FINAL JEE–MAIN EXAMINATION – APRIL, 2024

(Held On Monday 08th April, 2024) TIME : 9 : 00 AM to 12 : 00 NOON

NOON

MATHEMATICS

TEST PAPER WITH SOLUTION

SECTION-A

1. The value of k  for which the integral

In =

1

kn

0

(1 – x ) dx,



n , satisfies 147 I20 = 148 I21

is :

(1) 10 (2) 8

(3) 14 (4) 7

Ans. (4)

Sol. In =

1

kn

0

(1 x ) .1 dx



In = (1 – xk)n.x – nk

1

k n 1 k 1

0

(1 x ) .x .dx







In = nk

1

k n k n 1

0

[(1 x ) (1 x ) ]dx



  



In = nkIn – nkIn

n

n1

Ink

I nk 1





21

20

I21k

I 1 21k



147

148



 k = 7

2. The sum of all the solutions of the equation

(8)2x –16



(8)x + 48 = 0 is :

(1) 1 + log6(8) (2) log8(6)

(3) 1 + log8(6) (4) log8(4)

Ans. (3)

Sol. (8)2x – 16



(8)x + 48 = 0

Put 8x = t

t2 – 16 + 48 = 0

t = 4 or t = 12

8x = 4 8x = 12

x = log8x x = log812

sum of solution = log84 + log812

= log848 = log8(6.8)

= 1 + log86

3. Let the circles C1 : (x – )2 + (y – )2 =

2

1

r

and

C2 : (x – 8)2 +

2

15

y2









=

2

r

touch each other

externally at the point (6, 6). If the point (6, 6)

divides the line segment joining the centres of the

circles C1 and C2 internally in the ratio 2 : 1, then

(+ ) + 4

 

22

12

rr

equals

(1) 110 (2) 130

(3) 125 (4) 145

Ans. (2)

Sol.

r1

r2

C2

C1()

P

2 : 1

C2

() C1

P

(6, 6)

16 6

3



and

15 6

3



(, )



(2, 3)

Also, C1C2 = r1 + r2



2

22

15

(2 8) 3 2r r

2



    





r2 =

5

2

r1 = 2r2 = 5

() + 4

22

12

(r r )

= 5 + 4

25 25

4









= 130

4. Let P(x, y, z) be a point in the first octant, whose

projection in the xy-plane is the point Q. Let

OP =  ; the angle between OQ and the positive

x-axis be ; and the angle between OP and the

positive z-axis be , where O is the origin. Then

the distance of P from the x-axis is :

(1)

22

1 sin cos   

(2)

22

1 cos sin   

(3)

22

1 sin cos   

(4)

22

1 cos sin   

Ans. (1)

Sol. P(x, y, z), Q(x, y, O) ; x2 + y2 + z2= 2

ˆˆ

OQ xi yj

cos =

22

x

xy

cos =

2 2 2

z

x y z

sin2=

22

2 2 2

xy

x y z





distance of P from x-axis

22

yz



22

x



2

x

1



=

22

1 cos sin   



5. The number of critical points of the function

f(x) = (x – 2)2/3 (2x + 1) is :

(1) 2 (2) 0

(3) 1 (4) 3

Ans. (1)

Sol. f(x) = (x – 2)2/3 (2x + 1)

f'(x) =

2

3

(x – 2)–1/3 (2x + 1) + (x – 2)2/3 (2)

f'(x) = 2 ×

1/3

(2x 1) (x 2)

3(x 2)

  



1/3

3x 1 0

(x 2)





Critical points x =

1

3

and x = 2

6. Let f(x) be a positive function such that the area

bounded by y = f(x), y = 0 from x = 0 to x = a > 0

is e–a + 4a2 + a – 1. Then the differential equation,

whose general solution is y = c1f(x) + c2 , where c1

and c2 are arbitrary constants, is :

(1)

2

x

2

d y dy

(8e 1) 0

dx

  

(2)

2

x

2

d y dy

(8e 1) 0

dx

  

(3)

2

x

2

d y dy

(8e 1) 0

dx

  

(4)

2

x

2

d y dy

(8e 1) 0

dx

  

Ans. (3)

Sol.

a

a2

0

f(x)dx e 4a a 1



   



f(a) = –e–a + 8a + 1

f(x) = –e–x + 8x + 1

Now y = C1f(x) + C2

1

dy C f '(x)

dx 

= C1(e–x + 8) …..(1)

2

dy

dx

= –C1e–x  –ex

2

dy

dx

Put in equation (1)

2

xx

2

dy d y

e (e 8)

dx dx



  

2

x

2

d y dy

(8e 1) 0

dx

  

7. Let f(x) = 4cos3x +

33

cos2 x – 10. The number

of points of local maxima of f in interval (0, 2) is:

(1) 1 (2) 2

(3) 3 (4) 4

Ans. (2)

Sol. f(x) = 4cos3(x) +

33

cos2(x) – 10 ; x  (0, 2)

f'(x) = 12cos2x[–sin(x)] +

33

(2cos(x))[–sin(x)]

f'(x) = –6sin(x) cos(x)[2cos(x) +

3

]

–

+

–

+

–

+

–

2

0

local maxima at x =

57

,

66



8. Let A =

2 a 0

1 3 1

0 5 b







. If A3 = 4A2 – A – 21I, where

I is the identity matrix of order 3 × 3, then 2a + 3b

is equal to :

(1) –10 (2) –13

(3) –9 (4) –12

Ans. (2)

Sol. A3 – 4A2 + A + 21 I = 0

tr(A) = 4 = 5 + 6  b = –1

|A| = –21

–16 + a = –21  a = –5

2a + 3b = –13

9. If the shortest distance between the lines

1ˆˆ ˆ

L : r (2 )i (1 3 ) j (3 4 )k,         

2ˆˆ

ˆ

L : r 2(1 )i 3(1 )j (5 )k,      

is

m

n

, where gcd (m, n) = 1, then the value of

m + n equals.

(1) 384 (2) 387

(3) 377 (4) 390

Ans. (2)

Sol.

