JEE Knocout Test Solution Doc 30th March

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JEE Knockout Test Solution Exam Date : 30/03/2014

1. Time duration for this test is 180 minutes. This test consists of 90 questions. The

maximum 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 correct

response.

3. One-fourth (1/4) marks will be deducted for indicating incorrect response of each

question. No deduction from the total score will be made if no response is indicated for

an item in the answer sheet.

4. This test contains 90 Multiple Choice Questions with single correct answer. Each

Question 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 and

marks for wrong response will be deducted accordingly as per the instruction 3 above.

Part A – Mathematics

Question 1: Find the value of 15 15 15 15

2 3 4 152 3 14 .C C C C

(1) 1413(2 ) 1 (2) 13 × 2

15

(3) 13 × 214

(4) 13 × 214

– 1

Solution

15 15 15 152 3 4 151 2 3 ... 14C C C C

1515

1

( 1) r

r

r C

15 1515 15

1 1

r r

r r

r C C

1514 15

1

1

15(2 1)r

r

r Cr

= 15 × 214

– 215

+ 1

= 13 × 214

+ 1


Correct Option: (1)

Question 2: Evaluate: 3 2

2 2

3 9.

( 1)( 3)

x x xdx

x x

(1) 2 1ln 3 3tanx x c (2) 2 11

ln 3 tan2

x x c

(3) 2 11ln 3 3tan

2x x c (4)

2 1ln 3 tanx x c

Solution 3 2

2 2

3 2

2 2 2 2

2

2 2 2

2 2

3 9

( 1)( 3)

3( 3)

( 1)( 3) ( 1)( 3)

( 1) 3

( 1)( 3) 1

3

3 1

x x xI

x x

x x x

x x x x

x x

x x x

x

x x

and

1 2 3

xI dx

x

Substituting x2 + 3 = t, we get

2x dx = dt Therefore,

1

2

1

2

1ln

2

1ln 3

2

dtI

t

t c

x c

and

2 2

3

1I dx

x

= 3 tan–1

x + c Therefore,

1 2

2 11ln 3 3tan

2

I I I

x x c

Correct Option: (3)

Question 3: For any ordered pairs (a, b) and (c, d) of real numbers, define a relation, denoted by ‘R’, as

follows: (a, b) R (c, d) if a < c or b > d. Then, R is


(1) reflexive. (2) symmetric.

(3) transitive. (4) neither reflexive nor symmetric nor transitive.

Solution

1. (3, 5) R (3, 5) is not related; therefore, R is not reflexive because the expressions 3 < 3 and 5 > 5

are false.

2. (3, 5) R (15, 1) is true; however, (15, 1) R (3, 5) is not true. Therefore, R is not symmetric.

3. (3, 2) R (5, 10), where 3 < 5; and (5, 10) R (2, 9), where 10 > 9.

This implies that (3, 2) R (2, 9) is false because the expressions 3 < 2 and 2 > 9 are false.

Therefore, R is not transitive.

Correct Option: (4).

Question 4:

Statement 1: Let p and q be two logical statements, then compound statement ( p q) (p q) is

tautology.

Statement 2: ( p q) (p q) True ∀ 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.

Solution

p q p q p q q p ( ) ( )p q q p

T T F F T T T

T F F T F T T

F T T F T F T

F F T T T T T

( ) ( )p q q p = T for all possible inputs. Hence, it is tautology.

Correct Option: (1)

Question 5: Let

1sin , 0

( )

0 , 0

nx xf x x

x

such that f(x) is continuous at x = 0; f ' (0) is real and finite; and 0

lim ( )x

f x

does not exist. This holds true

for which of the following values of n?


(1) 0 (2) 1

(3) 2 (4) 3

Solution

Given f (x) is continuous at x = 0. This implies that

(0 ) (0 ) (0) 0f f f

0 0

1 1lim(0 ) sin lim(0 ) sin 0

0 0

n n

h hh h

h h

This holds true for 1or 2 or 3n (1)

Given f ' (0) is real and finite, that is,

0

0

(0 ) (0)(0 ) lim

sin(1 / ) 0lim

h

n

h

f h ff

h

h h

h

1

0

1lim sinn

hh

h

which is real and finite. This implies that

1 1n or 2

This holds true for 2n or 3 (2)

Let us consider

1( ) sinnf x x

x

Therefore,

1 21 1( ) sin ( ) cosn nf x nx x

x x

0 0lim ( ) lim ( )

x xf x f x

if n – 2 = 0. That is, if n = 2 0

1lim cos

x x

does not exist.

Note: For n = 3, 0 0

lim ( ) lim ( ) 0x x

f x f x

.

Thus, this condition holds true for n = 2 (3)

Using the intersection of the values of n in Equations (1), (2) and (3), we get the value of n = 2.

Correct Option: (3)


Question 6: The shortest distance between lines ; 0y x z and ; 1y x z is

(1) 0 (2) 1

(3) 2 (4) 1

2

Solution

If we draw y x and y x in xy-plane and if we raise the line y x by one unit in the positive z-

direction, we get ; 1.y x z

Therefore, the shortest distance between the two lines is unity.

Correct Option: (2)

Question 7: If 2(sin cos ) (sin 4 cos4 )y x x x x , then

(1) 0y x R (2) 0y x R

(3) 2 2y x R (4) 2 2y for some x R

Solution

We know that

2 sin cos 2x x

When ,4

x

we have

sin cos 2x x

When ,4

x

we have

2 1 0y

which implies that options (1) and (2) are incorrect.

Now, at ,4

x

we have

sin cos 2x x

That is, 2(sin 4 cos4 ) 2x x . Therefore, 2 2y for any x R .

which implies that option (4) is incorrect.


Note: The maximum value of sin cos x x is 2 , for x = / 4 and the maximum value of 2(sin 4 cos 4 ) x x is 2 , for x = / 16.

Correct Option: (3)

Question 8: Find the radius of the circle escribed to the triangle ABC (shown in the figure below) on the

side BC if NAB 30 ; BAC 30 ; AB AC 5.

(1) (10 2 5 3 5)(2 3)

2 2

(2)

(10 2 5 3 5)(2 3)

2 2

(3) (10 2 5 3 5)

(2 3)2 2

(4)

(10 2 5 2 1)( 3 1)

2 3

Solution

Since we need to compute the radius of an escribed circle, we would be needing the lengths of all the

sides of the given triangle ABC.

From the question, we already know AB = AC = 5.

For finding the length of side BC, let us draw a line AD which is the bisector of angle BAC, as shown in the figure below.


BAD DAC 15

Therefore,

BD BD 3 1sin15 and sin15

AB 5 2 2

Therefore,

5( 3 1)BD 5sin15

2 2

We also know that BC = 2BD

Therefore,

5( 3 1)BC

2

Now, we know that the required radius

r1 = tan2

As

tan2 2

AB BC CA A

5( 3 1)5 5

2(tan15 )

2


10 2 5 3 5

(2 3)2 2

Correct Option: (1)

Question 9: For a function y = f (x), we have Df = and

2 2( 1)( 1)dy

x x x x xdx

Then, y is strictly increasing for

(1) x (2) x ≤ 0

(3) x ≥ 0 (4) ( 1, 0) (1, )x

Solution

For strictly increasing function, we have

0 0 if 0 at some discrete pointsdy dy dy

dx dx dx

Let us consider that

g(x) = x2 + x + 1

The coefficient of x2 is 1, which is positive.

D = b

2 – 4ac = 1− 4 = −3 = Negative

Therefore,

( ) 0g x x

Let us also consider that

f(x) = x2 – x + 1

The coefficient of x

2 is 1, which is positive.

D = b2 – 4ac = 1− 4 = −3 = Negative

Therefore, 2 1 0x x x

If 0dy

dx , we have

2 2( 1)( 1) 0x x x x x

or x > 0 or x ≥ 0 .

Correct Option: (3)


Question 10: For an ellipse, 2 2( 1) ( 3)

19 4

x y ,the locus of midpoint of the chords which pass

through the point (1, 3) is a (an)

(1) parabola. (2) ellipse.

(3) hyperbola. (4) point.

Solution

Concept: If a chord passes through centre of the ellipse, then the mid-point is the centre itself.

The centre of the ellipse 2 2

2 21

x y

a b is (0, 0), where ,a b .

The centre of ellipse 2 2( 1) ( 3)

19 4

x y is (1, 3).

Correct Option: (4)

Question 11: In a binomial distribution with 10 trials and probability of success 1

3, what is the standard

deviation of the distribution in each trial?

