FINAL JEE–MAIN EXAMINATION – APRIL, 2024

(Held On Tuesday 09th April, 2024) TIME : 3 : 00 PM to 6 : 00 PM

MATHEMATICS

TEST PAPER WITH SOLUTION

SECTION-A

1.

 

1

2x

x0

e1 2x

Lim x





is equal to :

(1) e (2)

2

e



(3) 0 (4)e–e2

Ans. (1)

Sol.

1ln(1 2x)

2x

x0

ee

Lim x







=

 

ln 1 2x 1

2x

x0

e1

Lim( e) x







=

 

2

x0

ln 2x

1 2x

Lim( e) 2x









= (–e) × (–1)

4

22

= e

2. Consider the line L passing through the points

(1, 2, 3) and (2, 3, 5). The distance of the point

11 11 19

,,

3 3 3







from the line L along the line

3x 11 3y 11 3z 19

2 1 2

  



is equal to :

(1) 3 (2) 5

(3) 4 (4) 6

Ans. (1)

Sol.

x 1 y 2 z 3

2 1 3 2 5 3

  



  



x 1 y 2 z 3

1 1 2

  



= 

B

B(1 + , 2 + , 3 + 2)

D.R. of AB = <

3 8 3 5 6 10

,,

3 3 3

     

>

B

5 8 13

,,

3 3 3







3 8 2

3 5 1

 



3 – 8 = 6 – 10

3 = 2

=

2

3

AB =

36 9 36 9 3

33

 

3. Let

 

xx

2

00

1 dt y(t)dt

y' t





, 0  x  3, y  0,

y (0) = 0. Then at x = 2, y" + y + 1 is equal to :

(1) 1 (2) 2

(3)

2

(4) 1/2

Ans. (1)

Sol.

 

2

1y

y'(x) x



1 –

22

dy y

dx









22

dy 1y

dx









2

dy dx

1y



OR

2

dy dx

1y



sin–1y = x + c, sin–1y = –x + c

x = 0, y = 0  c = 0

sin–1 y = x, as y  0

sinx = y



dy cosx

dx 

2

dy sin x

dx 

 – sinx + sinx + 1 = 1

4. Let z be a complex number such that the real part

of

z 2i





is zero. Then, the maximum value of

|z –(6+8i)| is equal to :

(1) 12 (2) 

(3) 10 (4) 8

Ans. (1)

Sol.

z 2i z 2i 0

z 2i z 2i







zz 2iz 2iz 4( 1)   

zz 2zi 2zi 4( 1) 0     

2|z|2 = 8  |z| = 2

 

maximum 10 2 12

z6 8i   



5. The area (in square units) of the region enclosed by

the ellipse x2 + 3y2 = 18 in the first quadrant below

the line y = x is :

(1)

3

34



(2)

3

(3)

3

34



(4)

31

Ans. (2)

Sol.

22

xy1

18 6



y = x

22

x 3x 1

18 18



 4x2 = 18  x2 =

9

2

32 2

3

2

18 x dx

3





=

32

21

3

2

1x 18 x 18 x

sin

322

32











=

13 3 3

99

326

2 2 2





    





Required Area

1 9 1

18 9 3

22 3

64





   





3

6. Let the foci of a hyperbola H coincide with the foci

of the ellipse E :

 

2

2y1

x1 1

100 75





and the

eccentricity of the hyperbola H be the reciprocal of

the eccentricity of the ellipse E. If the length of the

transverse axis of H is  and the length of its

conjugate axis is , then 32 + 22 is equal to :

(1) 242

(2) 225

(3) 237

(4) 205

Ans. (2)

Sol.

F12

(1, 1)

F1

e1 =

75 5 1

1100 10 2

  

e2 = 2

F1 (6, 1), F2 (–4, 1)

2ae2 = 10  a =

52a 5

2

 = 5

4 = 1 +

222

2

bb 3a

a

b =

5

32



= 5

3

32 + 22 = 3 × 25 + 2 × 25 × 3

= 225

7. Two vertices of a triangle ABC are A(3, –1) and

B (–2, 3), and its orthocentre is P(1, 1). If the

coordinates of the point C are (, ) and the centre

of the circle circumscribing the triangle PAB is

(h, k), then the value of ( + ) + 2 (h + k) equals :

(1) 51 (2) 81

(3) 5 (4) 15

Ans. (3)

Sol.

