 34 km (4) 68 km

Chapter:Communication Systems


Solution

The range of broadcast can be calculated using relation

6

3

2

2 180 6.4 10

48 10 m

48km

d hR

Correct Option:(1)

Question30.Read the two statements S1 and S2 are given below and answer the correct option

S1: A normal human eye can clearly see all the objects just beyond a certain minimum distance.

S2: The human eye has the capacity to adjust suitably the focal length of its lens to a certain extent.

(1) S1 is True, S2 is True and S2 is a correct explanation for S1.

(2) S1 is True, S2 is True but S2 is NOT a correct explanation for S1.

(3) S1 is True but S2 is False.

(4) S1 is False but S2 is True.

Chapter:Optical Instruments


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Solution

A normal human eye can clearly see all the objects just beyond a certain minimum distance which is 25

cm for normal eye and it is known as near point. Hence statement 1 is correct.

Accommodation is the adjustment to the thickness of the lens in the eye to ensure that the image on the

retina is sharp. When thickness of the lens changes the focal length of the lens also changes. Thus, the

human eye has the capacity to adjust suitably the focal length of its lens to a certain extent. Therefore

statement 2 is also correct.

Statement 1 and statement 2 are two independent statements. Therefore, statement 2 is not a correct

explanation of statement 1.

Correct Option:(2)


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Part CChemistry

Question 1:Which of the following reactions does not contribute to air pollution in the troposphere?

(1)4 10 2 2 2C H (g) + 6O (g) CO(g) + 2CO (g) + 5H O(l)

(2)Lightning

2 2

2 2

N (g) O (g) 2NO

2NO(g) + O (g) 2NO (g)

(3)4 22HCHO CH (g) + CO (g)

(4)3 2Cl + O (g) ClO + O (g)

Chapter:Environmental Chemistry


Solution

The first reaction takes place during incomplete combustion of fossil fuels and releases oxides of carbon

into the troposphere. The second set of reactions represents formation of oxides of nitrogen in the

troposphere. The third reaction represents formation of hydrocarbons due to incomplete combustion of

fuels or anaerobic decomposition of organic matter. The fourth reaction between chlorine radical and

ozone takes place in the stratosphere and leads to depletion of ozone in the stratosphere. Chlorine free

radical is obtained from chlorofluorocarbons released into the atmosphere.

2 2 2F CCl (g) F C Cl(g) + Cl(g)h

Correct Option: (4)

Question 2:The four elements Ca, Mg, P and Cl are to be arranged in the order of their increasing

atomic sizes. Which of the following orders is correct?

(1) Cl < P < Mg < Ca (2) P < Cl < Ca < Mg

(3) Ca < Mg < Cl < P (4) Mg < Ca < P < Cl

Chapter:Classification of Elements and Periodicity of Properties


Solution


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Element Atomic

number

Number of

electrons

Cl 17 17

P 15 15

Mg 12 12

Ca 20 20

As theZeffincreases along a period, the atomic radius decreases, so the order is Mg < P < Cl. Down the

group, there is increase in the number of shells, hence the radius increases. So, the order is Mg < Ca.

Correct Option: (4)

Question 3:During the concentration of sulphide ores by froth floatation process, the separation of

sphalerite and galena is achieved by which of the following substances used as depressant?

(1) Potassium xanthate (2) Sodium cyanide

(3) Copper sulphate (4) Pine oil

Chapter:General Principles and Processes of Isolation of Elements


Solution

Collector, that is, xanthate acts only on solid particles through adsorption. The two ores show different

reactivity towards NaCN:

2 2 4(Soluble complex)

ZnS 4NaCN Na S Na Zn(CN)

PbS NaCN No reaction

Correct Option: (2)

Question 4:Which one of the following is an electron-deficient species?

(1) B2H6 (2) NH3 (3) 4AlH

(4) CH4

Chapter:p-Block Elements


Solution

After having 3 center2 electron bond also, B2H6is still electron deficient which can be evidenced by the

following spontaneous reaction:

2 6 2 6B H 2THF B H THF


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The octet/duplet is complete for the central atom in the rest of the hydrides, so they are not electron

deficient.