(1) 10

3 (2)

20

9

(3) 20

3 (4)

20

3

Solution

For the binomial distribution with n = 10 and 1

3p , we have the variance as follows:

npq = 1 2 20

103 3 9

Standard deviation is

20

Variance3

Correct Option: (4)

Question 12: Consider a parabola 2 16 .y x P is a point on the parabola, M is the foot of the perpendicular

drawn from P onto the directrix and S is the focus. If SPM is equilateral, then the focal radius SP is equal to

(1) 16 (2) 8

(3) 8 3 (4) 4


Solution

We have 2 4(4) ; 4y x a

and 2(4 ,8 )P t t is any point on parabola. The equation of directrix is x = –4 and 8my t . It is given that

SP PM MS 2 2 2

2 2 2 2 2 2 2

( ) ( ) ( )

(4 4) (8 ) (4 4) ( 4 4) ( 0)m

SP PM MS

t t t y

2 2 2 2 2 2 2(4 4) (8 ) 8 8 (8 )mt t y t

Now, 2 2 2(4 4) 8t

2 24 4 8 or 4 4 8t t

2 23 or 1 rejected ( Negative)t t

Therefore, 2 2 2 2 2 28 (8 ) 8 (1 ) 8 (4)SP t t

16SP

Correct Option: (1)


Question 13: Evaluate 3

4 21

.3

dx

x x

(1) 18 10 3

243

(2)

18 10 3

243

(3) 18 10 3

243

(4) 10 18 3

243

Solution

We have

4 2 3

dxJ

x x

Substituting 1

xt

, x > 0, t > 0, we get

2

2 32

4 2 2

/(since , 0)

(1 / )( 1 3 / ) 1 3

dtdx

t

dt t t dtt t t

t t t t

Substituting 1 + 3t

2 = u, where u > 1, we get

6tdt du

Therefore,

3/2

1 [( 1) / 3]

6

1 1

18

1

18 (3 / 2) (1 / 2)

uJ du

u

u duu

u u

Now, let us find the value of the given integral as follows:


3

4 21

1/3 3

21

4/3

4

4/3 4/34 4

3

1 3

1 1

18

1 1[ ] [ ]

27 9

1 4 2 1 2. 8 2

27 3 93 3

18 10 3

243

dxI

x x

t dt

t

u duu

u u u

Note: 1

xt

. Therefore,

3 1/3

1 1

( ) ( )f x dx g t dt

Using 21 3u t , we have 4/3

4

( )h u du

Correct Option: (1)

Question 14: The orthocentre of the triangle with vertices (5, 0), (0, 0), 5 5 3

,2 2

is

(1) (2, 3) (2) 5 5

,2 2 3

(3)

5 5,

6 2 3

(4)

5 5,

2 3

Solution

The given triangle is equilateral. Therefore, the orthocentre of the triangle is same as centroid of the

triangle. Thus, the orthocenter, that is, the centroid is given by

5 0 (5 / 2) 0 0 (5 3 / 2) 5 5, ,

3 3 2 2 3

Correct Option: (2)


Question 15: A circle touches the line y x at point (4, 4) on it. The length of the chord on the line

0x y is 6 2 . Then one of the possible equations of the circle is

(1) 2 2 30 0x y x y (2) 2 2 2 18 32 0x y x y

(3) 2 2 2 18 32 0x y x y (4) 2 2 2 22 32 0x y x y

Solution

Family of circles touching the line y x at the point (4, 4) is

2 2( 4) ( 4) ( ) 0x y y x

We need to find the member of this family which has length of chord 6 2 on 0x y .

For different 's , we get different circles:

2 2 8 8 32 0x y x y y x

2 2 ( 8 ) ( 8 ) 32 0 (1)x y x y

Now,

OA DC 4 2

6 2DE 3 2 (given)

2

Therefore,

2 2 2(3 2) (4 2)R

2250 100 10

2

Substituting 10 in Eq. (1), we get

2 2 2 18 32 0x y x y


[Substituting 10 ; in Eq. (1); we get 2 2 18 2 32 0x y x y , which does not exist in the given

options.]

Note: From Eq. (1), we get 2 2 2 2

2 2 2

(Radius)

( 8) ( 8)32

4 4 2

R g f c

Correct Option: (2)

Question 16: If the letters of the word ASSASSINATION are arranged at random, the probability that no

two ‘S’ occur together is

(1) 42

143 (2)

9!4!

13!

(3) 21

143 (4)

10! 4!1

13!

Solution

Total ways to arrange the word ASSASSINATION is

13!

4! 3! 2! 2!

For no two ‘S’ occur together, let us first arrange AAAIINNTO that can be done in 9!

3! 2! 2! ways which

would create 10 gaps:

_ _ _ _ _ _ _ _ _ _A A A I I N N T O

On these 10 gaps, four S can be put in 10

4C ways. We need to select four places out of ten to put S. Since

in each gap we put only one S; so these remain separate even if other gaps remain empty. Hence,

Favourable cases9! 10!

3! 2! 2! 4! 6!

Probability[9! / 3! 2! 2!] [10! / 4! 6!] 9! 10! 9 8 7 42

[13! / 4! 3! 2! 2!] 6! 13! 13 12 11 143

Correct Option: (1)


Question 17: Let , anda b c be such that 1 and 2.a b c The projection of b along a is equal to

that of c along a . Also, andb c are perpendicular to each other. Find .a b c

(1) 1 (2) 3

(3) 9 (4) 5

Solution

We have 1 and 2a b c and

ˆ ˆ and 0 ( (1)cos ,etc.)b a c a b c b a b

Then,

22 22 22( ) ( )a b c a b c a b a c b c

= 1 + 4 + 4 + 2(0) = 9

3a b c

Correct Option: (2)

Question 18: A potato is put in a 200°C oven and heats up at a rate proportional to the temperature

difference between the object and its surroundings. The potato is at 20°C when it is put in the oven. After 30 min, the temperature of the potato is 120°C. Then, the constant of proportionality K is (time is in min)

(1) ln(3 / 2)

15 (2)

ln (9 / 4)

60

(3) 3

4ln2

(4) 10log (3 / 2)

15

Solution

The potato heats up according to the differential equation

( 200), 0dT

K T Kdt

where t is time in minutes, T is the present temperature and dT

dt is the rate of increase in temperature with

time. Therefore, solving by variable separable method, we get


120 30

20 0

120 30020

2

200

200

ln 200 [ ]

ln80 ln180 (30)

ln180 ln80 ln (9 / 4) ln (3 / 2) 2ln (3 / 2)

30 30 30 30

dTK dt

T

dTK dt

T

T K t

K

K

Therefore, 3

ln 152

K

, where time is in min.

Correct Option: (1)

Question 19: If

cos 2 sin tan

( ) 1 1

1 2 1

x x x x x x

f x x

x

, find 20

( )lim .x

f x

x

(1) 0 (2) 1

(3) –1 (4) Does not exist

Solution

cos 2 sin tan

( ) 1 1

1 2 1

x x x x x x

f x x

x

From first row, taking x common, we have

cos 2sin tan

1 1

1 2 1

x x x

x x

x

Now, R3 → R3 – R2 is

cos 2sin tan

1 1

0 0

x x x

x x

x

Evaluating using third row, we have

[ (cos tan )]x x x x

2 (cos tan )x x x


2 (tan cos )x x x

Hence,

2

( )tan cos

f xx x

x

and thus

20

( )lim 1x

f x

x

Correct Option: (3)

Question 20: The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.8. Find the median of the

data.

(1) 5 (2) 6

(3) 7 (4)8

Solution

8 5 106

5

a b

7a b

Therefore,

2 2 64 25 10036 6.8

5

a b

2 2 25a b

3, 4 or 4, 3a b a b

Therefore, the median is 5.

Correct Option: (1)

Question 21: The system of homogeneous equations ( 1) ( 1) 0x y z ,

( 1) ( 2) 0x y z , ( 1) ( 2) 0x y z has non-trivial solutions for

(1) exactly three real values of λ (2) exactly two real values of λ

(3) exactly one real value of λ (4) infinitely many real values of λ


Solution

For non-trivial solution, or 0A D , That is,

1 1

1 2 0

1 2

Now, R2 → R2 – R1; R3 → R3 – R1 gives

1 1

1 1 3 0

1 1 1

Also R3 → R3 + R2 gives

1 1

1 1 3 0

0 0 4

Evaluation using third row, we get

4 – – –1 0

1

2

which is exactly the real value of λ.

Correct Option: (3)

Question 22: If and are the complex cube roots of unity and z be such that |z| = 4 and |z + 1| is

maximum, then the area of triangle formed by G, and (where G is the centroid of triangle formed by

z, and ) is

(1) 3

4 (2)

3 3

2

(3) 3 3

4 (4)

3

2

Solution

We know that | z| = 4 represents a point z on a circle with radius 4 and centre at the origin (0, 0). The distance of any point z from (–1, 0) on argand diagram is

| z + 1| = | z – (–1)|


| z + 1| = distance AP. Clearly, | z + 1| is maximum when z ia at point B, that is,

z = 4 + 0i = 4

Let us consider the triangle formed by ω, and z = 4 + 0i.

G = Centroid of a triangle formed by ω, and (4, 0), which implies that G ≡ (1, 0).


Area of the triangle shown is

1 1 3 3 1(Base)(Height) 1

2 2 2 2 2

3 3

4

Correct Option: (3)

Question 23: The minimum value of the expression 2 1 52 2 ,

2

x x

xx is

(1) 7 (2) 1/7(7.2)

(3) 8 (4) 1/3(3.10)

Solution

2 1 52 2

2

x x

x

2 2 1 1 1 1 12 2 2

2 2 2 2 2

x x x

x x x x x

1/82 12 2

5

2 2 (5 / 2 ) 12 (2 ) 1

8 (2 )

x x xx x

x

2 1 52 2 8

2

x x

x

Correct Option: (3)


Question 24: Which of the following statements is always correct?