C

D

Shortes distance (CD) =

AB p q

| p q |





=

ˆ ˆ ˆ ˆ

ˆˆ

(0i 2 j 2k).( 15i 7j 9k)

355

    

=

0 14 18

355



=

32

355

 m + n = 32 + 355 = 387

10. Let the sum of two positive integers be 24. If the

probability, that their product is not less than

3

4

times their greatest positive product, is

m

n

,

where gcd(m, n) = 1, then n – m equals :

(1) 9 (2) 11

(3) 8 (4) 10

Ans. (4)

Sol. x + y = 24, x, y N

AM > GM  xy



144

xy



108

Favorable pairs of (x, y) are

(13, 11), (12, 12), (14, 10), (15, 9), (16, 8),

(17, 7), (18, 6), (6, 18), (7, 17), (8, 16), (9, 15),

(10, 14), (11, 13)

i.e. 13 cases

Total choices for x + y = 24 is 23

Probability =

13 m

23 n



n – m = 10

11. If sinx = –

3

5

, where  < x <

3

2



,

then 80(tan2x – cosx) is equal to :

(1) 109 (2) 108

(3) 18 (4) 19

Ans. (1)

Sol. sinx =

3

5



,  < x <

3

2



tanx =

3

4

cosx =

4

5



80(tan2x – cosx)

=

94

80 16 5









= 45 + 64 = 109

12. Let I(x) =

22

6dx

sin x(1 cot x)



. If I(0) = 3, then

I

12









is equal to :

(1)

3

(2)

33

(3)

63

(4)

23

Ans. (2)

Sol. I(x) =

2

2 2 2

6dx 6 cosec x dx

sin x(1 cot x) (1 cot x)







Put 1 – cotx = t

cosec2x dx = dt

I =

2

6dt 6 c

t







I(x) =

6c

1 cot x



, c = 3

I(x) =

6

31 cot x



,

6

I3

12 1 (2 3)







 



6 6( 3 1)

I 3 3 3 3 2

12 2

31





    

 



13. The equations of two sides AB and AC of a

triangle ABC are 4x + y = 14 and 3x – 2y = 5,

respectively. The point

4

2, 3









divides the third

side BC internally in the ratio 2 : 1. The equation

of the side BC is :

(1) x – 6y – 10 = 0 (2) x – 3y – 6 = 0

(3) x + 3y + 2 = 0 (4) x + 6y + 6 = 0

Ans. (3)

Sol.

3x –2y = 5

4x + y = 14

A

B (x1, 14 – 4x1)

C

2 : 1

•

21

2x x 2

3



,

 

2

1

3x 5

2 14 4x

2

3











=

4

3



2x2 + x1 = 6, 3x2 – 4x1 = – 13

x2 = 1, x1 = 4

So, C(1, – 1) , B(4, –2)

m =

1

3



Equation of BC : y + 1 =

1

3



(x – 1)

3y + 3 = – x + 1

x + 3y + 2 = 0

14. Let [t] be the greatest integer less than or equal to

t. Let A be the set of al prime factors of 2310 and

f : A  be the function f(x) =

3

2

x

log x 5

























.

The number of one-to-one functions from A to the

range of f is :

(1) 20 (2) 120

(3) 25 (4) 24

Ans. (2)

Sol. N = 2310 = 231 × 10

= 3 × 11 × 7 × 2 × 5

A = {2, 3, 5, 7, 11}

f(x) =

3

2

x

log x 5





















f(2) = [log2 (5)] = 2

f(3) = [log2 (14)] = 3

f(5) = [log2 (25 + 25)] = 5

f(7) = [log2 (117)] = 6

f (11) = [log2 387] = 8

Range of f : B = {2, 3, 5, 6, 8}

No. of one-one functions = 5! = 120

15. Let z be a complex number such that |z + 2| = 1

and lm

z 1 1

z 2 5













. Then the value of

 

Re z 2

is :

(1)

6

5

(2)

16

5



(3)

24

5

(4)

26

5

Ans. (4)

Sol. |z + 2| = 1, Im

z 1 1

z 2 5













Let z + 2 = cos + isin

1cos isin

z2

   



 

z 1 1

1

z 2 z 2





= 1 – (cos – isin)

= (1 – cos) + isin

Im

z1

z2











= sin, sin =

1

5

cos = ±

1

125



= ±

26

5

26

Re(z 2) 5



16. If the set R = {(a, b) ; a + 5b = 42, a, b  }

has m elements and

 

mn!

n1

1i







= x + iy, where

I =

1

, then the value of m + x + y is :

(1) 8 (2) 12

(3) 4 (4) 5

Ans. (2)

Sol. a + 5b = 42, a, b N

a = 42 – 5b, b = 1, a = 37

b = 2, a = 32

b = 3, a = 27

b = 8, a = 2

R has "8" elements  m = 8

8n!

n1

(1 i ) x iy



  



for n



4, in! = 1

 (1 – i) + (1 – i2!) + (1 – i3!)

= 1 – I + 2 + 1 + 1

= 5 – I = x + iy

m + x + y = 8 + 5 – 1 = 12

17. For the function f(x) = (cosx) – x + 1, x  ,

between the following two statements

(S1) f(x) = 0 for only one value of x is [0, ].

(S2) f(x) is decreasing in

0, 2









and increasing in

,

2











.

(1) Both (S1) and (S2) are correct

(2) Only (S1) is correct

(3) Both (S1) and (S2) are incorrect

(4) Only (S2) is correct

Ans. (2)

Sol. f(x) = cosx – x + 1

f'(x) = –sinx – 1

f is decreasing



xR

f(x) = 0

f(0) = 2, f() = –

f is strictly decreasing in [0, ] and f(0).f() < 0

only one solution of f(x) = 0

S1 is correct and S2 is incorrect.

18. The set of all , for which the vector

ˆˆˆ

a ti 6 j 3k   

and

ˆˆ ˆ

b ti 2j 2 tk   

are

inclined at an obtuse angle for all t  is :

(1) [0, 1) (2) (–2, 0]

(3)

4,0

3









(4)

4,1

3









Ans. (3)

Sol.

ˆˆˆ

a ti 6j 3k   

ˆˆ ˆ

b ti 2j 2 tk   

so

a.b 0, t R  

t2 – 12 + 6t < 0

t2 + 6t – 12 < 0,

tR

< 0, and D < 0

362 + 48 < 0

12(3 + 4) < 0

40

3

  

also for a = 0,

a.b 0

hence a

4,0

3









19. Let y = y(x) be the solution of the differential

equation (1 + y2)etanxdx + cos2x(1 + e2tanx)dy = 0,

y(0) = 1. Then y

4









is equal to :

(1)

2

e

(2)

2

1

e

(3)

1

e

(4)

2

e

Ans. (3)

Sol. (1 + y2)

tanx

e dx

+ cos2x(1 + e2tan x) dy = 0

2 tan x

2 tan x 2

sec x e dy

dx C

1 e 1 y







tan–1(etan x) + tan–1 y = C

for x = 0, y = 1, tan–1 (1) + tan–11 = C

C =

2



tan–1(etan x) + tan–1 y =

2



Put x = , tan–1 e + tan–1 y =

2



tan–1 y = cot–1 e

y =

1

e

20. Let H :

22

xy

1

ab



be the hyperbola, whose

eccentricity is

3

and the length of the latus

rectum is

43

. Suppose the point (, 6),  > 0

lies on H. If  is the product of the focal distances

of the point (, 6), then 2 +  is equal to :

(1) 170 (2) 171

(3) 169 (4) 172

Ans. (2)

Sol. H :

22

yx1

ba



,

e3

2

a

e 1 3

b

  

2

a2

b



a2 = 2b2

length of L.R. =

2

2a 43

b

a =

6

P(, 6) lie on

22

yx1

36



2

12 1

6





266  

Foci = (0, ±be) = (0, 3) & (0, –3)

Let d1 & d2 be focal distances of P(, 6)

22

1

d (6 be)   

,

22

2

d (6 be)   

1

d 66 81

,

2

d 66 9

= d1 d2 =

147 75

= 105

2+ = 66 + 105 = 171

SECTION-B

21. Let A =

21

11









. If the sum of the diagonal

elements of A13 is 3n, then n is equal to ______.