C (, )

A(3,–1)

B (–2, 3)

D

MAB =

4

5

 MDP =

5

4

Equation of PC is y – 1 =

 

5x1

4

.......(1)

MAP =

21

2



 MBC = + 1

Equation of BC is y – 3 = (x + 2) .........(2)

On solving (1) and (2)

x + 4 =

 

5x1

4

 4x + 16 = 5x – 5   = 21

   = y = x + 5 = 26

 +  = 47

Equation of  bisector of AP

y – 0 = (x – 2) ............ (3)

Equation of  bisector of AB

y – 1 =

51

x

42









............(4)

On solving (3) & (4)

(x – 3)4 = 5x –

5

2

x =

19 h

2



y =

23 k

2



2(h + k) = –42

8. If the variance of the frequency distribution is 160,

then the value of c  N is

x

c

2c

3c

4c

5c

6c

f

2

1

(1) 5 (2) 8

(3) 7 (4) 6

Ans. (3)

Sol.

x

C

2C

3C

4C

5C

6C

f

2

1

(2 2 3 4 5 6)C 22C

x77

    



Var (x) =

 

22 2 2 2 2

c2 2 3 4 5 6

7

    

–

2

22c

7







=

22

92c 484

c

7 49



=

 

22

c 60c

644 484

49 49





2

160 c

160 c 7

49



  

9. Let the range of the function

f (x) =

1

2 sin3x cos3x

, x  IR be [a, b].

If  and  are respectively the A.M. and the G.M.

of a and b, then





is equal to :

(1)

2

(2) 2

(3)



(4) 

Ans. (1)

Sol. f(x)

1

2 sin3x cos3x

11

,

2 2 2 2









a b 1 ab

2

2 ab ba





 





=

12 2 2 2

22 2 2 2















=

   

2 2 2 2

22







=

2

10. Between the following two statements :

Statement-I : Let

ˆ ˆ ˆ

a i 2j 3k  

and

ˆ ˆ ˆ

b 2i j k  

. Then the vector

r

satisfying

a r a b  

and

a.r 0

is of magnitude

10

.

Statement-II : In a triangle ABC, cos2A + cos2B

+ cos2C  –

3

2

.

(1) Both Statement-I and Statement-II are incorrect

(2) Statement-I is incorrect but Statement-II is

correct

(3) Both Statement-I and Statement-II are correct

(4) Statement-I is correct but Statement-II is

incorrect

Ans. (2)

Sol.

ˆ ˆ ˆ

a i 2j 3k  

ˆ ˆ ˆ

a 2i j k  

a r a b  

;

a.r 0



 

arb



0



 

arb

 

 

a.a a.r a.b

 

14 = – 7   = – 2

arb

2





a

rb2



=

ˆˆ

2b a 3i k

22





Statement (I) is incorrect

cos2A + cos 2B + cos 2c  –

3

2

2A + 2B + 2C = 2

cos2A + cos2B + cos 2C

= – 1 – 4 cosA·cosB·cosC

 –1 – 4 ×

111

222



= –

3

2

Statement (II) is correct.

11.

   

 

3

/2 1/3 1/3

x

2

x

x2

sin cos dt

2t t

lim x2

























is equal

to :

(1)

2

9

8



(2)

2

11

10



(3)

2

3

2



(4)

2

5

9



Ans. (1)

Sol.

 

2

x2

0 .3x

sin(2x) cos(x)

lim 2x2

















=

 

2

x2

3x

2sin xcosx cos x

lim 2x2

















=

2

x2

2sin xsin sin

xx

22

lim 3x

22

xx

22





   

   



   



   







   





   



   



=

 

2

13

1122



   



   

   

=

2

9

8



12. The sum of the coefficient of x2/3 and x–2/5 in the

binomial expansion of

9

2/3 2/5

1

xx

2











is :

(1) 21/4 (2) 69/16

(3) 63/16 (4) 19/4

Ans. (1)

Sol. Tr + 1 =

 

r

2/5

9r

92/3

rx

Cx2









=

 

2r 2r

635

r

9r1

Cr

2















for coefficient of x2/3, put

2r 2r 2

63 5 3

  

r = 5

 Coefficient of x2/3 is =

5

951

C5







For coefficient of x–2/5, put

2r 2r 2

63 5 5

   

 r = 6

Coefficient of x–2/5 is 9C6

6

1

2







Sum = 9C5

5

1

2







+ 9C6

6

1

2







=

21

4

13. Let B =

13

15







and A be a 2 × 2 matrix such that

AB–1 = A–1. If BCB–1 = A and C4 + C2 + I = O,

then 2 –  is equal to :

(1) 16 (2) 2

(3) 8 (4) 10

Ans. (4)