(1) 2 4

0 0

1 1lim lim

x yx y

(2) 0

lim timesxnn

x x x n

(3) 2 3

0

lim 0n

xnn

x x x x

(4) 22

| | 0,7

where [·] is greatest integer function and | · | represents modulus function.

Solution

For option (1): If 1/4( )y x , then 2

0 0

1 1lim lim

x x xx is wrong. Therefore, the answer would

definitely depend on how x and y are related.

Hence, option (1) is incorrect.

For option (2): If 1

xn

, then

00 0

1lim times = lim lim 1

x xxx xn

n

x x x n n x nn

Therefore, the answer would definitely depend on how x and n are related. Hence, option (2) is incorrect.

For option (3): We have

2 3

00limit lim

1

00 (as )

1 0

xx

xx x x

x

n

Hence, option (3) is correct.

For option (4): 22

7 , which is wrong because

22 22and

7 7 .

Therefore, 22

7 is a very small positive fraction.

That is, 22

(0, 1).7


Thus, 22

07

.

Hence, option (4) is incorrect.

Correct Option: (3)

Question 25: The number of solutions of the equation x1 + x2 + x3 + x4 + x5 = 101, where xi’s are odd

natural numbers, is

(1) 1054C (2) 52

5C

(3) 52

4C (4) 50

4C

Solution

1 2 3 4 5 101x x x x x

xi are odd natural numbers. Let 2 1, 1,2,3,4,5i ix y i

1 2 3 4 52 1 2 1 2 1 2 1 2 1 101y y y y y

1 2 3 4 5 48y y y y y

yi ≥ 0. Therefore, the number of solutions is 48 5 1 52

5 1 4C C .

Note: The formula is 11

n rrC . It is the same as number of ways to divide 48 identical objects into five

places.

Correct Option: (3)

Question 26: The locus of 2 2( )( )( ) 0xy x y x y may consist of

(1) four planes. (2) three planes and one line.

(3) two planes and two lines. (4) one plane and three lines.

Solution

We have 2 2( )( )( ) 0xy x y x y

This implies that 0x or 0y

or 0x y

or 2 2 0x y

.

Let us consider x = 0, which implies that this is yz-plane.

Let us consider 0y , which implies that this is xz-plane.


Let us consider y x , that is, z can take any value. This implies that this is a plane perpendicular to xy-plane

and it contains the line y x in the xy-plane

Let us consider 2 2 0x y , which implies that 0x and 0y . This represents z-axis.

Correct Option: (2)

Question 27: Suppose f and g are two differentiable functions with the values as listed in the following

table:

x f (x) g (x) f ' (x) g ' (x)

2 3 4 5 –2

and ( ) 1( ) ln ( ) sin (sin ( ))f xh x e g x . Then find the value of h'(2).

(1) 5π – 26 (2) 5π – 14

(3) –26 (4) –14

Solution

We have

( )ln ( ) ( ) : ( ) .f xe f x x f x

and

1 1

1

sin [sin ( )] sin [sin(4)] [ (2) 4]

sin [sin( 4)]

4

g x g

Note

sin 4 sin[ (4 )]

sin(4 )

sin( 4)

and 4 ,2 2

1sin [sin ( )] ( );g x g x in the neighbourhood of x = 2. Therefore,


( ) ( ) [ ( )]

( ) ( )[ ( )] [ ( ) ( )]

(2) 5( 4)

–

3( )

14

2

5

h x f x g x

h x f x g x f x g x

h

Correct Option: (2)

Question 28: Find the Cartesian equation of the straight line passing through 2 3 i j k and 4 3 .i j k

(1) 3 3

12 2

x zy

(2)

2 31

3 2

x zy

(3) 2 2

13 3

x zy

(4)

3 31

3 3

x zy

Solution

The vector equation of the straight line passing through 2 3a i j k and 4 3b i j k is

(1 ) , p r a rb r

That is

(1 )(2 3 ) ( 4 3 3 )p r i j k r i j k

(2 2 4 ) (1 3 ) (3 3 )p r r i r r j r r k

(2 6 ) (1 2 ) (3 4 ) , p r i r j r k r

It implies

2 6 ; 1 2 ; 3 4

2 1 3

6 2 4

2 1 3

3 1 2

x r y r z r

x y zr

x y zr

Correct Option: (2)


Question 29: If a ≠ 0 and the equation ax2 + bx + c = 0 has two roots α and β such that α < –3 and β > 2,

which of the following is always true?

(1) ( ) 0a a b c (2) ( ) 0a a b c

(3) 9a – 3b + c > 0 (4) (9 3 )(4 2 ) 0a b c a b c

Solution

We have the equation ax2 + bx + c = 0 has two roots α and β such that α < –3 and β > 2.

If a > 0, then we have the following graphical representation:

Then, for all [ 3,2], ( ) 0x f x , we have the following graphical representation:

This implies that

( 1) 0 and (1) 0f f


0 and 0a b c a b c

( ) 0a a b c

If a < 0, then for all [ 3,2], ( ) 0x f x . This imply that

( 1) 0 and (1) 0f f

0and 0a b c a b c

( ) 0a a b c

Correct Option: (2)

Question 30: If 2 2( ) min , sin , ( 2 ) ,2

xf x x x

the area bounded by the curve

( ), -axis, 0 and 2y f x x x x is given by

(Note: x1 is the point of intersection of the curves x2 and sin

2

x; x2 is the point of intersection of the curves

sin2

x and 2( 2 )x )

(1) 1 2

1 2

22 2

0

sin ( 2 ) sin2 2

x x

x x

x xdx x dx x dx dx

(2) 1 2

1 2

22 2

0

sin ( 2 )2

x x

x x

xx dx dx x dx

, where 1 0,3

x

and 2

5,2

3x

(3) 1 2

1 2

22 2

0

sin ( 2 )2

x x

x x

xx dx dx x dx

, where 1 ,3 2

x

and 2

3,2

2x

(4) 1 2

1 2

22 2

0

sin ( 2 )2

x x

x x

xx dx dx x dx

, where 1

2,

2 3x

and 2 ( ,2 )x


Solution

At 21, 1x y x , we have

1sin sin 1

2 2

xy

Thus, 1 1x .

At

22 3.14

, 13 3

x y x

, we have

1sin 1

6 2y

Thus, 13

x

. Therefore, 1 0,3

x

because at (0, 2π), the graph intersects at 0 and x1 only. Similarly,

2 2 ,23

x

by symmetry about x = π.

Note: At 2 11 1, we have ( ) sin

2

xx x x

; At 2 2

2 2, we have ( 2 ) sin2

xx x x

.

Correct Option: (2)


Part B – Physics

Question 1: An RLC series circuit is driven by a sinusoidal emf with angular frequency . If is

increased without changing the amplitude of the emf, the current amplitude increases. If the inductance is

L, the capacitance is C and the resistance R, it implies that

(1) 1

LC

(2) 1

LC

(3) L R (4) L R

Solution

The peak current in an RLC circuit is given by

00

0

2 2( )L C

VI

Z

V

R X X

Since 0I increases when is increased while keeping 0V the same, it implies that the impedance Z

decreases. This means that L CX X decreases as is increased. But, an increase in will cause

inductive reactance ( )LX L to increase and capacitive reactance ( 1/ )CX C to decrease. We,

therefore, conclude that C LX X .

Hence, 1

LC

.

Correct Option: (2)

Question 2: A thin movable plate is separated from two fixed plates P1 and P2 by two highly viscous

liquids of coefficients of viscosity 1 and 2 as shown in the figure below, where 2 = 9 1. Area of

contact of movable plate with each fluid is the same. If the distance between two fixed plates is h, then the

distance h1 of movable plate from upper plate such that movable plate can be moved with a constant

velocity by applying a minimum force on it is (Assume only linear velocity distribution in each liquid.)


(1) 4

h (2)

2

h

(3) 6

h (4)

3

h

Solution

Force required to pull the movable plate is given by

1 2

1 1

1 1

1 1

1

1 1

( )

9( )

1 9

( )

v vF A

h h h

v vA

h h h

Avh h h

For minimum force to pull the movable plate,

1

0dF

dh

1 2 2

1 1 1

1 90

( )

dFAv

dh h h h

2 2

1 1

9 1

( )h h h

1 1

3 1

( )h h h

1 13h h h

14

hh

Correct Option: (1)


Question 3: A binary compound AB3 is initially prepared from atoms A and B which are radioactive with

decay constants λ and 2λ , respectively. The time, at which the compound will have equal number of

atoms of A and B, is

(1) 1.1

(2)

0.69

(3) 0.34

(4)

3

Solution

A A

2 2

B B A

( ) (0)

( ) (0) 3 (0)

t

t t

N t N e

N t N e N e

When NA = NB at time t = T,

1 3

ln3 1.1

Te

T

Correct Option: (1)

Question 4: Consider the circuits shown in the figure given below.

A B


C D

Each resistor has resistance R , battery has emf E and internal resistance r . Assuming ,r R rank the

circuits in order of the total power dissipated, from the greatest to least.