B2H6: NH3:

4AlH

: CH4:

Correct Option: (1)

Question 5:When spherically symmetrical field of ligands surrounds the central metal ion, then which of

the following options is correct for the change in energy order of d-orbitals of the central metal ion?

(1) t2g> eg (2) eg> t2g (3) eg= t2g (4) Cannot be predicted

Chapter:Coordination Compounds


Solution

If a symmetrical field of negative charges surrounds a metal ion, the d-orbitals remain degenerate.

Correct Option: (3)

Question 6: Which of the following pair of elements of lanthanoids do not follow the regular trend of

decrease in radius from Ce to Lu?

(1) Eu and Gd (2) Sm and Tm (3) Pr and Er (4) Eu and Yb

Chapter:dandf-Block Elements


Solution

The number of electrons participating in the metallic bonding is three for all the lanthanoid elements

except Eu (At. no. 63) and Yb (At. no. 70) as the outermost electrons are favored to participate in

bonding. In these two cases, one electron from 5d orbital is shifted to 4f orbital to get half-filled and

fully filled stable electronic configuration. Hence, these two elements use only two electrons in metallic

bonding, and atomic radii do not show the expected decrease.

Correct Option: (4)


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Question 7:Identify the correct designation for this tripeptide:

(1) Cys-Met-Glu (2) Cys-Met-Asp (3) Met-Cys-Glu (4) Asp-Cys-Met

Chapter: Biomolecules


Solution

Based on the respective structures of amino acids, the correct designation is Cys-Met-Glu.

Correct Option: (1)

Question 8:Which of the following ketones has the largest equilibrium constant for the addition of

water?

(1) (2)

(3) (4)

Chapter:Aldehydes and Ketones


Solution

The equilibrium constant for hydration depends on stability of produced hydrate and ketone. Since the

stability of ketone increases with the increasing electron density on carbonyl carbon atom, its stability is

increased by electron donating group.

In contrast, the hydrates stability is increased by electron withdrawing group, but stability of ketone is

decreased.


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placing triple bond at carbonyl carbon. In this way, we cannot place a triple bond in the compound given

in option (3). So, this ketone cannot be formed by alkynes on hydration.

Correct Option: (3)

Quick Tip:Just look for minimum two hydrogen atoms at alpha position. If they exist, then alkyne

formation is possible with the same skeleton and we can also form a ketone from it.

Question 10:Potassium phthalimide on reaction with compound (A) followed by hydrolysis forms

isopentyl amine. The compound A will be

(1) 3 3 3CH CH(CH ) CH(Br) CH (2) 3 3 2 3CH C(CH )(Br) CH CH

(3) 2 3 2 3Br CH CH(CH ) CH CH (4) 3 3 2 2CH CH(CH ) CH CH Br

Chapter:Alkyl and Aryl halides


Solution

The given reaction sequence is for Gabriel phthalimide synthesis. The synthesis is successful only if

haloalkane is primary. Secondary haloalkanes produce elimination product, alkenes.

RNH2is iso-pentyl amine,

So A will be

Correct Option: (4)

Quick Tip:TheNH2group will be attached to that carbon atom at which B atom is present.

Question 11:Which of the following molecules would you expect to be aromatic?

NK + RBr

O

O

+ H O/H2

+

KBr

C OH

C OH

O

O

+ RNH2

(A)

CH3

CH CH2

CH2

NH2

CH3

CH3

CH CH2

CH2

Br

CH3


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(1) I, II and III only (2) I, II, IV only (3) II, III, IV only (4) All are aromatic

Chapter: Hydrocarbons: Aromatic


Solution

According to Huckels rule, two conditions should satisfy for aromaticity. The planar cyclic compound

with (4n+ 2) pi electrons are aromatic compounds.

I. This carbon ring contains 6 pi electrons with all carbon atoms in sp2hybridized state. The angle strain

in this ring is not much so it is also planar and aromatic.