Sol. BCB–1 = A

 (BCB–1) (BCB–1) = A.A

 BCI CB–1 = A2

 BC2B–1 = A2

 B–1(BC2B–1)B = B–1(A.A)B

From equation (1)

C2 = A–1.A.B

C2 = B

Also AB–1 = A–1

 AB–1.A = A–1 A = I

 A–1(AB–1A) =A–1I

B–1A = A–1

Now characteristics equation of C2 is

|C2–I| = 0

|B – I| = 0



13

15



= 0

 (1 – ) (5 – l) –3 = 0  (2 – 6 +5) – 3 = 0

 2 – 6 + 2 = 0

 2 – 6B + 2I = 0

 C4 – 6C2 + 2I = 0

 = – 6

 = 2

 2 –  = 4 + 6 = 10

14. If loge y = 3 sin–1x, then (1 – x)2 y" –xy' at x =

1

2

is equal to :

(1) 9e/6 (2) 3e/6

(3) 3e/2 (4) 9e/2

Ans. (4)

Sol.

 

1

ln y 3sin x





2

11

y' 3

y1x













2

3y

y' 1x



1

at x 2





362

3e

y' 2 3e

3

2





 







 

2

1

1 x y' y 2x

2 1 x

y'' 3 1x



  















 

2

xy

1 x y'' 3 3y 1x



  







1

at x 2



,

11

3sin 3

26

2

y e e e



    

   

   

  

 

2

22

1

atx 2

1e

2

1 x y'' 3 3e 3

2













  





21

3e 3 3









 

21

atx 2

1 x y'' xy' 



222

11

3e 3 2 3e 9e

2

3





   





15. The integral

3/4

–1

1/4

1x

cos 2cot 1x













dx is equal

to:

(1) –1/2 (2) 1/4

(3) 1/2 (4) –1/4

Ans. (4)

Sol. I =

3/4 1

1/4

1x

cos 2cot dx

1x















3/4

1

1/4

1x

cos 2 dx

tan 1x















21

3/4

2

1/4 1

1x

1 tan tan 1xdx

1x

1 tan tan 1x

























=

3/4

1/4

1x

11xdx

1x

11x







  







  



=

3/4

1/4

2x dx

2





=

 

3/4

3/4 2

1/4

x

dx

x2











=

191

216 16









= –

1

4

16. Let a, ar, ar2, ........be an infinite G.P. If

n

n0

ar 57







and

3 3n

n0

ar







= 9747, then a + 18r is

equal to :

(1) 27 (2) 46

(3) 38 (4) 31

Ans. (4)

Sol.

n

n0

ar 57







a + ar + ar2 +  = 57

a57

1r





............ (I)

3 3n

n0

a r 9747







3 3 3 3 6

a a r a r ........... 9746      

3

a9746

1r



............ (II)

3

33

3

a

(I) 57

(1 r) 19

a

(II) 9717

1r



  



On solving,

23

r andr (rejected)

32



a = 19

2

a 18r 19 18 31

3

     

17. If an unbiased dice is rolled thrice, then the

probability of getting a greater number in the ith

roll than the number obtained in the (i–1)th roll,

i = 2, 3, is equal to :

(1) 3/54 (2) 2/54

(3) 5/54 (4) 1/54

Ans. (3)

Sol. Favourable cases =

63

C

Total out comes = 63

Probability of getting greater number than previous

one =

63

3

C20 5

r 216 54



18. The value of the integral

 

2

e

1

log dx

x x 1







is :

(1)

e9 4 5

5 2 log 12





 





(2)

e9 4 5

2 5 log 12





 





(3)

e7 4 5

5 2 log 12





 





(4)

e7 4 5

2 5 log 12





 





Ans. (2)

Sol.



22

e

1

I 1.log x x 1 dx



  





22

2

e2

1

x

1x1

xlog x x 1 dx

x x 1













    











2

e2

1

x

xlog x x 1 dx

x1



    





2

22

e1

xlog x x 1 x 1 

    

 

e

2log 2 5 5  

 

e

log 1 2 2    

   

2

ee

log 2 5 5 log 2 1 2     

   

2

ee

log 2 5 5 log 2 1 2     

 

2

e

25

2 5 log 21







   







e9 4 5

2 5 log 21





   







19. Let , ;  > , be the roots of the equation

x2–

2x 3 0

. Let Pn = n–n, n  N. Then

 

11 3 10 2

P10 +

 

11 2 10

P11 – 11P12 is

equal to :

(1)

9

10 2P

(2)

9

10 3P

(3)

9

11 2P

(4)

9

11 3P

Ans. (2)

Sol.