(1) D, C, B, A (2) D, B, C, A

(3) C, B, D, A (4) C, D, B, A

Solution

Power consumed in circuit A is

2

A3

EP

R r

Power consumed in circuit B is

2

B

EP

R r

Power consumed in circuit C is

2

C2 ( / 2)

EP

R r

Power consumed in circuit D is

2

D( / 3)

EP

R r

Since r R , we have

2

( / 3)

E

R r>

2E

R r>

2

2 ( / 2)

E

R r>

2

3

E

R r

DP > BP > CP > AP

Correct Option: (2)


Question 5: The ratio of x-coordinate to the y-coordinate of the center of mass of a system of three

particles of 2 kg, 3 kg and 4 kg kept at the vertices of an equilateral triangle of side 1 m (as shown in the

figure) will be

2 kg 4 kg

O

3 kg

x

y

(1) 11

3 3 (2)

3 3

11

(3) 1

3 (4)

3

1

Solution

Assuming O at origin,

1/22

21 1 3OC m and AC 1 m

2 2 2

The x-coordinate of center of mass will be

xcm =

1 1 2 2 3 3

1 2 3

m x m x m x

m m m=

12(0)+3 +4(1)

112= m

2+3+4 18

and the y-coordinate of the center of mass will be

ycm =

1 1 2 2 3 3

1 2 3

m y m y m y

m m m =

2(0)+ 3 3 2 + 4(0) 3 3= m

2 + 3 + 4 18


The ratio of x-coordinate to the y-coordinate of the center of mass of system will be

cm

cm

11

3 3

x

y

Correct Option: (1)

Question 6: A satellite is in circular Earth orbit at an altitude of 200 km above Earth’s surface. At that

height, the acceleration due to gravity is 29.20ms . If the radius of the Earth is 66.4 10 m , what is the

orbital speed of the satellite?

(1) 143.4 ms (2) 17.79 kms

(3) 14.34 kms (4) 1767 ms

Solution

Let m be the mass of the satellite and h be the height of satellite above the surface of Earth. The

gravitational force acting on satellite provides the centripetal force, therefore

gravitation centripetal

2

6 3

1

1

( )

( )

9.20(6.4 10 200 10 )

7792ms

7.79 kms

F F

mvmg

R h

v g R h

Correct Option: (2)

Question 7: The following question consists of two statements, 1 and 2. Of the four choices given, choose

the one that best describes the two statements.

Statement 1: Two balls A and B, moving towards each other, undergo head-on collision and bounce back

from each other. After the collision, each ball’s velocity is equal in magnitude and opposite in direction to

what it was before. The total momentum of the two-ball system is zero.

Statement 2: The momentum of a system is conserved in the absence of any external force.

(1) Statement 1 is true, Statement 2 is false

(2) Statement 1 is true, Statement 2 is true; Statement 2 is the correct explanation of Statement 1

(3) Statement 1 is true, Statement 2 is true; Statement 2 is not the correct explanation of Statement 1

(4) Statement 1 is false, Statement 2 is true


Solution

In the absence of an external force, the momentum of a system is conserved. So, Statement 2 is true. During

the head-on collision, no external force acts on the two-ball system. Taking care of the direction of each

ball’s momentum by appropriate sign, we can write

1 1 2 2 1 1 2 2

1 1 2 2

( ) ( ) ( ) ( )

0

m v m v m v m v

m v m v

But, 1 1 2 2m v m v is the total momentum of the system before collision. Thus, Statement 1 is true.

Statement 2 is the correct explanation of Statement 1.

Correct Option: (2)

Question 8: A series of transmitter/receiver antenna towers, each of height 50 m, are planned to be

arranged in a circle at latitude of 60o on the moon, for operation in the line of sight mode. Given the

radius of the moon to be 1734 km, the number of towers needed is

(1) 358 (2) 207

(3) 155 (4) 57

Solution

The radius of the circle at a latitude of 60o = R sin 30

o = R/2 = 867 km

Let n towers of height h be needed, then

2 ( / 2)2 2

2 2

1734000

2 2 50

207

RRh

n

Rn

h

Correct Option: (2)

Question 9: For a glass prism, the index of refraction for visible light ranges from 1.524 at the blue end

of the spectrum to 1.509 at the red end. What is the dispersive power of the prism, assuming that the

refractive index for yellow light is the mean of the two given indices?

(1) 0.029 (2) 0.034

(3) 0.018 (4) 0.038


Solution

If nV, nR and nY represent refractive indices of violet, red and yellow light respectively then the dispersive

power can be calculated as

V R

Y 1

1.524 1.509

1.524 1.5091

2

0.015

0.516

0.029

n n

n

Correct Option: (1)

Question 10: A solid spherical region of radius R having a spherical cavity with diameter R, as shown in

the figure given below, has a total charge Q. The electric field at a point P is

(1)

22

0

2 1 1

7 2 2

Q

x x R

(2)

22

0

8 1 1

21 2 2

Q

x x R

(3)

22

0

8 1 1

7 2 2

Q

x x R

(4)

22

0

2 1 1

21 2 2

Q

x x R

Solution

Charge density of the configuration is

33

3

6

74

3 2

Q Q

RRR

(1)

Using the superposition principle, the electric field at point P will be


3

3

22

0

0

3

22

0

3

22

0

443 23

44

2

1 1

38

2

1 1(2)

3 2 2

RR

Ex R

x

R

x Rx

R

x x R

Substituting Eq. (1) in Eq. (2) we get

3

22

0

3

23 2

0

22

0

1 1

3 2 2

6 1 1

7 3 2 2

2 1 1

7 2 2

RE

x x R

Q R

R x x R

Q

x x R

Correct Option: (1)

Question 11: A parallel plate capacitor is charged completely and then disconnected from the battery. If

the separation between the plates is reduced by 50% and the space between the plates is filled with a

dielectric slab of dielectric constant 10, then the potential difference between the plates

(1) decreases by 95% (2) increases by 95%

(3) decreases by 50% (4) increases by 50%

Solution

The capacitance of parallel plate capacitor is

0 AC

d

where A is the area of the plates and d is the separation between the plates.

If the separation between the plates is reduced by 50% and the space between the plates is filled with a

dielectric slab of dielectric constant 10, the new capacitance becomes


0 0' 20/ 2

A AC K

d d

' 20C C

Since charge q is constant, we have

constantq CV

The relation between potential and capacitance is

1V

C

Therefore,

'20

VV

Therefore, % decrease in potential is

20 100

VV

V

= 95%.

Correct Option: (1)

Question 12: The current in a solenoid is decreased to one-half of its original value. Which one of the

following statements is true concerning the self-inductance of the solenoid?

(1) The self-inductance does not change.

(2) The self-inductance increases by a factor of 2.

(3) The self-inductance decreases by a factor of 2.

(4) The self-inductance decreases by a factor of 4.

Solution

The self-inductance of a solenoid is given by

2

0 N AL

l

where N is the number of turns, A is the cross-sectional area, and l is the length of the solenoid.

We see from this relation that the inductance is independent of the current. Thus, the self-inductance does

not change when the current is decreased.


Correct Option: (1)

Question 13: Two small balls of the same mass, having equal charge q, are suspended by two insulating

strings of equal length L from the same hook in an elevator. The elevator is falling freely. The tension in

the string at equilibrium with respect to frame attached with elevator will be

(1) 2

2

0

1

4 4

q

L (2)

22

2 2

2

0

1

4 4

qm g

L

(3) mg (4) 2

22

2

0

1

4

qmg

L

Solution

In case of freely falling elevator,the effect of gravity is cancelled by the pseudo force (w.r.t. the

elevator) therefore, the angle made by string with each other is 180º and the tension in the string at equilibrium is

2 2

2 2

0 0

1 1

4 (2 ) 4 4

q qF

L L

Correct Option: (1)

Question 14: A helicopter is lifting a 2100 kg jeep. The steel suspension cable is 48 m long and has a

radius of 35.0 10 m. Find the amount the cable is stretched when the jeep is hoisted up with an

acceleration of 21.5ms . (Given that:

11 22.0 10 NmY )

(1) 1.2 cm (2) 3.6 cm

(3) 5.1 cm (4) 7.3 cm

Solution

Tension in the cable

( )T m g a

Stress in the cable

2

( )T m g a

A r

According to Hooke’s law,


2

( )

( / )

m g aY

r L L

So,

2

2

3 2 11 2

2

( ).

(2100kg)(9.8 1.5)(ms )(48m)

(5.0 10 m) (2.0 10 Nm )

7.3 10 m

m g a LL

r Y

Correct Option: (4)

Question 15: Neglecting air resistance, a 1.0kg projectile has an escape velocity of about 111 kms at the

surface of the Earth. The corresponding escape velocity for a 2.0 kg projectile is

(1) 13.5kms (2) 15.5kms

(3) 110kms (4)

111kms

Solution

Escape velocity of the projectile at the surface of the Earth is given by

eesc

e

2GMv

R (1)

where eM is the mass of the Earth, and eR is radius of the Earth.

From Eq. (1) we conclude that escape velocity is independent of the projectile’s mass, so escape velocity

for a 2.0 kg projectile is also111 kms .

Correct Option: (4)

Question 16: A can is filled with liquid up to 97% of its capacity. The temperature of the can and liquid

is 0.0 o C . The material from which the can is made has a coefficient of linear expansion of5 o 12.8 10 ( C) . At a temperature of 100.0 o C , the liquid inside the can expands to fill it to the brim.