Ans. (7)

Sol. A =

21

11









A2 =

2 1 2 1 3 3

1 1 1 1 3 0

  

     



     

     

A3 =

3 3 2 1 3 6

3 0 1 1 6 3

  

     



     



     

A4 =

3 6 2 1 0 9

6 3 1 1 9 9

  

     



     



     

A5 =

0 9 2 1 9 9

9 9 1 1 9 18

   

     



     



     

A6 =

9 9 2 1 27 0

9 18 1 1 0 27

   

     



     



     

A7 =

62

26

27 0 54 27 3 2 –27

0 27 27 27 27 3



   

    

   

  

    

37 = 3n  n = 7

22. If the orthocentre of the triangle formed by the

lines 2x + 3y – 1 = 0, x + 2y – 1 = 0 and

ax + by – 1 = 0, is the centroid of another triangle,

whose circumecentre and orthocentre respectively

are (3, 4) and (–6, –8), then the value of |a – b| is

_____.

Ans. (16)

Sol. 2x + 3y – 1 = 0

x + 2y – 1 = 0

ax + by – 1 = 0

2 1

0(3, 4)

(–6, –8)

H

G

(6, 6)

6 6 8 8

,

33









= (0, 0)

(0,0)

2x+3y–1=0

x+2y–1=0

m =

(–1, 1)

ax + by – 1 = 0

1 0 a 1

1 0 b



   

  



  

–a = b

  ax – ay – 1 = 0

ax – a

2x

11

3









2a a

xa 33









a3

x5a





a3

2 3y 1 0

5a





  





2a 6

13a 6

5a

y3 3 5a









y =

a2

5a



a2

5a 2

a3

5a



















a – 2 = 2a + 6

a = –8

b = 8

–8x + 8y – 1 = 0

|a – b| = 16

23. Three balls are drawn at random from a bag

containing 5 blue and 4 yellow balls. Let the

random variables X and Y respectively denote the

number of blue and Yellow balls. If

X

and

Y

are

the means of X and Y respectively, then 7

X

+ 4

Y

is equal to ______.

Ans. (17)

Sol.

5 4 5 4 5 4 5 4

0 1 1 2 2 1 3 0

99

33

Blue balls 0 1 2 3 4 5

Pr ob. 0 0

...

C . C C C C C C C

CC

5 4 5 4 5 4

1 2 2 1 3 0

9

3

.

C C C C 2 C C 3

7x 7

C

   



30 80 30 7

84



=

140 70 35

12 6 3



5 4 5 4 5 4

2 1 1 2 0 3

yellow 0 1 2 3 4

0

C C C C C C

40 60 12 112 16

4y 4

84 21 3



   

24. The number of 3-digit numbers, formed using the

digits 2, 3, 4, 5 and 7, when the repetition of digits

is not allowed, and which are not divisible by 3, is

equal to ______.

Ans. (36)

Sol. 2, 3, 4, 5, 7

total number of three digit numbers not divisible

by 3 will be formed by using the digits

(4, 5, 7)

(3, 4, 7)

(2, 5, 7)

(2, 4, 7)

(2, 4, 5)

(2, 3, 5)

number of ways = 6 × 3! = 36

25. Let the positive integers be written in the form :

7

8

9

10

4

2

3

1

5

6

.

If the kth row contains exactly k numbers for every

natural number k, then the row in which the

number 5310 will be, is ______.

Ans. (103)

Sol. S = 1 + 2 + 4 + 7 + ….. + Tn

S = 1 + 2 + 4 + ……

Tn = 1 + 1 + 2 + 3 + …… + (Tn – Tn–1)

Tn =

n1

1 [2 (n 2) 1]

2





   





Tn = 1 +

n(n 1)

12





n = 100 Tn = 1 +

100 99

2



= 4950 + 1

n = 101 Tn = 1 +

101 100

2



= 5050 + 1 = 5051

n = 102 Tn = 1 +

102 101

2



= 5151 + 1 = 5152

n = 103 Tn = 1 +

103 102

2



= 5254

n = 104 Tn = 1 +

104 103

2



= 5357

26. If the range of f() =

42

sin 3cos ,

sin cos

   

  

is

[], then the sum of the infinite G.P., whose first

term is 64 and the common ratio is





, is equal to

______.

Ans. (96)

Sol.

42

sin 3cos

f( ) sin cos

  

   

2

42

2cos

f( ) 1 sin cos



     

2

42

2cos

f( ) 1

cos cos 1



  

   

22

2

f( ) 1

cos sec 1

  

   

min.

f( ) 1

f()max. = 3

64

S 96

1 1 / 3





27. Let

n2n

r

r0

(4r 2r 1) C



   



and

n

r

r0

C1

r 1 n 1





  









. If 140 <

2



< 281,

then the value of n is ______.

Ans. (5)

Sol.

n

2n

r

r0

(4r 2r 1). C



   



n n n

2 n 1 n 1 n

r 1 r r

r 0 r 0 r 0

nn

4 r . . C 2 r. . C C

rr



  

   

  

n

n1

r1

r0

4n C







nn

n 1 n

r 1 r

r 0 r 0

2n C C









 = 4n(n – 1) . 2n–2 + 4n.2n–1 + 2n.2n-1 + 2n

 = 2n-2[4n(n-1) + 8n + 4n + 4]

 = 2n–2 [4n2 + 8n + 4]

 = 2n(n + 1)2

n

r

r0

C1

r 1 n 1



  





n1

n

r1

r0

C1

n 1 n 1











 

n 1 n 1

1 n 1

11 C .... C

n1





   



n1

2

n1





n 1 2

3

n1

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

2





   



140 < (n + 1)3 < 281

n = 4  (n + 1)3 = 125

n = 5  (n + 1)3 = 216

n = 6  (n + 1)3 = 343

n = 5

28. Let

ˆˆ ˆ

a 9i 13j 25k,  

ˆˆ ˆ

b 3i 7j 13k  

and

ˆˆ

ˆ

c 17i 2 j k  

be three given vectros. If

r

is a

vector such that

r a (b c) a   

and

r.(b c) 0

,

then

2

| 593r 67a |

(593)



is equal to _______ .

Ans. (569)

Sol.