2

x 2x 3 0  





n 2 n 1 n

2 3 0



     

n 2 n 1 n

and 2 3 0



     

Subtracting

     

n 2 n 2 n 1 n 1 n n

2 3 0

   

        

n 2 n 1 n

P 2P 3P 0



   

Putn 10

12 11 10

P 2P 3P 0  

n9

11 10 9

P 2P 3P 0  

   

10 11 11 10 11

11 3.P 2P P 10 2P P   

 

99

0 10 3P 10 3P   

20. Let

ˆ ˆ ˆ

a 2i j k   

,

ˆˆ

b i k  

,

ˆˆ

c j k  

,

where  and  are integers and  = –6. Let the

values of the ordered pair (, ) for which the area

of the parallelogram of diagonals

ab

and

bc

is

21

2

, be (1, 1) and (2, 2).

Then

22

11

 

– 22 is equal to

(1) 17 (2) 24

(3) 21 (4) 19

Ans. (4)

Sol. Area of parallelogram =

12

1| d d

2

A =

1 21

(a b) (b c)

22

   

so,

ˆˆˆ

a b i j 2k    

ˆˆ

b c i j   

   

ˆˆ

ˆ

i j k

a b b c 1 2

10

    



=

ˆˆ

ˆ

i( 2 ) j(2) k( )     

22

(a b) (b c) 4 4 ( ) 21          

2 2 2

4 4 2 21        

22

5 12 17    

22

5 29   

and  = –6

and given i are integers

so,

 = –3,  = 2

or

 = 3,  = –2

   

11

, 3,2   

   

22

, 3, 2   

22

1 1 2 2 9 4 6 19         

SECTION-B

21. Consider the circle C : x2 + y2 = 4 and the parabola

P : y2 = 8x. If the set of all values of , for which

three chords of the circle C on three distinct lines

passing through the point (, 0) are bisected by the

parabola P is the interval (p, q), then (2q – p)2 is

equal to _____.

Ans. (80)

Sol.

(x1,y1)

(2t2,4t)

(,10)

T = S1

xx1 + yy1 =

22

11

xy

x1 =

22

11

xy

(2t2) = 4t4 + 16t2

 = 2t2 + 8

2

8t

2

 

Also, 4t4 + 16t2 – 4 < 0

t2 = –2 +

5

4 2 5  

   (8, 4 +

25

)

 (2q – p)2 = 80

22. Let the set of all values of p, for which

f(x) = (p2 –6p + 8) (sin22x – cos22x) + 2(2 – p)x + 7

does not have any critical point, be the interval

(a, b). Then 16ab is equal to _____ .

Ans. (252)

Sol. f(x) = – (p2 –6p + 8) cos 4n + 2(2–p)n + 7

f1(x) = +4(p2– 6p + 8) sin 4x + (4–2p)  0

sin4x 

2p – 4

4(p 4)(p 2)

sin4x 

2(p – 2)

4(p 4)(p 2)

p  2

sin4x 

1

2(p 4)



11

2(p 4) 



on solving we get



79

p,

22







Hence a =

79

,b

22



 16ab = 252

23. For a differentiable function f : IR  IR, suppose

f '(x) = 3f(x) + , where   IR, f(0) = 1 and

x

lim



f (x) = 7. Then 9f (–loge3) is equal to______.

Ans. (61)

Sol.

dy

dx

– 3y = 

If =

3dx

e



= e–3x

 y – e–3x =

3x

e dx





y e–3x =

3x

ec

3







(* e3x)

3x

y C e

3



  



on substituting x = 0, y = 1

x  –, y = 7

we get y = 7 – 6e3x

9f(–loge3) = 61

24. The number of integers, between 100 and 1000

having the sum of their digits equals to 14, is

______.

Ans. (70)

Sol. N = a b c

(i) All distinct digits

a + b + c = 14

a 1

b, c {0 to 9}

by hit & trial : 8 cases

(6, 5, 3) (8, 6, 0) (9, 4, 1)

(7, 6, 1) (8, 5, 1) (9, 3, 2)

(7, 5, 2) (8, 4, 2)

(7, 4, 3) (9, 5, 0)

(ii) 2 same, 1 diff a = b ; c

2a + c = 14

by values :

(3,8)

(4,6) Total

(5, 4) 3! 51

2!

(6,2)

(7,0)











= 14 cases

(iii) all same :

3a = 14

a =

14

3

× rejected

0 cases

Hence, Total cases :

8 × 3! + 2 × (4) + 14

= 48 + 22

= 70

25. Let A = {(x, y) : 2x + 3y = 23, x, y  N} and

B = {x : (x, y)  A}. Then the number of one-one

functions from A to B is equal to _____.