The coefficient of volume expansion of the liquid is

(1) 4 o 13.38 10 ( C) (2) 4 o 13.96 10 ( C)

(3) 4 o 13.08 10 ( C) (4)

4 o 12.96 10 ( C)


Solution

If V is the volume of the can at 0.0 o C , then the increase in its volume at 100.0 o C is

Can (3 )(100 0) 300V V V

where is the coefficient of linear expansion of can.

The initial volume of the liquid is 0.97V. The change in volume of the liquid is

L 0.97 ( )(100 0) 97V V V

where is the coefficient of volume expansion of liquid.

At 100.0 o C , both can and liquid have the same volume. Thus

Can L0.97V V V V

(1 300 ) 0.97 (1 100 )V V

5

4 o 1

0.03 300

97

0.03 300 2.8 10

97

3.96 10 ( C)

Correct Option: (2)

Question 17: The reflection of perpendicularly incident white light by a soap film in air has an

interference maximum at 600 nmand a minimum at 450 nm , with no minimum in between. If refractive

index 1.33n for the film, the film thickness is

(1) 552 nm (2) 445 nm

(3) 338 nm (4) 286 nm

Solution


Since the refractive index of the film is greater than that of air, ray 1 will undergo a phase change of

on reflection. Ray 2, on the other hand, will not undergo any phase change upon reflection. (See figure.)

Let t be the thickness of the film. For constructive interference, ray 1 and ray 2 should be in phase, that is,

path difference should be an integral multiple of wavelength. Path difference is 2t due to travelling in the

medium and film / 2 due to reflection at a boundary. Let it be m times the wavelength. Wavelength will

be reduced by a factor of refractive index n. Therefore,

film film

film

12

2

2 1

4

2 1 600(1)

4

t m

mt

m

n

where t is the thickness of the film.

So, by substituting m = 1, 2, 3… in Eq. (1), we get

150 450 750

, , , ...tn n n

(2)

For destructive interference:

film film

film

1 12

2 2

2

450(3)

2

t m

mt

m

n

So, by substituting m = 1, 2, 3… in Eq. (3) ,we get

225 450 675, , , ...t

n n n (4)

Comparing Eq. (2) and Eq. (4) , we conclude that

450

450

1.33

338 nm

tn


Correct Option: (3)

Question 18: Combination of logical gates shown below yields

A

B

X

(1) NOT gate (2) OR gate

(3) XOR gate (4) NAND gate

Solution

Here two NAND gates are being used as input for another NAND gate. Truth table for this would be

A B NAND(A,B)=C NAND(A,B)=D NAND(C,D)

0 0 1

1

1 0

0 1 1 0

1 0 1 1 0

1 1 0 0 1

So the output is equal to OR gate.

Correct Option: (2)

Question 19: The magnetic field due to current distribution at any point is given by ˆB Ar k , where

r is the position vector of the point, and A is a constant. The net current through a circle of radius R , in

the x–y plane and centered at the origin is given by

(1)

2

0

AR

(2)

0

2 AR

(3)

3

0

4

3

AR

(4)

2

0

2 AR


Solution

Let us consider the given circle as an Amperian loop – it lies in the x–y plane, with center at the origin,

and has a radius of R .

Now, the magnetic field at any point on this loop will be ˆB AR k , and will have a direction tangential

to the circle (at the circumference).

Hence, .B dl for this circular loop will be

. (2 )B dl B R .

Applying Ampere’s law, we get

0

0

0

0

2

0

.

(2 )

(2 )

( )(2 )

2

B dl I

B R I

B RI

AR R

AR

Correct Option: (4)

Question 20: An ideal gas with pressure P, volume V, and temperature T is expanded isothermally to a

volume 2V and pressure P1. If the same gas is expanded adiabatically to a volume 2V, the final pressure is

P2. If the ratio of the specific heats for the gas is 1.67, then the ratio P2/P1 is nearly

(1) 1/2 (2) 1/34

(3) 1/32

(4) 5/32

Solution

For the isothermal process,

1 1 2 2

1

1

(2 )

(1)2

PV PV

P V PV

PP

For the adiabatic process,


2

2

constant

(2 )

(1/ 2) (2)

PV

P V PV

P P

From Eq. (1) and Eq. (2) we get

12

1

5/3 1

2/3

1/3

(1 / 2)

1

2

2

4

P

P

Correct Option: (2)

Question 21: Two bodies P and Q move along the same straight line. Both start at 0x at time 0t .

Their velocities are plotted over the same time interval as shown below in figure.

What information does this graph provide about whether P and Q collide?

(1) P and Q collide at an instant 0t t

(2) P and Q collide at an instant 0t t

(3) P and Q do not collide in the interval 00 t t

(4) Nothing about their collision can be concluded from the graph

Solution

Both P and Q start together from the same point. Hence, their displacements will be equal to the areas

under their respective v t graphs. From the given graphs it is clear that in the interval 00 t t , the area

under Q is greater than that under P. So, Q’s displacement is greater than that of P. Hence, P and Q will

not collide in the interval 00 t t .

Correct Option: (3)


Question 22: For a constant volume process, the heat capacity of gas A is greater than the heat capacity

of gas B. If both the gases absorb the same amount of energy as heat, at constant volume, then

(1) The temperature of A increases by a greater extent than it does for B

(2) The temperature of B increases by a greater extent than it does for A

(3) The internal energy of A increases by a greater extent than it does for B

(4) The internal energy of B increases by a greater extent than it does for A

Solution

If nA and nB represents the number of moles of gas A and B respectively. For a constant volume process,

work done 0W . Thus, Q U . Since Q is same for both the gases, hence the change in internal

energy is same for both. So,

A B

A B

A A A B B B

A B B

B A A

U U

Q Q

n C T n C T

T n C

T n C

Since the heat capacity of A is greater than that of B, A A B Bn C n C . Hence, the temperature of B

increases by a greater extent than it does for A, that is, A BT T .

Correct Option: (2)

Question 23: A converging lens of focal length 20 cm is placed in contact with a diverging lens of focal

length 30 cm . The focal length of this combination is

(1) 60 cm (2) 25 cm

(3) 12 cm (4) 10 cm

Solution

Power of the converging lens

1

1

1 1

0.20P D

f

Power of the diverging lens

2

2

1 1

0.30P D

f

Power of the combination,


1 2

1 1

0.20 0.30

0.10

(0.20)( 0.30)

1

0.60

P P P

D

Hence, focal length of the combination,

10.60 m = 60 cmf

D

Correct Option: (1)

Question 24: If the work function for a certain metal is 1.8 eV, what is the stopping potential for

electrons ejected from the metal when light of wavelength 400 nm shines on the metal?

(1) 0.95 V (2) 1.31 V

(3) 2.08 V (4) 3.44 V

Solution

Photon energy

34 8

9

19

(6.63 10 )(3 10 )

400 10

4.97 10 J

hcE

Work function

19

19

1.8eV

1.8(1.6 10 C)(1V)

2.88 10 J

Therefore, maximum kinetic energy

max

19

19

(4.97 2.88) 10 J

2.09 10 J

K E


Hence, stopping potential will be

max0

19

19

2.09 10 J

1.6 10 C

1.306V 1.31V

KV

e

Correct Option: (2)

Question 25: A particle is performing SHM about mean position O as shown in the figure given below.

Its amplitude is 5 m and time period of oscillation is 4 s. The minimum time taken by the particle to move

from point A to B is nearly

O A B

3m

4m

(1) 0.18 s (2) 0.20 s

(3) 0.15 s (4) 0.24 s

Solution

The displacement equation of SHM is

1

sin

sin

sin

x A t

xt

A

xt

A

Substituting values given in the question, we have

1

1

3sin 37 37 rad

5 180t

(1)

and 1

2

4sin 53 53 rad

5 180t

(2)

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


2 1

2 1

( ) 53 37180

153 37

180

16180

8

2

8 4

180 180

8

45

0.18s

t t

t t

T

T

Correct Option: (1)

Question 26: A tuning fork of frequency 600 Hz is moving towards a reflecting wall with a speed of

30 ms–1

. The velocity of sound in air is 330 ms–1

. The observer is between the source and the wall and is

also moving towards the wall. The beat frequency heard by the observer is 4 Hz. Now if the tuning fork is

replaced with a new tuning fork of unknown frequency, the beat frequency heard by the observer

becomes 3 Hz. The frequency of the new tuning fork is

O

(1) 450 Hz (2) 425 Hz

(3) 400 Hz (4) 475 Hz

Solution

Apparent frequency of source


011

s

c v

c v

Apparent frequency of reflected sound

021

s

c v

c v

Beats frequency

11

s

202 v

c v

Initial beats frequency is 4 Hz, therefore

0 1

s

1

24Hz

2 6004Hz

330 30

v

c v

v

1

1

1

1

2 6004 Hz

300

4 4 Hz

1ms

v

v

v

The observer moves with a velocity 1 ms–1

towards the wall.

When the tuning fork is replaced with a new tuning fork of frequency new , the new beats frequency is

new

ne

1

w

s

2'

2 13 Hz

330 30

v

c v

new23 Hz

300

new = 450 Hz

The frequency of new tuning fork is 450 Hz.