II. Though nitrogen atoms can avail lone pairs for aromaticity, but in this case it is not required. Ring

contains 6 pi electrons and all atoms in sp2

hybridized state. So, the compound is aromatic.

III. It contains 8 pi electrons, so it is not an aromatic compound.

IV. Ring contains 6 pi electrons with all carbon atoms sp2hybridized, so it is an aromatic compound.

Correct Option: (2)

Question 12:Which of the following is a copolymer?

(1)

(2)

(3)


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(4)

Chapter: Polymers


Solution

Homopolymers are formed by polymerization of the same type of monomers. Copolymers are formed

when a mixture of more than one monomeric species is allowed to polymerize. For example, Buna-N is a

copolymer of acrylonitrile and 1,3-butadiene. From the given options, option (2) represents a

copolymer.

Correct Option: (2)

Question 13:Which of the following carbocations will not rearrange?

(1) (2) (3) (4)

Chapter: Some Basic Principles in Organic Chemistry


Solution

The rearrangement of a carbocation is possible only if it can form more stable carbocation. Sometimes,

it can also release angle strain present in relatively smaller or bigger rings. Hence, (1), (2) and (4)

rearrange as shown below. The compound in (3) cannot rearrange because more stable carbocation is

not possible by shifting alkyl or halide group.

(1) (2)


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(4)

Correct Option: (3)

Question 14: The correct order of stability of the following free radicals is

(1) III > I > II > IV (2) I > II > IV > III (3) IV > II > I > III (4) I > III > II > IV

Chapter: Some Basic Principles in Organic Chemistry


Solution

The free radicals are electron-deficient species, whose stability is enhanced by electron donating groups.

Therefore in (III), allylic free radical is most stable since electron density can be transferred by both

inductive as well as resonance effects. As (I), (II) and (IV) are 3, 2and 1free radicals, respectively, the

correct order of stability is III > I > II > IV.

Correct Option: (1)

Question 15: Which of the following sets of quantum numbers is not representing the electrons which

are eliminated from Fe to convert it into Fe3+

?

(1) n= 4, l= 0, m= 0, s= 1/2 (2) n= 4, l= 0, m= 0, s= 1/2

(3) n= 3, l= 2, m= 0, s= 1/2 (4) n= 3, l= 1, m= 0, s= 1/2

Chapter:Atomic Structure


Solution

Ground state electronic configuration of Fe: 1s22s

22p

63s

23p

63d

64s

2

The electronic configuration of Fe3+: 1s22s

22p63s

23p63d

54s0

Hence, two electrons are removed from the fourth shell and the third shell. Two types of orbitals (dand

s) are involved. The value of lfor s-orbital is 0 and for d-orbital, l = 2.


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Fourth set of quantum numbers contains l= 1, that is,p-orbital. Thep-orbitals are not involved in

ionization.

Correct Option: (4)

Quick Tip: Just look for the electrons removed from the atom, and from which type orbitals are involved

in this process.

Question 16: For which of the following equilibrium, the value of Kpis not greater than the value of KC?

(1)2

2C(s) O (g) 2CO(g) (2)2 4 22C(s) 2H O(g) CH (g) CO (g)

(3)4 2 4 2CuSO 5H O(s) CuSO (s) 5H O(g) (4) 2 2Mg(OH) (s) MgO(s) H O(g)

Chapter:Chemical Equilibrium


Solution

( ) np C

K K RT ; where n= product side gaseous moles reactant side gaseous moles.

(1) 2 1 1 ( )p Cn K K RT (2)02 2 0 ( )

p C p Cn K K RT K K

(3) 55 0 5 ( )p Cn K K RT (4) 1 0 1 ( )p Cn K K RT

Correct Option: (2)

Question 17: The correct order of increasing covalent character of the following is

(1) KCl < CaCl2< AlCl3< SiCl4 (2) SiCl4< AlCl3< CaCl2< KCl

(3) AlCl3< CaCl2< KCl < SiCl4 (4) CaCl2< SiCl4< KCl < AlCl3

Chapter:Chemical Bonding and Molecular Structure


Solution

According to Fajans rule, as the charge on the cation increases, its effective nuclear charge as well as

polarizing power increases. Hence, covalent character will also increase. The increasing order should be

KCl < CaCl2< AlCl3< SiCl4.