Ans. (24)

Sol. 2x + 3y = 23

x = 1 y = 7

x = 4 y = 5

x = 7 y = 3

x = 10 y = 1

A B

(1, 7) 1

(4, 5) 4

(7, 3) 7

(10, 1) 10

The number of one-one functions from A to B is

equal to 4!

26. Let A, B and C be three points on the parabola

y2 = 6x and let the line segment AB meet the line L

through C parallel to the x-axis at the point D. Let

M and N respectively be the feet of the

perpendiculars from A and B on L.

Then

2

AM BN

CD









is equal to _______.

Ans. (36)

Sol.

t1

M

C

N

t3

L

(at1

2, 2at3)

t2

B

A

D

(at1

2, 2at3)

AB AD

mm



13

2

12 1

2a(t t )

2

tt at









22

1 1 1 3 1 2 2 3

at a{t t t t t t t }     



 

1 3 2 3 1 2

a(t t t t t t )   



13

AM 2a(t t )



23

BN 2a(t t )





2

3

CD at  



2

3 1 3 2 3 1 2

CD at a(t t t t t t )   



 

2

3 1 3 2 3 1 2

a t t t t t t t   



 

3 3 1 2 3 1

a t (t t ) t (t t )   



3 2 3 1

CD a (t t )(t t )  



2

1 3 2 3

3 2 3 1

2a(t t ) ·2a(t t )

AM·BN

CD a(t t )(t t )









 







   

29

16a 16 36

4

  



27. The square of the distance of the image of the point

(6, 1, 5) in the line

x 1 y z 2

3 2 4





, from the

origin is ______.

Ans. (62)

Sol.

I

M

L

A(6,1,5)

Let M(3 + 1, 2, 4+ 2)

AM ·b 0

 9 – 15 + 4 – 2 + 16 – 12 = 0

 29 = 29

  = 1

M (4, 2, 6), I = (2, 3, 7)

Required Distance =

4 9 49

=

62

Ans. 62

28. If

1 1 1

....

1 2 1012



  



     



–

1 1 1 1

....

2 1 4 3 6 5 2024 2023



   



   



=

1

2024

, then  is equal to-

Ans. (1011)

Sol.

1 1 1

...

1 2 2012



  



     



1 1 1 1 1 1 1

...

1 2 3 4 2023 2024 2024



     

       



     

     





1 1 1

...

1 2 2012



  



     



1 1 1 1 1

...

1 2 3 4 2023



     







1 1 1 1

2 ...

2024 2 4 2022





     







1

2024





1 1 1

...

1 2 2012



  



     





1 1 1

...

1 2 2023



   







1 1 1 1

...

2024 1 2 1011



    







1

2024





 

1 1 1

...

1 2 2012

  

     

1 1 1

...

1012 1013 2023

   

  = 1011

29. Let the inverse trigonometric functions take

principal values. The number of real solutions of

the equation 2 sin–1 x + 3 cos–1 x =

2

5



, is ______.

Ans. (0)

Sol.

11

2

2sin x 3cos x 5









1

cos 2

x5



 





1

cos 3

x5





Not possible

Ans. 0

30. Consider the matrices : A =

25

3m









, B =

20

m







and X =

x

y







. Let the set of all m, for which the

system of equations AX = B has a negative solution

(i.e., x < 0 and y < 0), be the interval (a, b).

Then

b

a

8 dm

A



is equal to ________.

Ans. (450)

Sol.

25

A3m









,

20

Bm







x

Xy







2x – 5y = 20 …(1)

3x + my = m …(2)



2m 60

y2m 15





y < 0 

15

m ,30

2









25m

x2m 15



x < 0 

15

m ,0

2











15

m ,0

2









|A| = 2m + 15

Now,

 

00

2

15

15 2

2

8 (2m 15)dm 8 m 15m 



  





225 225

842















225

8 450

4

  

PHYSICS

TEST PAPER WITH SOLUTION

SECTION-A

31. A nucleus at rest disintegrates into two smaller

nuclei with their masses in the ratio of 2:1. After

disintegration they will move :-

(1) In opposite directions with speed in the ratio of

1:2 respectively

(2) In opposite directions with speed in the ratio of

2:1 respectively

(3) In the same direction with same speed.

(4) In opposite directions with the same speed.