Correct Option: (1)


Question 27: Two blocks rest on a horizontal frictionless surface, as shown in the figure.

The surface between the top and bottom blocks is rough such that there is no slipping between the two

blocks. A 30N horizontal force is applied to the bottom block. What is the frictional force between the

two blocks?

(1) 10 N (2) 20 N

(3) 25 N (4) 30 N

Solution

Since there is no slipping between the blocks, their common acceleration is

2302ms

(10 5)a

If the frictional force is f , then it is the only force acting on the upper block.

2(5kg)(2ms ) 10Nf ma

Correct Option: (1)

Question 28: If the velocity of light c, Planck’s constant h and time t are taken as basis of fundamental

units, then the dimension of force will change to

(1) hc–1

t–2

(2) hc–1

t2

(3) hc–1

t–1

(4) h–1

c–1

t–2

Solution

Dimensions of h and c are


2 1[ ]h ML T

and 1[ ]c LT

Let the force be expressed in terms of fundamental units as

F = ha c

b t

d

Taking dimensions on both sides, we get

[MLT–2

] = [ML2T–1

]a [LT

–1]

b [T]

d = [M

a L

2a+b T–a-b+d

] (1)

Comparing both sides of Eq. (1), we get

a = 1 (2)

2a + b = 1 (3)

and – a – b + d = – 2 (4)

Substituting Eq. (2) in Eq. (3), we get

2 + b = 1

b = –1 (5)

Substituting Eq. (2) and Eq. (5) in Eq. (4), we get

– 1 – (–1) + d = – 2

d = – 2 (6)

Hence force can be expressed as

F = h c–1

t–2

Correct Option: (1)

Question 29: In Davisson–Germer experiment, the correct relation between angle of diffraction and

glancing angle is

(1) 0902

(2) 090

2

(3)2

(4)

Solution


In the Davisson–Germer apparatus as shown above, the relation between angle of diffraction and the

glancing angle is given by

2

= 90.

Therefore, 902

.

Correct Option: (1)

Question 30: A uniform density rigid L-shaped thin frame POQ (hinged at O) of mass 21 kg is

undergoing planar motion as shown in the figure given below. The arrows show the velocities at the

terminal points P and Q. The kinetic energy of the frame at this instant is

(1) 182 J (2) 224 J

(3) 672 J (4) 91 J

Solution

Intersection point of perpendiculars to velocities of any two points would be the point from where

instantaneous axis of rotation (IAR) passes. Hence IAR passes through O. Let be the angular velocity

of the frame, then velocity at point Q can be expressed as

Q OQv

8 4

2 rad/s

Moment of Inertia IARI about IAR is

IAR IAR(OP) IAR(OQ)

2 2

OP OQ

OP OQ

3 3

I I I

M M

Finding mass using unitary method gives us OP 9 kgM and OQ 12 kgM . So


2 2

IAR

2

3 49 12

3 3

27 64

91 kgm

I

The kinetic energy will be

R

2

2

A

1

2

191 2

2

182 J

IK I

Correct Option: (1)

Part C – Chemistry

Question 1: The alkali metals can be easily detected by flame test because they have (1) low ionization enthalpy. (2) high conductance.

(3) low density. (4) low electron gain enthalpy.

Solution

Because of low ionization enthalpy, the alkali metals can be easily detected by flame test. When any salt of the metal is heated in a Bunsen burner flame, the heat excites one of the orbital electrons to higher

energy level. When this electron falls back to its original energy level, it gives out the extra energy

absorbed. For alkali metals, this energy falls in the visible light range. Correct Option: (1)

Question 2: In which of the following pairs, the two species are not isostructural?

(1)

4PCl and SiCl4 (2) 4BFand SF4 (3) SF6 and

3

6AlF (4) CO2 and 2NO

Solution (1) Tetrahedral, tetrahedral

(2) Tetrahedral, see-saw

4BF has 4 bond pairs and 0 lone pairs on central atom. Hence, boron is sp

3 hybridized and tetrahedral in

shape.


SF4 has 4 bond pairs and 1 lone pair on the central atom. Hence, sulphur is sp

3d hybridized and see-saw in

shape.

(3) Octahedral, octahedral

(4) Linear, linear

O=C=O O N O

Correct Option: (2)

Question 3: Which of the oxides given below is the most basic? (1) Li2O (2) Na2O (3) K2O (4) MgO

Solution

Li, Na and K belong to the same group. The basic character in the group increases from top to bottom in a group. So, K2O is the most basic. As far as MgO is concerned, it is less basic than Na2O as Mg comes after Na in a period. On moving from left to right across a period, the metallic character of the element decreases and the oxides of elements become less basic.

Correct Option: (3)

Question 4: Which of the following relationship is not correct regarding their atomic radius? (1) Ag ≈ Au (2) Pd > Ni (3) W >> Mo (4) La > Y

Solution

Due to lanthanoid contraction, the radius of elements of first transition series is less than that of radius of corresponding second transition series elements but that of third transition series elements is nearly same

as the second transition series, that is

Radius First transition series. < Radius Second transition series and Radius Second transition series. ≈ Radius Third transition series The exception to this behavior is observed in case of Sc < Y < La. Lanthanoid contraction is not observed

in case of La because 4f14

is not present in its electronic configuration. Hence,

Option (1) is correct: rAg = 1.34 Å and rAu = 1.34 Å.

Option (2) is correct: rPd = 1.28 Å and rNi = 1.15 Å. Option (3) is incorrect: rW = 1.30 Å and rMo = 1.29 Å, that is, both are almost the same.

Option (4) is correct: rLa = 1.69 Å and rY = 1.62 Å.

Correct Option: (3)


Question 5: Which of the following is not usually associated with eutrophication? (1) Reduced photosynthesis

(2) Increased heavy metal concentrations

(3) Increased nutrient concentrations

(4) Reduced dissolved oxygen Solution Eutrophication is a process in which a body of water develops a high concentration of nutrients

such as nitrogen and phosphorus. Their increased concentration leads to growth of aquatic plants in

general and in production of blue-green algae. This algal mat reduces the availability of light below its surface, thus greatly reducing photosynthesis. The bacteria and algae die and their decomposition

increases the BOD of water, thus the dissolved oxygen content of water is reduced.

Heavy metal concentration is not affected by eutrophication. Correct Option: (2)

Question 6: In the following equilibrium: 2 2

4 4Stable form Metastable form

[ON Ru(NO) OH] [NO Ru(NO) (OH)]

which of the following statements is correct?

(1) Coordination numbers of the above two complexes are not the same.

(2) NO is acting as an ambidentate ligand in metastable form.

(3) NO is acting as an ambidentate ligand in both stable and metastable forms. (4) The complexes show coordination isomerism.

Solution

The coordination numbers of both the complexes is 6. NO can act as ambidentate ligand but both kinds of linkages are observed (i.e., Ru–N and Ru–O) in

metastable form only.

The complexes show linkage isomerism. Correct Option: (2)

Question 7: Which of the following statements is incorrect for the reaction given below? conc NaOH

ArCHO HCHO

(1) The reaction involves hydride transfer from one aldehyde to another. (2) ArCHO is reduced in the reaction, which results in the major products.

(3) Aldol addition product is not possible for the given reactants.

(4) One of the major products of the reaction is ethanol.

Solution

Since both aldehydes lack alpha hydrogen, aldol condensation is not possible. In this case, Cannizzaro’s

reaction will take place. Two aldehydes without alpha hydrogens will give crossed Cannizzaro’s products. However, HCHO transfers hydride more easily as compared to ArCHO, so the major products of the

reaction are ArCH2OH and HCOONa. conc NaOH

2ArCHO HCHO ArCH OH HCOONa

Correct Option: (4)

Quick Tip: In crossed Cannizzaro reaction, HCHO always gets oxidized and ArCHO always gets reduced.

Question 8: In the test for halogens, sodium fusion extract is boiled with concentrated nitric acid to

(1) increase the solubility of silver halides formed. (2) increase the precipitation of silver halides formed.

(3) supplement the concentration of nitrate ions.


(4) decompose sodium cyanide and sodium sulphide, if formed during Lassaigne’s test.

Solution In sodium fusion test, halogens are converted into soluble salts of sodium, which on reaction with silver

nitrate precipitate out. However, nitrogen and sulphur, if present in the compound, will form sodium

cyanide and sodium sulphide and then precipitate out as silver cyanide and silver sulphide, respectively

on addition of silver nitrate. Therefore, sodium extract is boiled with concentrated nitric acid first to decompose sodium cyanide and sodium sulphide formed during Lassaigne’s test, so that they do not

interfere with silver nitrate test.

Correct Option: (4)

Question 9: What is the order of rate of electrophilic substitution reaction for the following compounds?

(I) (II) (III) (IV)

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

Solution The electrophilic substitution reaction takes place through two-step mechanism in which the first step is

the rate determining step. The first step is the attack of electrophile on benzene ring and can be

accelerated by increasing electron density on benzene ring.

The electron density in benzene ring is increased or decreased according to the substituent on the ring. According to electron displacement effects, the groups attached to the given compounds have the

following electron withdrawing or electron donating tendencies:

– I effect order: F > Cl and + I effect order: t-Butyl group > Methyl group

The reaction rate will be more in case of methyl group because it is further stabilized by hyperconjugation

which is not possible in t-butyl group.