Correct Option: (1)

Question 18: The correct order of increasing CO bond length of2

3 2CO, CO , CO

is

(1)2

2 3CO CO CO (2) 23 2CO CO CO


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(3)2

2 3CO CO CO (4) 23 2CO CO CO

Chapter:Chemical Bonding and Molecular Structure


Solution

All bonds in 23CO are equivalent because it is resonance hybrid.

So, the bond order of 23CO is 4/3.

Bond order in CO is 3 and of CO2it is 2.

Since bond length is inversely proportional to the bond order, bond length order is2

2 3CO CO CO

Correct Option:(1)

Question 19: The reaction of tin metal with acid can be written as

2+

2 o

2 Sn /SnSn(s) 2H (aq) Sn (aq) H (g); 0.14 VE

Assume that2+[Sn ] 1M and the partial pressure of hydrogen gas is 1 atm, then the

(1) cell reaction is spontaneous at pH = 5.

(2) cell reaction is non-spontaneous at standard conditions.

(3) cell reaction is spontaneous at pH = 2.

(4) cell reaction is spontaneous for all pH values.

Chapter:Electrochemistry


Solution

2

2Sn(s) 2H (aq) Sn (aq) H (g)

Nernst equation for this cell reaction: 22

Ho

cell cell 2

[Sn ]ln

[H ]

pRTE E

nF

At standard temperature; 22

Ho

cell cell 2

[Sn ]0.0591ln

[H ]

pE E

n

+ 2+2

o o o

cell H /H Sn /Sn 0 ( 0.14 V) 0.14 VE E E

C

O O

O

C

O O

O

C

O O

O


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For pH = 5,cell 5 2

0.059 10.14 log 0.16 V

2 (10 )E

For standard state;o

cell cell 0.14 VE E

For pH = 2, cell 2 20.059 1

0.14 log 0.14 0.12 0.02 V2 (10 )E

Correct Option: (3)

Question 20: Select the correct statement about compressibility factor of one mole gas which is kept

inside a 22.4 L vessel at 273 K.

(1) ForZ> 1 the pressure of gas will be less than 1 atm.

(2) ForZ> 1 the pressure of gas will be more than 1 atm.

(3) If b dominates, pressure will be less than 1 atm.(4) If a dominates, pressure will be greater than 1 atm.

Chapter:Gaseous State


Solution

Using ideal gas law:1 0.082 273

1atm22.4

nRTp

V

So if the gas is ideal, then its pressure is 1 atm. However, it is a real gas, so at different conditions its

pressure may vary from the ideal pressure.

If its compressibility factorZ> 1, this indicates that repulsive forces dominate inside the gas. These

repulsive forces imply that the pressure should be more than ideal pressure, that is, more than 1 atm

pressure.

If b dominates, then mm( ) 1pV pb

p V b RT ZRT RT

This equation implies thatZwill be more than one always, and pressure should be more than 1 atm.

If a dominates, then

mm2

m m

1pVa a

p V RT ZV RT RTV

This equation implies thatZ< 1, so the pressure should be less than 1 atm as the attraction forces will

dominate here.

Correct Option: (2)


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Question 21: The precipitate of CaF2(Ksp= 1.7 1010

) is observed when equal volumes of the following

are mixed.