Ans. (1)

Sol. By conservation of momentum

pi = pf

O = m1 u1 + m2 u2

1

2

u1

u2







as

1

2

m2

m1



move in opposite direction with speed ratio 1 : 2

32. The following figure represents two biconvex

lenses L1 and L2 having focal length 10 cm and

15 cm respectively. The distance between L1 & L2

is :

L1

L

(1) 10 cm (2) 15 cm

(3) 25 cm (4) 35 cm

Ans. (3)

Sol.

f1

f2

D = f1 + f2 = 25 cm

Paraxial parallel rays pass through focus and ray

from focus of convex lens will become parallel

33. The temperature of a gas is –78° C and the average

translational kinetic energy of its molecules is K.

The temperature at which the average translational

kinetic energy of the molecules of the same gas

becomes 2K is :

(1) –39°C (2) 117°C

(3) 127°C (4) –78°C

Ans. (2)

Sol. K.E =

1

nf RT

2

Ti = –78°C –78°CK

 T

To double the K.E energy temp also

become double

Tf = 390 K

Tf = 117°C

34. A hydrogen atom in ground state is given an

energy of 10.2 eV. How many spectral lines will

be emitted due to transition of electrons ?

(1) 6 (2) 3

(3) 10 (4) 1

Ans. (4)

Sol. Hydrogen will be in first excited state therefore it

will emit one spectral line corresponding to

transition b/w energy level 2 to 1

35. The magnetic field in a plane electromagnetic

wave is By = (3.5 × 10–7) sin (1.5 × 103x + 0.5

×1011t)T. The corresponding electric field will be

(1) Ey = 1.17 sin (1.5 × 103x + 0.5 × 1011t)Vm–1

(2) Ez = 105 sin (1.5 × 103x + 0.5 × 1011t)Vm–1

(3) Ez = 1.17 sin (1.5 × 103x + 0.5 × 1011t)Vm–1

(4) Ey = 10.5 sin (1.5 × 103x + 0.5 × 1011t)Vm–1

Ans. (2)

Sol. E0 = B0C

E0 = 3 × 108

× (3.5 × 10–7

) sin (1.5 × 103

x + 0.5 ×1011t)

E0 = 105 sin (1.5 × 103x + 0.5 × 1011t)Vm–1

Data inconsistent while calculating speed of wave.

You can challenge for data.

36. A square loop of side 15 cm being moved towards

right at a constant speed of 2 cm/s as shown in

figure. The front edge enters the 50 cm wide

magnetic field at t = 0. The value of induced emf

in the loop at t = 10 s will be :

2cm/s

15cm

×

B=1.0T

50cm

(1) 0.3 mV (2) 4.5 mV

(3) zero (4) 3 mV

Ans. (3)

Sol. At t = 10 sec complete loop is in magnetic field

therefore no change in flux

Bv

d

e0

dt





e = 0 for complete loop

37. Two cars are travelling towards each other at speed

of 20 m s–1 each. When the cars are 300 m apart,

both the drivers apply brakes and the cars retard at

the rate of 2 m s–2 . The distance between them

when they come to rest is :

(1) 200 m (2) 50 m

(3) 100 m (4) 25 m

Ans. (3)

Sol.

A

20 m/s

B

300 m

BA

u 40 m s

BA

a 4 m s

Apply (v2 = u2 + 2as)relative

O = (40)2 + 2(–4)(S)

S = 200 m

Remaining distance = 300 – 200 = 100 m

38. The I-V characteristics of an electronic device

shown in the figure. The device is :

I

V(volt)

5(A)

(1) a solar cell

(2) a transistor which can be used as an amplifier

(3) a zener diode which can be used as voltage regulator

(4) a diode which can be used as a rectifier

Ans. (3)

Sol. Theory

Zener diode used as voltage regulator

39. The excess pressure inside a soap bubble is thrice

the excess pressure inside a second soap bubble.

The ratio between the volume of the first and the

second bubble is :

(1) 1 : 9 (2) 1 : 3

(3) 1 : 81 (4) 1 : 27

Ans. (4)

Sol.

P1

r1

P2

r2

10

1

4T

PP r



20

2

4T

PP r



1 0 2 0

P P 3(P P )  

12

4T 4T

3

rr



r2 = 3r1

3

1

3

22

4r

V1

3

4

V 27

r

3







40. The de-Broglie wavelength associated with a

particle of mass m and energy E is

h / 2mE

.

The dimensional formula for Planck's constant is :

(1) [ML–1T–2] (2) [ML2T–1]

(3) [MLT–2] (4) [M2L2T–2]

Ans. (2)

Sol.

h

2mE



or E = h

2 2 1

[ML T ] h[T ]





21

h [ML T ]





41. A satellite of 103 kg mass is revolving in circular

orbit of radius 2R. If

4

10 R

6J

energy is supplied to

the satellite, it would revolve in a new circular

orbit of radius :

(use g = 10m/s2, R = radius of earth)

(1) 2.5 R (2) 3 R

(3) 4 R (4) 6 R

Ans. (4)

Sol.