+R effect: F > Cl since lone pairs of fluorine are in 2p orbitals which result in more effective overlapping

between fluorine and carbon atoms. So, in the case of fluoride group, the reaction rate will be more, even

though it is more electron withdrawing than chloride group. Hence, Cl deactivates the ring more than F and methyl group activates it more than t-butyl group.

Correct Option: (3)

Question 10: Consider the following pairs of reaction reagents. In which of the following pairs, the

second nucleophile would be more nucleophilic in a polar aprotic solvent?

(1) F and I

(2) CH3CH2SH and CH3CH2OH

(3) CH3NH and CH3NH2 (4) CH3O

and CH3COO

Solution

In polar aprotic solvents, nucleophilicity is proportional to the basicity of reagents. In other words, more

solvated nucleophile in protic solvents shows more nucleophilicity in polar aprotic solvents. So, the order of nucleophilicity in each case is:

F > I

CH3CH2SH < CH3CH2OH

CH3NH > CH3NH2 CH3O

> CH3COO


In the case of second option, CH3CH2OH is more basic than CH3CH2SH.

Correct Option: (2) Quick Tip: Compare the basicity of the given species. More is the basicity, greater will be the

nucleophilicity.

Question 11: Which of following statements is not true about o-, m-, and p-nitrophenol? (1) Boiling point of o-nitrophenol is lower than p-nitrophenol.

(2) Nitration of phenol with concentrated nitric acid gives p-nitrophenol as the major product.

(3) p-nitrophenol is a stronger acid than o-nitrophenol. (4) m-nitrophenol is the weakest acid among them.

Solution

The acidic strength order: p-nitrophenol > o-nitrophenol > m-nitrophenol. Due to intramolecular hydrogen bonding in o-nitrophenol, its boiling point is less than p-nitrophenol.

Nitration of phenol with concentrated nitric acid leads to formation of 2,4,6-trinitrophenol, commonly

known as picric acid.

Correct Option: (2)

Question 12: The correct order of the boiling point of comparable molecular weight of acid and its

derivatives is (1) Primary amides > carboxylic acids > nitriles > esters

(2) Carboxylic acids > primary amides > nitriles > esters

(3) Carboxylic acids > nitriles > primary amides > esters (4) Primary amides > carboxylic acids > esters > nitriles

Solution

Boiling point of primary amides is the maximum due to dipole–dipole interactions and intermolecular

hydrogen bonding.

Carboxylic acids also have strong hydrogen bonding due to which they form dimers and their boiling

points are higher than that of corresponding nitriles and esters but lower than primary amides.

Nitriles show strong dipole–dipole interactions, so, their boiling points are more than esters but lower

than acids.


Correct Option: (1)

Question 13: Which of the following alkyl halides forms a substitution product in an SN1 reaction that is

the same as the substitution product formed in an SN2 reaction?

(1) (2)

(3) (4)

Solution In SN2 reaction, back side attack of nucleophile takes place and the whole reaction occurs in a single step.

SN1 reaction occurs in two steps in which carbocation is produced as intermediate.

So, rearrangement can take place in SN1 reaction which will change the skeleton of the compound, and hence a product with different skeleton will be obtained.

In the compound given in option (4), its carbocation will not rearrange. Therefore, both SN2 and SN1

products will be the same in this case.

Correct Option: (4) Quick Tip: Both SN1 and SN2 mechanisms produce the same compound if stereochemistry is ignored.

However, different products can be obtained if the intermediate rearranges. So, we need to form

carbocation by removing the leaving group and check whether this carbocation can rearrange. Rearrangement will produce different compounds.

Question 14: Which of the following polymer does not contain hydrogen bonding? (1) Nylon-6,6 (2) Polyurethanes (3) Kevlar (4) Dacron

Solution

Fibres are long-chain polymers in which chains are linked by strong intermolecular forces. The

intermolecular forces are in general strong hydrogen bonds, dipole–dipole and van der Waals forces. The polymers given in options are fibres, but Dacron is a polyester which contains strong dipole–dipole

linkages instead of hydrogen bonding.

Correct Option: (4)

Question 15: Arrange the following carbon acids in the increasing order of their pKa values?


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

Solution Each of these compounds contains methylene (–CH2–) group clamped between two electron withdrawing

groups. So, after deprotonation from methylene group, negative charge can be delocalized and conjugate

base can be stabilized. If stronger electron withdrawing group is attached to this methylene group, it will give more stable conjugate base which is produced from a stronger acid.

Since compound (IV) contains –NO2 group which is most electron withdrawing through – M effect and –

I effect, its presence makes the compound (IV) most acidic. Compound (III) contains –COOEt group and

is least electron withdrawing. The –OEt group shows mesomeric effect (+M) and inductive effect (–I) but +M effect is more dominant than – I effect. In the case of compound (I) and compound (II), group -

COCH3 and –CHO are more electron withdrawing than –COOEt. Therefore acidic order is IV > II > I >

III. Since pKa = log Ka, so increasing order is IV < II < I < III. Correct Option: (1)

Question 16: Among the following, the paramagnetic compound is

(1) Na2O2 (2) O3 (3) N2O (4) NO2 Solution

Na2O2 contains 22O ions, the molecular orbital configuration is

2 * 2 2 * 2 2 2 2 * 2 * 21 1 2 2 2 2 2 2 2z x y x ys s s s p p p p p

As there are no unpaired electrons, so it is not paramagnetic.

Similarly, O3 and N2O are even electron species. This implies non-existence of unpaired electrons and no paramagnetism.

NO2 is an odd-electron molecule, that is, there exists at least one unpaired electron, so it is paramagnetic.

Correct Option: (4)

Question 17: Which of the following is best resonance structure for azide ion, 3N ?

(1) (2)

(3) (4)

Solution Nitrogen belongs to the second period, so the valence shell can have 8 electrons only. Therefore, the structure given in option (3) is not possible as nitrogen cannot form five bonds.

The formal charges for the structures are 1 1 1

:N N N:

0 1

2:N N N:

0 0 1

:N N N:

3

2 2:N N N:

Out of the other structures, Option (1) species has sufficient bonds, that is, the octet of every atom is

complete. Also, it has the least formal charge, so it is the best Lewis structure.

Correct Option: (1)


Quick Tip: The most stable structure is the one which contains maximum bonding pair and least formal

charge allocation. Also, the structure with least charge separation of unlike charges and maximum charge separation of like charges is favoured.

Question 18: What volume of wet methane (saturated with water vapour) would you have to collect at

29C and 1 atm pressure to be sure that the sample contains 244 ml of dry methane at 950 torr. [Given

that vapour pressure of water at 29C = 30 torr.] (1) 317.5 ml (2) 187.5 ml (3) 305 ml (4) 244 ml

Solution

The moles of methane gas required is 950 244

molpV

nRT RT

Same number of moles of methane should be present in wet methane, which is saturated with water

vapour. So, wet methane contains water vapour and the partial pressure of water vapour in it should be

equal to the vapour pressure of water for given conditions.

Therefore, partial pressure of methane in wet methane = 760 30 = 730 torr. Thus,

4CH

950 244317.5 ml

730

nRT RTV

p RT

Correct Option: (1) Quick Tip: Options (2) and (4) can be neglected easily since

pmix > pdry methane > pwet methane

Question 19: Silver cyanide, AgCN is one of the important salts used in the extraction of silver metal. It

is sparingly soluble in water and its solubility is approximately 2 106

mol/l at 25C. If the dissociation

constant of HCN at 25C is 5 1010

, what is the solubility product of AgCN at this temperature?

(1) 133.6 10 (2) 124 10 (3) 143.24 10 (4) 123.6 10

Solution

The reaction is

AgCN(s) Ag (aq) CN (aq)

Since CN comes from weak acid, so its hydrolysis will take place, therefore

2

6 6 6

CN (aq) H O HCN(aq) OH (aq)

At equilibrium 2 10 (1 ) 2 10 2 10 (where is the degree of hydrolysis)h h h h

6 2

6

[HCN][OH ] (2 10 )

[CN ] (2 10 )(1 )

wh

a

K hK

K h

or 14 6 2

5

10

10 2 102 10

5 10 (1 )

h

h

or 2

210 10 10 0 0.91(1 )

hh h h

h

Therefore, 6 7[CN ] 2 10 (1 0.91) 1.8 10 mol/l

and 6 7 13

sp [Ag ][CN ] 2 10 1.8 10 3.6 10K

Correct Option: (1)

Question 20: Which of the following statements is not correct about the Freundlich adsorption isotherm?

(1) At low pressure, the amount of adsorbate depends linearly on pressure.

(2) At high pressure, the amount of adsorbate is independent of pressure. (3) The slope of log(x/m) vs. log(p) may vary from zero to one.


(4) The most probable value of (1/n) is around one.

Solution

According to Freundlich isotherm, 1/nxkp

m

; where n > 1. At low pressure, the isotherm is

approximately linear, so (x/m) depends linearly on pressure and at high pressure (x/m) becomes constant.

So, (x/m) is almost independent of pressure.

The slope of log( / ) vs. log( )x m p curve is 1/n. Since the value of 1/n varies from 0 to 1, the slope may

vary from 0 to 1. But, in most cases the value of 1/n lies between 0.1 and 0.5.