ˆˆ ˆ

a 9i 13j 25k  

ˆˆ ˆ

b 3i 7j 13k  

ˆˆ

ˆ

c 17i 2j k  

ˆˆ ˆ

b c 20i 5j 12k   

ˆˆ ˆ

b c 14i 9j 14k    

 

r b c a 0   

 

r b c a   

r a b c   

But

r.(b c) 0



   

a b c . b c 0    



a.b b.b c.b a.c b.c c.c 0      

c.c b.b 294 227 67

389 204 593

a.b a.c

  

   







67

r b c a

593

  



593r 67a 593(b c)  



2

| b c | 569



29. Let the area of the region enclosed by the curve

y = min{sinx, cosx} and the x-axis between x = –

to x =  be A. Then A2 is equal to _______.

Ans. (16)

Sol. y = min{sinx, cosx}

x–axis x– x=



(1

(2



/4 0

/4

0sin x (cosx)







=

1

12



3 /4 3 /4

(sin x cos x) ( cosx sin x)

 



    



3 /4

(cosx sin x)





 

11

10 22



     





11

122

   

/2 /2

/4

1

cosxdx (sin x) 1 2





  



A = 4

A2 = 16

30. The value of

3 10

2

x0

1 cosx cos2x cos3x..... cos10x

lim 2 x











is

______.

Ans. (55)

Sol.

2 2 2 2

2

x0

x 4x 9x 100x

1 1 1 1 ..... 1

2! 2! 2! 2!

lim 2 x





     

    



     



     





By expansion

2 2 2 2

2

x0

x 1 4x 1 9x 1 100x

2 1 1 1 . 1 . ..... 1 .

2 2 2 3 2 10 2

lim x





      

    



      



      



2 2 2 2

2

x0

x 2x 3x 10x

1 1 1 1 ..... 1

2 2 2 2

lim 2 x





     

    



     



     





2

x0

1 2 3 10

2 1 1 x .....

2 2 2 2

lim x





     





1 2 3 10

2 .....

2 2 2 2



   





1 + 2 + …… + 10 =

10 11 55

2



PHYSICS

TEST PAPER WITH SOLUTION

SECTION-A

31. Three bodies A, B and C have equal kinetic energies

and their masses are 400 g, 1.2 kg and 1.6 kg

respectively. The ratio of their linear momenta is :

(1)

1: 3 : 2

(2)

1: 3 : 2

(3)

2 : 3 :1

(4)

3 : 2 :1

Ans. (1)

Sol. KE =

2

P

2m

Pm

Hence, PA : PB : PC

=

400 : 1200 : 1600

=

1: 3 : 2

32. Average force exerted on a non-reflecting surface

at normal incidence is 2.4 × 10–4N. If 360 W/cm2 is

the light energy flux during span of 1 hour 30

minutes. Then the area of the surface is:

(1) 0.2 m2 (2) 0.02 m2

(3) 20 m2 (4) 0.1 m2

Ans. (2)

Sol. Pressure =

IF

CA



4

48

360 2.4 10

A

10 3 10









 A = 2 × 10–2 m2 = 0.02 m2

33. A proton and an electron are associated with same

de-Broglie wavelength. The ratio of their kinetic

energies is:

(Assume h = 6.63 × 10–34 J s, me = 9.0 × 10–31 kg

and mp = 1836 times me)

(1) 1 : 1836 (2) 1 :

1

1836

(3) 1 :

1

1836

(4) 1 :

1836

Ans. (1)

Sol.  is same for both

h

P

same for both

P 2mK

Hence,

1

Km



pe

ep

KE m1

KE m 1836

  

34. A mixture of one mole of monoatomic gas and one

mole of a diatomic gas (rigid) are kept at room

temperature (27°C). The ratio of specific heat of

gases at constant volume respectively is:

(1)

7

5

(2)

3

2

(3)

3

5

(4)

5

3

Ans. (3)

Sol.

v mono

v dia

3R

(C ) 3

2

5

(C ) 5

R

2



35. In an expression a × 10b :

(1) a is order of magnitude for b  5

(2) b is order of magnitude for a  5

(3) b is order of magnitude for 5 < a  10

(4) b is order of magnitude for a  5

Ans. (2)

Sol. a × 10b

if a  5 order is b

a > 5 order is b + 1

36. In the given circuit, the terminal potential difference

of the cell is :

3V

1

4

(1) 2 V (2) 4 V

(3) 1.5 V (4) 3 V

Ans. (1)

Sol.

3V

1

4



3V

1

2

3

i 1A

12





v = E – ir

= 3 – 1 × 1 = 2V

37. Binding energy of a certain nucleus is 18 × 108 J.

How much is the difference between total mass of

all the nucleons and nuclear mass of the given

nucleus:

(1) 0.2 g (2) 20 g

(3) 2 g (4) 10 g

Ans. (2)

Sol. mc2 = 18 × 108

m × 9 × 1016 = 18 × 108

m = 2 × 10–8 kg = 20 g

38. Paramagnetic substances:

A. align themselves along the directions of external

magnetic field.

B. attract strongly towards external magnetic field.

C. has susceptibility little more than zero.

D. move from a region of strong magnetic field to

weak magnetic field.

Choose the most appropriate answer from the

options given below:

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

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

Ans. (4)

Sol. A, C only

39. A clock has 75 cm, 60 cm long second hand and

minute hand respectively. In 30 minutes duration

the tip of second hand will travel x distance more

than the tip of minute hand. The value of x in

meter is nearly (Take  = 3.14) :

(1) 139.4 (2) 140.5

(3) 220.0 (4) 118.9

Ans. (1)

Sol. xmin =  × rmin

=

60 m.

100



xsecond = 30 × 2 × rsecond

=

75

30 2 100

 

x = xsecond – xmin

= 139.4 m

40. Young's modulus is determined by the equation

given by Y = 49000

2

m dyne

cm

where M is the mass

and  is the extension of wire used in the

experiment. Now error in Young modules(Y) is

estimated by taking data from M- plot in graph

paper. The smallest scale divisions are 5 g and 0.02

cm along load axis and extension axis respectively.

If the value of M and  are 500 g and 2 cm

respectively then percentage error of Y is :

(1) 0.2 % (2) 0.02 %

(3) 2 % (4) 0.5 %

Ans. (3)

Sol.

Ym

  



5 0.02

500 2



= 0.01 + 0.01

Y0.02

Y





Y

% 2%

Y



41. Two different adiabatic paths for the same gas

intersect two isothermal curves as shown in P-V

diagram. The relation between the ratio

a

d

V

and the

ratio

b

c

V

is:

P

a

b

c

Vc

Vb

Vd

Va

d

V

(1)

1

ab

dc

VV









(2)

ab

dc

VV



(3)

ab

dc

VV



(4)

2

ab

dc

VV







Ans. (3)

Sol. For adiabatic process

TV–1 = constant

11

a a d d

T V T V

 

  

1

ad

da

VT











11

b b c c

T V T V

 

  

1

bc

cb

VT











ab

dc

VV



dc

ab

TT











42. Two planets A and B having masses m1 and m2 move

around the sun in circular orbits of r1 and r2 radii

respectively. If angular momentum of A is L and that

of B is 3L, the ratio of time period

A

B

T







is:

(1)

3

2

1

r







(2)

3

1

2

r







(3)

3

2

1

m

1

27 m







(4)

3

1

2

m

27 m







Ans. (3)

Sol.