M

R

2R

m

Total energy =

GMm

2(2R)



if energy =

4

10 R

6

is added then

4

GMm 10 R GMm

4R 6 2r





where r is new radius of revolving and

2

GM

gR



42

mgR 10 R mgR

4 6 2r

   

(m = 103 kg)

3 4 3 2

10 10 R 10 R 10 10 R

4 6 2r

   

   

1 1 R

4 6 2r

   

r = 6R

42. The effective resistance between A and B, if

resistance of each resistor is R, will be

R

A

B

R

(1)

2R

3

(2)

8R

3

(3)

5R

3

(4)

4R

3

Ans. (2)

Sol. From symmetry we can remove two middle

resistance.

New circuit is

R

A

B

R



R

A

B

2R



R

A

B

2R

3



A

B

8R

3

43. Five charges +q, +5q, –2q, +3q and –4q are

situated as shown in the figure. The electric flux

due to this configuration through the surface S is :

–4q

+5q

+3q

q

–2q

S

(1)

0

5q



(2)

0

4q



(3)

0

3q



(4)

0

q



Ans. (2)

Sol. As per gauss theorem,

 =

 

in

00

q 2q 5q

q  





0

4q



44. A proton and a deutron (q= +e, m = 2.0u) having

same kinetic energies enter a region of uniform

magnetic field

B

, moving perpendicular to

B

. The

ratio of the radius rd of deutron path to the radius rp

of the proton path is :

(1) 1 : 1 (2)

1: 2

(3)

2 :1

(4)

1: 2

Ans. (3)

Sol. In uniform magnetic field,

 

2m K.E

m

R qB qB





Since same K.E

m

Rq





p

deutron d

proton p d

q

Rm

R m q







21



 dp=

2 :1



45. UV light of 4.13 eV is incident on a photosensitive

metal surface having work function 3.13 eV. The

maximum kinetic energy of ejected photoelectrons

will be :

(1) 4.13 eV (2) 1 eV

(3) 3.13 eV (4) 7.26 eV

Ans. (2)

Sol. Ephoton = (work function) + K.Emax

K.Emax

K.Emax = 1 eV

46. The energy released in the fusion of 2 kg of

hydrogen deep in the sun is EH and the energy

released in the fission of 2 kg of 235U is EU. The

ratio

H

U

E

is approximately :

(Consider the fusion reaction as

1

14

2

4 H 2e He 2v 6 26.7 MeV,



     

energy

released in the fission reaction of 235U is 200 MeV

per fission nucleus and NA = 6.023×1023)

(1) 9.13 (2) 15.04

(3) 7.62 (4) 25.6

Ans. (3)

Sol. In each fusion reaction, 4

1

1H

nucleus are used.

Energy released per Nuclei of

1

1H

=

26.7 MeV

4

 Energy released by 2 kg hydrogen (EH)

=

A

2000 26.7

N MeV

14



&

  Energy released by 2 kg Vranium (EV)

=

A

2000 N 200MeV

235 

So,

H

V

E26.7

235 7.84

E 4 200

  



 Approximately close to 7.62

47. A real gas within a closed chamber at 27°C

undergoes the cyclic process as shown in figure.

The gas obeys PV3 = RT equation for the path A to

B . The net work done in the complete cycle is

(assuming R = 8J/molK):

P(N/m2)

20

A

C

10

B

2

4

V(m3)

(1) 225 J (2) 205 J

(3) 20 J (4) –20 J

Ans. (2)

Sol. WAB =

PdV



(Assuming T to be constant)

=

3

RTdV

V



=

4

3

2

RT V dV





= 8 × 300 ×

22

1 1 1

242















= 225 J

WBC =

2

4

P dV 10(2 – 4) –20J



WCA = 0

Wcycle = 205 J

Note : Data is inconsistent in process AB.

So needs to be challenged.