(1)4 2+ 4

10 M Ca + 10 M F (2)2 2+ 3

10 M Ca + 10 M F

(3)5 2+ 3

10 M Ca + 10 M F (4)3 2+ 5

10 M Ca + 10 M F

Chapter:Ionic Equilibrium


Solution

The reaction is

2

2CaF Ca 2F

The solubility product is 2 2 10sp [Ca ][F ] 1.7 10K

The ionic product is2 2

I [Ca ][F ]K

(1)2 2 4 4 2 13

I [Ca ][F ] 0.5 10 (0.5 10 ) 1.25 10K

(2)2 2 2 3 2 9

I [Ca ][F ] 0.5 10 (0.5 10 ) 1.25 10K

(3)2 2 5 3 2 12

I [Ca ][F ] 0.5 10 (0.5 10 ) 1.25 10K

(4)2 2 3 5 2 14

I [Ca ][F ] 0.5 10 (0.5 10 ) 1.25 10K

In the case of (2), ionic product is greater than solubility product, so precipitation will be observed in this

case.

Correct Option: (2)

Question 22: Consider the decomposition of gaseous N2O5; 2 5 2 22N O 4NO (g) O (g) . At moderate

gas pressures, it follows the mechanism given below:

1

1

2 5 2 3N O NO +NOk

k

2

2 3 2 2

NO NO NO NO Ok

3

3 2NO NO 2NOk

Which of the following statements is correct?

(1) Overall order of the reaction is 2.0 (2) Order w.r.t. N2O5is 1.0

(3) Order w.r.t. NO3is 1.0 (4) Order cannot be predicted

Chapter: Chemical Kinetics


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Solution

The molecule NO3is an intermediate whose concentration is small and can be calculated by the steady

state approximation. Same can be said for NO.

Rate of production of NO = Rate of destruction of NO

22 2 3 3 3 2

3

[NO ][NO ] [NO][NO ] [NO] [NO ]k

k kk

For NO3, rate of production of NO3= rate of destruction of NO3

1 2 5 1 2 2 2 3 3[N O ] ( [NO ] [NO ] [NO])[NO ]k k k k 1 2 5

3

1 2 2 2

[N O ][NO ]

[NO ] 2 [NO ]

k

k k

for the steady state condition of NO3.The rate of production of oxygen, which is also the rate of reaction,

is

22 2 3

d[O ][NO ][NO ]

dk

t

Substituting the value of [NO3], we get

1 2 2 522 5

1 2

[N O ]d[O ][N O ]

d 2

k kk

t k k

where 1 2

1 22

k kk

k k

.

Correct Option: (2)

Question 23: Which of the following statements is correct about the defects in the crystals?

(1) Schottky defects are observed in the compounds in which there is a large difference in the size of

cation and anion, whereas Frenkel is observed when cation and anion are approximately of the same

size.

(2) Schottky defects disturb the ratio of cations and anions, whereas Frenkel defects maintain the ratio.

(3) Schottky defects lead to lowering in density, whereas Frenkel defects do not affect the density.

(4) Both interstitial defects and Frenkel defects cause increase in the density of solid.

Chapter: Solid State


Solution


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Schottky defect is observed in ionic solids in which cationic and anionic sizes are approximately the

same. It arises when equal number of cations and anions are missing from their position. It causes

decrease in the density of solid.

Frenkel defect is observed in ionic solids in which cationic and anionic size difference is large. It arises

when the smaller ion leaves its original place and moves into interstitial spaces. This causes no change in

cationanion ratio, and hence no change in the density.

Interstitial defect arises when a constituting particle is also placed into interstitial space. This causes an

increase in density.

Correct Option: (3)

Question 24: The solubility of pure oxygen in water at 20C and 1 atm pressure is 43 mg O2per liter of

water. When air is in contact with water and the air pressure is 585 torr at 20C, how much amount of

oxygen is dissolved in 500 ml water? The average concentration of oxygen in air is 21.1%(V/V).

(1) 7 mg (2) 3.5 mg (3) 16.5 mg (4) 33 mg

Chapter:Solutions


Solution

From Henrys law: 1 1gas H gas2 2

C pC K p

C p

Given that C1= 43 mg/l;p1= 1 atm = 760 torr; C2= ?;p2=21.1

585 123.435 torr 100

Substituting in the above expression, we get

2

123.44 43mg/L 7 mg/L

760C

Therefore,2

7Mass of O in 500 ml water 500 3.5 mg

1000

Correct Option: (2)

Question 25: A saturated solution is prepared by dissolving 0.2 g of polypeptide in water to give 500 mL

of solution. The solution has an osmotic pressure of 3.74 torr at 27C. What is the approximate

molecular mass of the polypeptide?