2

1

A1

rL

T 2m



……. (1)

2

B2

r3L

T 2m



……. (2)



2

A 1 1

B 2 2

T m r

3. .

T m r







23

A1

B2

Tr

   



   

   



4

23

1A

2B

rT

   



   

   

3

2A

1B

mT

1.

27 m T

   



   

   

43. A LCR circuit is at resonance for a capacitor C,

inductance L and resistance R. Now the value of

resistance is halved keeping all other parameters

same. The current amplitude at resonance will be

now:

(1) Zero (2) double

(3) same (4) halved

Ans. (2)

Sol. In resonance Z = R

V

IR



R  halved

 I  2I

I becomes doubled.

44. The output Y of following circuit for given inputs is :

A

B

Y

(1) A•B(A + B) (2) A • B

(3) 0 (4)

AB

Ans. (3)

Sol. By truth table

A

B

Y

0

1

0

1

0

1

0

45. Two charged conducting spheres of radii a and b

are connected to each other by a conducting wire.

The ratio of charges of the two spheres

respectively is:

(1)

ab

(2) ab

(3)

a

b

(4)

b

a

Ans. (3)

Sol. Potential at surface will be same

12

Kq Kq

ab



1

2

qa

qb



46. Correct Bernoulli's equation is (symbols have their

usual meaning) :

(1) P + mgh +

1

2

mv2 = constant

(2) P + gh +

1

2

v2 = constant

(3) P + gh + v2 = constant

(4) P +

1

2

gh +

1

2

v2 = constant

Ans. (2)

Sol.

2

1

P gh V

2

   

= constant

47. A player caught a cricket ball of mass 150 g

moving at a speed of 20 m/s. If the catching

process is completed in 0.1 s, the magnitude of

force exerted by the ball on the hand of the player

is:

(1) 150 N (2) 3 N

(3) 30 N (4) 300 N

Ans. (3)

Sol.

P mv 0

Ft 0.1







3

150 10 20 30 N

0.1







48. A stationary particle breaks into two parts of

masses mA and mB which move with velocities vA

and vB respectively. The ratio of their kinetic

energies (KB : KA) is :

(1) vB : vA (2) mB : mA

(3) mB vB : mA vA (4) 1 : 1

Ans. (1)

Sol. Initial momentum is zero.

Hence

AB

PP

 mAvB = mBVB

2

AA

2

BB

1mv

(KE) v

2

1

(KE) v

mv

2



BB

AA

(KE) v



49. Critical angle of incidence for a pair of optical

media is 45°. The refractive indices of first and

second media are in the ratio:

(1)

2 :1

(2) 1 : 2

(3)

1: 2

(4) 2 : 1

Ans. (1)

Sol. sinc =

R2

d1







sin 45° =

2

1



2

1

2







1

2

1







50. The diameter of a sphere is measured using a

vernier caliper whose 9 divisions of main scale are

equal to 10 divisions of vernier scale. The shortest

division on the main scale is equal to l mm. The

main scale reading is 2 cm and second division of

vernier scale coincides with a division on main

scale. If mass of the sphere is 8.635 g, the density

of the sphere is:

(1) 2.5 g/cm3 (2) 1.7 g/cm3

(3) 2.2 g/cm3 (4) 2.0 g/cm3

Ans. (4)

Sol. Given 9MSD = 10VSD

mass = 8.635 g

LC = 1 MSD – 1 VSD

LC = 1 MSD –

9

10

MSD

LC =

1

10

MSD

LC = 0.01 cm

Reading of diameter = MSR + LC × VSR

= 2 cm + (0.01) × (2)

= 2.02 cm

Volume of sphere =

33

4 d 4 2.02

3 2 3 2

   

  

   

   

= 4.32 cm3

Density =

mass

volume

=

8.635 1.998 2.00g

4.32 

SECTION-B

51. A uniform thin metal plate of mass 10 kg with

dimensions is shown. The ratio of x and y

coordinates of center of mass of plate in

n

9

.The

value of n is_______.

(0,2)

(1,2)

(2,2)

(1,1)

(2,1)

(3,2)

(0,0)

(3,0)

Ans. (15)

Sol. m1 =  × 5 = 10 Kg

CM(x1, y1)

(0, 0)

+

CM

m1 =  × 5 = 10Kg

m2 =  × 1 = 2Kg

(0, 0)

m3 =  × 6 = 12Kg

=

  m1 x1 + m2 x2 = m3 x3

10x1 + 2(1.5) = 12(1.5)  x1 = 1.5 cm

  m1 y1 + m2 y2 = m3 y3

10y1 + 2(1.5) = 12 × 1  y1 = 0.9 cm

1

x1.5 15

y 0.9 9



n = 15

CM

52. An electron with kinetic energy 5 eV enters a

region of uniform magnetic field of 3 T

perpendicular to its direction. An electric field E is

applied perpendicular to the direction of velocity

and magnetic field. The value of E, so that electron

moves along the same path, is_______ NC–1.

(Given, mass of electron = 9 × 10–31 kg, electric

charge = 1.6 × 10–19C)

Ans. (4)

Sol. For the given condition of moving undeflected, net

force should be zero.

qE = qVB

E = VB

2 KE B

m





19

6

31

2 5 1.6 10 3 10

9 10





  

  



= 4 N/C

53. A square loop PQRS having 10 turns, area 3.6 ×

10–3 m2 and resistance 100  is slowly and

uniformly being pulled out of a uniform magnetic

field of magnitude B = 0.5 T as shown. Work done

in pulling the loop out of the field in 1.0 s is

_________× 10–6 J.

× × × ×

×

× × × ×

×

× × × ×

×

× × × ×

×

× × × ×

×

P

Q

S

R



Ans. (3)

Sol.  = NBv

NB v

iRR





2 2 2

N B v

F N(i B) R



2 2 3

NB

WF Rt



   



A = 2

2 3 2

(10 10)(0.5) (3.6 10 )

W100 1



  



W = 3.24 × 10–6 J

54. Resistance of a wire at 0 °C, 100 °C and t °C is

found to be 10 , 10.2  and 10.95  respectively.

The temperature t in Kelvin scale is______.