Correct Option: (4)

Question 21: Which of the following solutions will show positive deviation from ideal behaviour?

(1) C2H5OH + H2O (2) CH3COCH3 + CHCl3 (3) H2O + HCl (4) H2O + HNO3

Solution

Positive deviation is shown by the solutions in which interactions between the molecules of its

components (solute and solvent) are reduced. It causes increase in vapour pressure and it is greater than the value calculated by using Raoult’s law.

In pure water, molecules are held by strong intermolecular force known as hydrogen bonding. But, it is

reduced when ethanol molecule comes in between two molecules of water. So, the solution will show positive deviation.

Correct Option: (1)

Question 22: Which of the following electron transitions in hydrogen atom will produce photon with

maximum wavelength?

(1) n = 5 to n = 1 (2) n = 4 to n = 2 (3) n = 6 to n = 4 (4) n = 6 to n = 5

Solution The energy of photon is calculated as the energy difference of the orbits,

2 1photon n nE E E E

Hence, photon

hc hcE E

E

So, emitted photon’s wavelength is inversely proportional to the energy difference of energy levels. Since

2 2

2 1

1 1E

n n

, moving away from nucleus the gap between successive energy levels is suppressed. So

for the transition from highest energy level to lower energy level will produce lowest energy photon.

For n = 6 to n = 5, transition is taking place from the highest energy level, n = 6 to just lower energy level,

n = 5. As the energy difference is minimum, so the wavelength of emitted photon will be maximum. Correct Option: (4)

Question 23: At 460C, the reaction, 2 2 3SO (g) NO (g) NO(g) SO (g); has KC = 85. A reaction

flask at 460C contains these gases at the following concentrations: [SO2] = 0.0025 M, [NO2] = 0.0035 M, [NO] = 0.025 M and [SO3] = 0.04 M.

Which of the following statements is correct? (1) Reaction is proceeding towards forward direction.

(2) Reaction is proceeding towards reverse direction.

(3) Equilibrium would be affected on addition of inert gas at constant pressure. (4) The value of Kp is greater than KC.

Solution

The reaction quotient is given by


3

2 2

[SO ][NO] 0.04 0.025114

[SO ][NO ] 0.0025 0.0035Q

Since Q > KC, this implies that the product concentration is more than the equilibrium concentration, so,

reverse reaction takes place.

The number of moles of reactants and products is the same; hence, change in pressure will not affect the

equilibrium. The addition of an inert gas at constant pressure will also not affect the equilibrium. The two equilibrium constants are related as:

( ) gn

p CK K RT

where ng = coefficient of gaseous products coefficient of gaseous reactants. Since coefficient of

reactants and products is the same, ng =0 and so Kp = KC. Correct Option: (2)

Question 24: 100 g of 0.1 M Ca(OH)2 aqueous solution of density 1.5 g/cm3 is mixed with 100 g of 0.1

M H2SO4 of density 1.1 g/cm3. Which of the following statements is not correct about the final mixture?

(1) Final solution is neutral with [H+] 10

7 M.

(2) The final calcium ion concentration, [Ca2+

] 0.04 M.

(3) The final sulphate ion concentration, 2

4[SO ] 0.06 M

(4) Ca(OH)2 is the limiting reagent in the reaction.

Solution

2 2 4 4 2Ca(OH) H SO CaSO 2H O

2

MassMoles of Ca(OH) taken Molarity Volume Molarity

density

1000.1× mmol 6.66 mmol

1.5

2 4

100Similarly, moles of H SO taken 0.1 mmol 9.09 mmol

1.1

Therefore, Ca(OH)2 is the limiting reagent and it cannot neutralize the acid completely. As the mmol of

H2SO4 > mmol of Ca(OH)2, so the resulting mixture is acidic.

The final volume of solution = 66.6 ml + 90.9 ml = 157.5 ml 2+

2 mmol of Ca ions 6.66[Ca ] 0.04 M

total volume 157.5

2

4

9.09[SO ] 0.06 M

157.5

2 (9.09 6.66)[H ] 0.03 M

157.5

Therefore, the final mixture is not neutral as equivalents of acid and base are not equal and the final

hydrogen ion concentration is 0.03 M. Correct Option: (1)

Question 25: Which of the following contains the maximum number of atoms?

(1) 16 g of oxygen atoms (2) 16 g of methane gas (3) 48 g of ozone (4) All contain the same number of atoms

Solution


A A

16Number of atoms in 16 g O atoms

16N N

4 A A

16Number of atoms in 16 g CH 5 5

16N N

A A

48Number of atoms in 48 g ozone 3 3

48N N

Correct Option: (2)

Quick Tip: 16 g of oxygen atoms = 1 mol of atoms

16 g CH4 = 5 mol of atoms

48 g of O3 = 3 mol of atoms

Question 26: A solid is formed with three types of atoms A, B and C. A forms fcc lattice, while B atoms

occupy all tetrahedral voids and C atoms occupy half of the octahedral voids. The formula of the solid is (1) A2B4C (2) A2B2C (3) AB2C2 (4) AB2C

Solution

The atoms of A per unit cell 1 1

8 6 48 2

The atoms of B per unit cell = 2 4 8

The atoms of C per unit cell = 1

4 22

Therefore formula is 4 8 2 2 4A B C A B C

Correct Option: (1)

Question 27: The standard enthalpy of formation of carbon disulphide, CS2(g), is 115.3 kJ/mol. Given

that o

f (S) 277 kJ/molH and o

f (C) 716.7 kJ/molH , where o

fH is the standard heat of formation of

the gaseous atoms from the elements in their standard state. Using the data, calculate the bond enthalpy of

C=S.

(1) 577.7 kJ/ mol (2) 1155.4 kJ/mol (3) 878.4 kJ/mol (4) 439.2 kJ/mol

Solution The chemical equation for formation of CS2 can be written as

o

f 2C(s) 2S(s) S C S(g); (CS (g)) 115.3 kJ/molH

It implies that o o o o

f 2 f f bond(CS (g)) (C) 2 (S) 2 (C=S)H H H H

Therefore, o

bond

o

bond

115.3 716.7 2 277 2 (C S)

(C S) 577.7 kJ/mol

H

H

Correct Option: (1)

Question 28: The decomposition of N2O5 follows the reaction; 2N2O5 (g) 4NO2(g) + O2(g) and it is a first order reaction. The reaction was allowed to take place in a closed vessel. After 30 minutes, the total


pressure was 284.5 mm Hg, and on complete decomposition the total pressure was 584.5 mm Hg.

Calculate the rate constant of the reaction.

(1) 8.7 103

min1

(2) 5.2 103

min1

(3) 5.2 103

s1

(4) 2.5 103

min1

Solution The reaction can be expressed as

2 5 2 22N O 4NO (g) O (g)

Initial pressure 0 0

After 30 min. 2 / 2

On completion 0 2 / 2

p

p x x x

p p

After the reaction is over, 1 mol of reactant changes to 5/2 mol of products.

Therefore, the initial pressure, p = 2/5 584.5 = 233.8 mm Hg

3Total pressureafter 30 min 2 284.5

2 2

x xp x x p

or 3

284.5 50.72

xp

Therefore, x = 33.8 and p – x = 200

3 12.303 2.303 233.8log log 5.2 10 min

30 200

pk

t p x

Correct Option: (2)

Question 29: For the reaction 2

2 2 2NiO (s) Cl (aq) H (aq) Cl (g) Ni (aq) H O , the standard cell

potential is 0.32 V. The value of Go will be

(1) 61.8 kJ (2) 61.8 kJ (3) 30.9 kJ (4) 123.6 kJ

Solution

The balanced chemical equation: 2

2 2 2NiO (s) 2Cl (aq) 4H (aq) Cl (g) Ni (aq) 2H O

We know that o oG nFE ; where n implies number of electrons involved in the reaction, and F is

96500 C/mol. Since NiO2 is converting from Ni4+

to Ni2+

, two electrons are involved per molecule of

NiO2.

Therefore, 4

o 6.18 102 mol 96500 C/mol 0.32 V 61.8 kJ; (C V J)

1000G

Correct Option: (1)

Question 30: Calculate the osmotic pressure of a solution obtained by mixing 100 ml of 3.4% solution

(w/V) of urea (mol. wt. 60) and 100 ml of 1.6% solution (w/V) of cane sugar (mol. wt. 342) at 20C. (1) 14.73 atm (2) 13.61 atm (3) 7.38 atm (4) 6.82 atm

Solution

For urea, the osmotic pressure is 11 1 1 1 1

1

wV n RT RT

M

For sugar, the osmotic pressure is 22 2 2 2 2

2

wV n RT RT

M

Since, 100 ml of urea solution are mixed with 100 mL of cane-sugar solution, and thus, the total volume

becomes 200 ml in which 3.4 g urea and 1.6 g sugar is present

1 1

200 3.40.0821 293 6.82 atm

1000 60 (where R = 0.0821 l atm/mol-K)

and 2 2

200 1.60.0821 293 0.56 atm

1000 342


So, total 1 2 7.38 atm

Correct Option: (3)

JEE Knocout Test Solution Doc 30th March

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Transcript of JEE Knocout Test Solution Doc 30th March