(1) 8559 (2) 2000 (3) 180 (4) 203

Chapter:Solutions


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We have o o og g

H U n RT U H n RT

Moles of N2O taken =176

4 mol44

Therefore,oMoles Change for one mole 4 ( )gU H n RT

or8.314 300

4 163 1 kJ 662 kJ1000

U

Correct Option: (1)

Question 28: Which of the following statements is not correct?

(1) Final temperature in reversible adiabatic expansion is greater than that in irreversible adiabatic

expansion.

(2) When heat is supplied to an ideal gas in isothermal process, kinetic energy of gas remains constant.

(3) When an ideal gas is subjected to adiabatic expansion, it gets cooled.

(4) Entropy increases when an ideal gas expands isothermally.

Chapter:Chemical Thermodynamics


Solution

For adiabatic process, VU w w C T V

wT

C

Since work done in reversible process is more than that in irreversible process, the temperature change

is more in reversible process, that is, the temperature decrease will be more.

Since kinetic energy of ideal gases is proportional to root of temperature, for isothermal it will not

change.

From the above relations, it is clear that when an ideal gas is subjected to adiabatic expansion, its

temperature decreases.

Entropy increases when an ideal gas expands isothermally because increasing the volume causes

increase in randomness.Correct Option: (1)

Question 29: Ammonium nitrate detonates above 300C according to the chemical equation given

below:

4 3 2 2 2NH NO N O H O


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Select the correct statement:

(1) The volume change is approximately 3.5 times the initial volume at STP.

(2) The volume change is approximately 1.5 times the initial volume taken at 323C.

(3) On explosion of 80 g of ammonium nitrate, 78.4 L of total volume is obtained at STP.

(4) 100 g of NH4NO3produces volume of 215.3 L of the total gases at 323C and 1 atm pressure.

Chapter: Some Basic Concepts of Chemistry


Solution

The balanced form of equation is:4 3 2 2 22NH NO (s) 2N (g) O (g) 4H O(g)

Since the reactant is in solid state, so the volume occupied by it is negligible.

The moles are approximately 3.5 times, but the volume is not 3.5 times. At STP, water is l iquefied so

volume is reduced and corresponds to the gaseous moles left.

Taking 80 g of NH4NO3(one mole) will produces 3.5 mol

But at STP, gaseous moles are 1.5 mol, so volume produced is approximately 1.5 22.4 L 33.6 L

100 g of NH4NO3contains100

1.25 mol80

Therefore, gaseous moles appeared at given condition 3.5 1.25 4.375 mol

The volume of gases,3.5 1.25 0.082 600

215.3 L1

nRTV

p

Correct Option: (4)

Question 30:The presence of halogen, in an organic compound, is detected by

(1) iodoform test. (2) Tollens reagent. (3) Beilsteins test. (4) Millons test.

Chapter:Purification and Characterization of Organic Compounds


Solution

In Beilstein test, a (clean) copper wire is heated in the non-luminous flame of the Bunsen burner until it

ceases to impart any green or bluish green color to the flame. The heated end of the wire is dipped into

the organic compound and is again introduced into the Bunsen flame. The appearance of a bluish green

or green flame due to the formation of volatile cupric halides shows the presence of halogens in the

given organic compound.

Correct Option:(3).


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Quick Tip:Compounds like urea, thiourea that do not contain halogen give positive Beilsteins test.

Fluoride ion is not detected by this test as copper fluoride is not volatile. Hence, it is not a very

dependable test, and thus, negative Beilstein test is more useful than a positive one.

JEE Knocout Test QP_Solution_Dec22 (1)

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Transcript of JEE Knocout Test QP_Solution_Dec22 (1)