Ans. (748)

Sol. R = R0(1 + T)

0

RT

R

 

Case-I

0 ºC  100 ºC

10.2 10 (100 0)

10

  

….. (1)

Case-II

0 ºC  t ºC

10.95 10 (t 0)

10

  

…... (2)

t 0.95 475º C

100 0.2

  

t = 475 + 273 = 748 K

55. An electric field,

ˆ ˆ ˆ

2i 6j 8k

E6





passes through

the surface of 4 m2 area having unit vector

ˆ ˆ ˆ

2i j k

ˆ

n6









. The electric flux for that surface

is _________ V m.

Ans. (12)

Sol.

EA  

ˆ ˆ ˆ ˆ ˆ ˆ

2i 6j 8k 2i j k

4

66

   

   



   

   

4(4 6 8) 12Vm

6

    

56. A liquid column of height 0.04 cm balances excess

pressure of soap bubble of certain radius. If density

of liquid is 8 × 103 kg m–3 and surface tension of

soap solution is 0.28 Nm–1, then diameter of the

soap bubble is________cm.

(if g = 10 ms–2)

Ans. (7)

Sol.

4S

gh R



34

4 0.28

R8 10 10 4 10





   

0.28 28

m cm

88



 R = 3.5 cm

Diameter = 7 cm

57. A closed and an open organ pipe have same

lengths. If the ratio of frequencies of their seventh

overtones is

a1

a









then the value of a is _____.

Ans. (16)

Sol. For closed organ pipe

c

v 15v

f (2n 1) 44

  

For open organ pipe

o

v 8v

f (n 1) 22

  

c

o

f15 a 1

f 16 a





 a = 16

58. Three vectors

OP,OQ

and

OR

each of magnitude

A are acting as shown in figure. The resultant of

the three vectors is

Ax

. The value of x is _____.

O

90°

45°

P

Q

R

Ans. (3)

Sol.

A

AA

ˆˆ

R A i A j

22

   

   

   

   

22

AA

R A A

22

   

   

   

   

3A

59. A parallel beam of monochromatic light of

wavelength 600 nm passes through single slit of

0.4 mm width. Angular divergence corresponding

to second order minima would be _____×10–3 rad.

Ans. (6)

Sol.

2

sin b





9

4

2 600 10

4 10







= 3 × 10–3 rad

Total divergence = (3 + 3) × 10–3 = 6 × 10–3 rad

60. In an alpha particle scattering experiment distance of

closest approach for the  particle is 4.5×10–14 m. If

target nucleus has atomic number 80, then

maximum velocity of -particle is _______× 105

m/s approximately.

(

9

0

19 10

4



SI unit, mass of  particle =

6.72 × 10–27 kg)

Ans. (156)

Sol.

2

min

4KZe

mr



9

19

27 14

4 9 10 80 1.6 10

6.72 10 4.5 10





  

  

  

= 9.759 × 1025 × 1.6 × 10–19

= 156 × 105 m/s

CHEMISTRY

TEST PAPER WITH SOLUTION

SECTION-A

61. Given below are two statements:

Statement I :

Cl

NO2

O2N

Compound-A

IUPAC name of Compound A is 4-chloro-1,

3-dinitrobenzene:

Statement II:

NH2

CH3

C2H5

Compound-B

IUPAC name of Compound B is

4-ethyl-2-methylaniline.

In the light of the above statements, choose the

most appropriate answer from the options given

below:

(1) Both Statement I and Statement II are correct

(2) Statement I is incorrect but Statement II is

correct

(3) Statement I is correct but Statement II is

incorrect

(4) Both Statement I and Statement II are incorrect

Ans. (2)

Sol. Statement I :

Cl

NO2

O2N

6

1

2

3

4

5

IUPAC name

 1-chloro-2, 4-dinitrobenzene

 statement-I is incorrect

Statement-II :

CH3

NH

1

2

3

4

5

6

C2H5

  4-ethyl-2-methylaniline

 statement-II is correct

62. Which among the following compounds will

undergo fastest SN2 reaction.

(1)

Br

(2)

Br

(3)

Br

(4)

Br

Ans. (3)

Sol. 1

Br

2

Br

3

Br

4

Br

fastest SN2 reaction give

Br

1°

Rate of SN2 is Me – x > 1° – x > 2° – x > 3° – x

63. Combustion of glucose (C6H12O6) produces CO2

and water. The amount of oxygen (in g) required

for the complete combustion of 900 g of glucose is:

[Molar mass of glucose in g mol–1 = 180]

(1) 480 (2) 960

(3) 800 (4) 32

Ans. (2)

Sol. C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O()

900

180

= 5 mol 30 mol

Mass of O2 required = 30 × 32 = 960 gm

64. Identify the major products A and B respectively in

the following set of reactions.

CH3

CH3COCl

Pyridine

B

Conc. H2SO4

A

OH



(1)

CH3

A=

and B=

CH3

OCOCH3

(2)

CH3

A=

and B =

CH3

OH

CH3CO

(3)

CH2

A=

and B =

CH3

COCH3

(4)

CH2

A=

and B =

CH3

OH

COCH3

Ans. (1)

Sol.

(B)

CH3

OCOCH3

CH3COCl

pyridine

CH3

OH

(Acetylation)

Conc. H2SO4



CH3

+ H2O

E1 Reaction

(A)

65. Given below are two statements : One is labelled as

Assertion A and the other is labelled as Reason R:

Assertion A : The stability order of +1 oxidation

state of Ga, In and Tl is Ga < In < Tl.

Reason R : The inert pair effect stabilizes the

lower oxidation state down the group.

In the light of the above statements, choose the

correct answer from the options given below :

(1) Both A and R are true and R is the correct

explanation of A.

(2) A is true but R is false.

(3) Both A and R are true but R is NOT the correct

explanation of A.

(4) A is false but R is true.

Ans. (1)

Sol. The relative stability of +1 oxidation state

progressively increases for heavier elements due to

inert pair effect.

Stability of A+1 < Ga+1 < In+1 < T+1

66. Match List I with List-II

List-I

(Name of the test)

List-II

(Reaction sequence involved)

[M is metal]

A

Borax bead

test

I.

MCO3 MO

32

Co(NO )





CoO. MO

B.

Charcoal

cavity test

II.

MCO3  MCl2  M2+

C.

Cobalt nitrate

test

III

MSO4

Na2B4O7



M(BO2)2  MBO2  M

D.

Flame test

IV

MSO4

Na2CO3



MCO3 

MO  M

Choose the correct answer from the option below :

(1) A-III, B-I, C-IV, D-II

(2) A-III, B-II, C-IV, D-I

(3) A-III, B-I, C-II, D-IV

(4) A-III, B-IV, C-I, D-II

Ans. (4)

Sol. Cobalt nitrate test

MCO3 MO

32

Co(NO )





CoO. MO

Flame test

MCO3  MCl2  M2+

Borax Bead test

MSO4

Na2B4O7



M(BO2)2  MBO2  M

Charcoal cavity test

MSO4

Na2CO3



MCO3  MO  M

67. Match List I and with List II

List-I (Molecule)

List-II(Shape)

A

NH3

I.

Square pyramid

B.

BrF5

II.

Tetrahedral

C.

PCl5

III

Trigonal pyramidal

D.

CH4

IV

Trigonal bipyramidal

Choose the correct answer from the option below :

(1) A-IV, B-III, C-I, D-II

(2) A-II, B-IV, C-I, D-III

(3) A-III, B-I, C-IV, D-II

(4) A-III, B-IV, C-I, D-II

Ans. (3)

Sol.

N

H

Br

F

Square pyramidal

Trigonal pyramidal

P

Cl

Trigonal bipyramidal

C

H

Tetrahedral

Cl

68. For the given hypothetical reactions, the

equilibrium constants are as follows:

X Y ; K1 = 1.0

Y Z ; K2 = 2.0

Z W ; K3 = 4.0

The equilibrium constant for the reaction

X W is

(1) 6.0 (2) 12.0

(3) 8.0 (4) 7.0

Ans. (3)

Sol.

XY

k1 = 1

YZ

k2 = 2

Z

k3 = 4

X

k1  k2  k3

k = 1 × 2 × 4

k = 8

69. Thiosulphate reacts differently with iodine and

bromine in the reaction given below :

22

2 3 2 4 6

2S O I S O 2I¯



  

22

2 3 2 2 4

S O 5Br 5H O 2SO 4Br¯ 10H

  

    

Which of the following statement justifies the

above dual behaviour of thiosulphate?

(1) Bromine undergoes oxidation and iodine

undergoes reduction by iodine in these

reactions

(2) Thiosulphate undergoes oxidation by bromine

and reduction by iodine in these reaction

(3) Bromine is a stronger oxidant than iodine

(4) Bromine is a weaker oxidant than iodine

Ans. (3)

Sol. In the reaction of S2O32– with I2, oxidation state of

sulphur changes to +2 to +2.5

In the reaction of S2O32– with Br2, oxidation state

of sulphur changes from +2 to +6.

 Both I2 and Br2 are oxidant (oxidising agent)

and Br2 is stronger oxidant than I2.

70. An octahedral complex with the formula

CoCl3nNH3 upon reaction with excess of AgNO3

solution given 2 moles of AgCl. Consider the

oxidation state of Co in the complex is 'x'. The

value of "x + n" is _____.

(1) 3 (2) 6

(3) 8 (4) 5

Ans. (3)

Sol.

 

3

3 5 2

Co(NH ) Cl Cl



+ excess AgNO3  2AgCl

(2 moles)

x + 0 – 1 – 2 = 0

x = +3

n = 5

 x + n = 8

71.

CHO

OH

H

OH

CH2OH

H

HO

H

The incorrect statement regarding the given

structure is

(1) Can be oxidized to a dicarboxylic acid with Br2

water

(2) despite the presence of – CHO does not give

Schiff's test

(3) has 4-asymmetric carbon atom

(4) will coexist in equilibrium with 2 other cyclic

structure

Ans. (1)

Sol.

(1)

CHO

OH

H

OH

CH2OH

H

HO

H

*

Br2

H2O

COOH

OH

H

OH

CH2–OH

H

HO

H

statement 1 is incorrect

(monocarboxylic acid)

(2) correct

(3) c.c. is 4 (correct)

(4)

OH

CH2–OH

H

OH

H

HO

OH

H

CHO

H–C–OH

HO–C–H

H–C–OH

CH2OH

-D-glucose

H

CH2–OH

OH

H

OH

H

HO

OH

H

-D-glucose

(correct)

72. In the given compound, the number of 2° carbon

atom/s is _____.

CH3–C(CH3)–CH–C(CH3)–CH3

H

(1) Three (2) One

(3) Two (4) Four

Ans. (2)

Sol.

CH3 – C – CH –C – CH3

H

3°

1°

2°

CH3

1°

CH3

1°

only one 2° carbon is present in this compound.

73. Which of the following are aromatic?

A.

B.

C.

D.

(1) B and D only

(2) A and C only

(3) A and B only

(4) C and D only

Ans. (1)

Sol.

A.

Non aromatic

B.

Aromatic

C.

Non aromatic

D.

Aromatic

74. Among the following halogens

F2, Cl2, Br2 and I2

Which can undergo disproportionation reaction?

(1) Only I2

(2) Cl2, Br2 and I2

(3) F2, Cl2 and Br2

(4) F2 and Cl2

Ans. (2)

Sol. F2 do not disproportionate because fluorine

do not exist in positive oxidation state however

Cl2, Br2 & I2 undergoes disproportionation.

75. Given below are two statements:

Statement I : N(CH3)3 and P(CH3)3 can act as

ligands to form transition metal complexes.

Statement II: As N and P are from same group, the

nature of bonding of N(CH3)3 and P(CH3)3 is always

same with transition metals.

In the light of the above statements, choose the

most appropriate answer from the options given

below:

(1) Statement I is incorrect but Statement II is

correct

(2) Both Statement I and Statement II are correct

(3) Statement I is correct but Statement II is

incorrect

(4) Both Statement I and Statement II are incorrect

Ans. (3)

Sol. N(CH3)3 and P(CH3)3 both are Lewis base and acts

as ligand, However, P(CH3)3 has a -acceptor

character.

76. Match List I with List II

List-I (Elements)

List-II(Properties in

their respective groups)

A

Cl,S

I.

Elements with highest

electronegativity

B.

Ge, As

II.

Elements with largest

atomic size

C.

Fr, Ra

III

Elements which show

properties of both

metals and non metal

D.

F, O

IV

Elements with highest

negative electron gain

enthalpy

Choose the correct answer from the options given

below :

(1) A-II, B-III, C-IV, D-I

(2) A-III, B-II, C-I, D-IV

(3) A-IV, B-III, C-II, D-I

(4) A-II, B-I, C-IV, D-III

Ans. (3)

Sol. Elements with highest electronegativity F, O

Elements with largest atomic size Fr, Ra

Elements which shows properties of both metal

and non-metals i.e. metalloids Ge, As

Elements with highest negative electron gain

enthalpy Cl, S

77. Iron (III) catalyses the reaction between iodide and

persulphate ions, in which

A. Fe3+ oxidises the iodide ion

B. Fe3+ oxidises the persulphate ion

C. Fe2+ reduces the iodide ion

D. Fe2+ reduces the persulphate ion

Choose the most appropriate answer from the

options given below:

(1) B and C only (2) B only

(3) A only (4) A and D only

Ans. (4)

Sol. 2Fe3+ + 2I–  2Fe2+ + I2

2Fe2+ + S2O82–  2Fe3+ + 2SO42–

Fe+3 oxidises I– to I2 and convert itself into Fe+2.

This Fe+2 reduces S2O82– to SO42– and converts

itself into Fe+3.

78. Match List I with List II

List-I (Compound)

List-II

(Colour)

A

Fe4[Fe(CN)6]3.xH2O

I.

Violet

B.

[Fe(CN)5NOS]4–

II.

Blood Red

C.

[Fe(SCN)]2+

III.

Prussian Blue

D.

(NH4)3PO4.12MoO3

IV.

Yellow

Choose the correct answer from the options given

below :

(1) A-III, B-I, C-II, D-IV

(2) A-IV, B-I, C-II, D-III

(3) A-II, B-III, C-IV, D-I

(4) A-I, B-II, C-III, D-IV

Ans. (1)

Sol. Fe4[Fe(CN)6]3 .xH2O  Prussian Blue

[Fe(CN)5NOS]4– Violet

[Fe(SCN)]2+ Blood Red

(NH4)3PO4.12MoO3  Yellow

79. Number of complexes with even number of

electrons in t2g orbitals is -

[Fe(H2O)6]2+, [Co(H2O)6]2+, [Co(H2O)6]3+,

[Cu(H2O)6]2+, [Cr(H2O)6]2+

(1) 1 (2) 3

(3) 2 (4) 5

Ans. (2)

Sol. [Fe(H2O)6]2+

Fe+2d6

t2g

eg

Electron in t2g = 4(even)

[Co(H2O)6]2+

Co+2d7

eg

t2g

Electron in t2g = 5(odd)

[Co(H2O)6]3+

Co+3d6

eg

t2g

Electron in t2g = 6(even)

[Cu(H2O)6]2+

Cu+2d9

eg

t2g

Electron in t2g = 6(even)

[Cr(H2O)6]2+

Cr+2d4

eg

t2g

Electron in t2g = 3(odd)

80. Identify the product (P) in the following reaction:

COOH

i) Br2/Red P

ii) H2O

(P)

(1)

COOH

Br

(2)

COBr

(3)

CHO

Br

(4)

COOH

Br

Ans. (1)

Sol. HVZ Reaction

COOH

i) Br2/Red P

ii) H2O

COOH

Br

SECTION-B

81. A hypothetical electromagnetic wave is show

below.

1.5pm

The frequency of the wave is x × 1019 Hz.

x = ______ (nearest integer)

Ans. (5)

Sol.  = 1.5 × 4 pm

= 6 × 10–12 meter

 = C

6 × 10–12 × = 3 × 108

 = 5 × 1019 Hz

82.

90L

10L

B

A

Consider the figure provided.

1 mol of an ideal gas is kept in a cylinder, fitted

with a piston, at the position A, at 18°C. If the

piston is moved to position B, keeping the

temperature unchanged, then 'x' L atm work is

done in this reversible process.

x = ______ L atm. (nearest integer)

[Given : Absolute temperature = °C + 273.15,

R = 0.08206 L atm mol–1 K–1]

Ans. (55)

Sol.  = –nRT ln

2

1

V







= –1 × 08206 × 29115 ln

100

10







= –55.0128

Work done by system  55 atm lit.

83. Number of amine compounds from the following

giving solids which are soluble in NaOH upon

reaction with Hinsberg's reagent is _____.

NH2

O

NH2

,

H2NNH–C–NH2

O

NH–CH3

N

H

NH2

H2N–C–NH2

O

NH2

OCH3

NH2

N

1

NH2

Ans. (5)

Sol. Primary amine give an ionic solid upon reaction

with Hinsberg reagent which is soluble in NaOH.

NH2

N

H

NH2

NH

OCH3

NH2

84. The number of optical isomers in following

compound is : _____.

Br

CH3

Ans. (32)

Sol.

Br

CH3

*

Total chiral centre = 5

No. of optical isomers = 25 = 32.

85. The 'spin only' magnetic moment value of MO42– is

____ BM. (Where M is a metal having least

metallic radii. among Sc, Ti, V, Cr, Mn and Zn).

(Given atomic number : Sc = 21, Ti = 22, V = 23,

Cr = 24, Mn = 25 and Zn = 30)

Ans. (0)

Sol. Metal having least metallic radii among Sc, Ti, V,

Cr, Mn & Zn is Cr.

Spin only magnetic moment of CrO42–.

Here Cr+6 is in d0 configuration (diamagnetic).

86. Number of molecules from the following which

are exceptions to octet rule is _____.

CO2, NO2, H2SO4, BF3, CH4, SiF4, ClO2, PCl5,

BeF2, C2H6, CHCl3, CBr4

Ans. (6)

Sol.

N

O

exception to

octet rule

B

F

exception to

octet rule

S

O

OH

HO

O

exception to

octet rule

O=C=O

complete

octet

C

H

complete

octet

H

Si

F

complete

octet

F

exception to

octet rule

H—C—C—H

H

complete

octet

F—Be—F

exception to

octet rule

exception

to octet rule

C

Cl

H

complete

octet

Cl

C

Br

complete

octet

Br

Cl

O

P

Cl

H

87. If 279 g of aniline is reacted with one equivalent of

benzenediazonium chloride, the mximum amount

of aniline yellow formed will be _____ g. (nearest

integer)

(consider complete conversion)

Ans. (591)

Sol.

NH2

+

N2+Cl¯

ESR

NH2

N

yellow dye

m.wt.=93

given wt.=279gm

moles =

279

93

=3

moles formed =3

m.wt. = 197

amount formed

=197 × 3 = 591 gm

88. Consider the following reaction

A + B  C

The time taken for A to become 1/4th of its initial

concentration is twice the time taken to become 1/2

of the same. Also, when the change of

concentration of B is plotted against time, the

resulting graph gives a straight line with a negative

slope and a positive intercept on the concentration

axis.

The overall order of the reaction is ____.

Ans. (1)

Sol. For 1st order reaction

75% life = 2 × 50% life

So order with respect to A will be first order.

[B]

t

So order with respect to B will be zero.

Overall order of reaction = 1 + 0 = 1

89. Major product B of the following reaction has

____ -bond.

CH2CH3

KMnO4–KOH



(A)

HNO3/H2SO4

(B)

Ans. (5)

Sol. Major product B is 

CH2CH3

KMnO4–KOH



(A)

HNO3

(B)

H2SO4

C–OH

O

N=O

O

C–OK

O

Total number of  bonds in B are 5

90. A solution containing 10g of an electrolyte AB2 in

100g of water boils at 100.52°C. The degree of

ionization of the electrolyte () is _____ × 10–1.

(nearest integer)

[Given : Molar mass of AB2 = 200g mol–1. Kb

(molal boiling point elevation const. of water)

= 0.52 K kg mol–1, boiling point of water = 100°C ;

AB2 ionises as AB2  A2+ + 2B¯]

Ans. (5)

Sol. AB2  A+2 + 2B

i = 1 + (3 – 1) 

i = 1 + 2

Tb = kb im

0.52 = 0.52 (1 + 2)

10

200

100

1000

1 = (1 + 2)

10

20

2 = 1 + 2

 = 0.5

Ans.  = 5 × 10–1