48. A 1 kg mass is suspended from the ceiling by a

rope of length 4m. A horizontal force 'F' is applied

at the mid point of the rope so that the rope makes

an angle of 45° with respect to the vertical axis as

shown in figure. The magnitude of F is :

F

1 Kg

T2

T1

(1)

0N

2



(2) 1 N

(3)

1N

10 2

(4) 10 N

Ans. (4)

Sol. T1sin45° = F

T1 cos45° = T2 = 1×g

 tan45° =

F

g

F = 10N

49. A spherical ball of radius 1×10–4 m and density 105

kg/m3 falls freely under gravity through a distance

h before entering a tank of water, If after entering

in water the velocity of the ball does not change,

then the value of h is approximately :

(The coefficient of viscosity of water is 9.8 × 10–6

N s/m2)

(1) 2296 m (2) 2249 m

(3) 2518 m (4) 2396 m

Ans. (3)

Sol.

 

2

BL

T

R

2g

V9

 





 

2

4

53

T6

10 10

2

V 10 10

99.8 10





  





VT = 224.5

when ball fall from height (h)

V 2gh

2

V

h 2518m

2g









50.

E

A

B

A B E

0 0 0

0 1 X

1 0 Y

1 1 0

In the truth table of the above circuit the value of X

and Y are :

(1) 1, 1 (2) 1, 0

(3) 0, 1 (4) 0, 0

Ans. (1)

Sol. For x

E

0=A

1=B

0

1

0

1

x = 1

For y

E

1=A

0=B

0

1

0

1

SECTION-B

51. A straight magnetic strip has a magnetic moment

of 44 Am2. If the strip is bent in a semicircular

shape, its magnetic moment will be ………. Am2.

(Given  =

22

7

)

Ans. (28)

Sol. Magnetic moment of straight wire = mx = 44





R

=R

Magnetic moment of arc

= m × 2 r

= m ×

2



=

44 2 88 28





52. A particle of mass 0.50 kg executes simple

harmonic motion under force F = –50(Nm–1)x. The

time period of oscillation is

xs

35

. The value of x is

……………..

(Given  =

22

7

)

Ans. (22)

Sol. m = 0.5 kg

F = – 50 (x)

ma = (–50x)

0.5 a = – 50x

a = (–100x)

W2 = 100 w = 10

T =

2 22 22

10 5 7 15 35



   

  

   



   

22

35 35



x 22

53. A capacitor of reactance

43

and a resistor of

resistance 4 are connected in series with an ac

source of peak value

8 2V.

The power dissipation

in the circuit is ……………..W.

Ans. (4)

Sol.

~

R = 4

Z =

22

R X L

Z =

 

2

4 4 3

= 8

Vrms =

 

V 8 2 8V

22



Irms =

rms

V81A

z8



Power dissipated = I2

rms × R = 1 × 4 = (4W)

54. An electric field

1

ˆ

(2xi)NC



exists in space. A

cube of side 2m is placed in the space as per figure

given below. The electric flux through the cube is

……………… Nm2/C.

X

2m

Y

Z

0

2m

Ans. (16)

Sol.

(0,0)

x=2

x=4

x

y

ˆ

E 2xi

E.A

2

in – 4 4 16 Nm / c    

2

out 8 4 32Nm / c   

dnet = in + out = –16 + 32 = 16 Nm2 /c

55. A circular disc reaches from top to bottom of an

inclined plane of length l. When it slips down the

plane, if takes t s. When it rolls down the plane

then it takes

1/2

2









t s, where  is ………..

Ans. (3)

Sol. For slipping

a = gsin

 =

1

2

at2

2

tgsin

 

For rolling

'2

2

g sin

ak

1R







R

k2









a' =

2gsin

3



 

2

1a ' t '

2





62

t' 2gsin 2 gsin











3



56. To determine the resistance (R) of a wire, a circuit

is designed below, The V-I characteristic curve for

this circuit is plotted for the voltmeter and the

ammeter readings as shown in figure. The value of

R is ………… .

R

V

10 k

E

mA

I(mA)

2

3

4

0

4

6

8

V(volt)

Ans. (2500)

Sol.

4

10 R

Req 10 R



E = 4V, I = 2mA

 

4

3

4

4 10 R

E

I 2 10

Req 10 R



   

20R = 40000 + 4R

16R = 40000

R = 2500

57. The resultant of two vectors

A

and

B

is

perpendicular to

A

and its magnitude is half that

of

B

. The angle between vectors

A

and

B

is

………….

Ans. ( 150 )

Sol.





Bcos =

B

2

 = 60°

So, angle between

A

&

B

is 90° + 60° = 150°

58. Monochromatic light of wavelength 500 nm is

used in Young's double slit experiment. An

interference pattern is obtained on a screen When

one of the slits is covered with a very thin glass

plate (refractive index = 1.5), the central maximum

is shifted to a position previously occupied by the

4th bright fringe. The thickness of the glass-plate is

……………….µm.

Ans. (4)

Sol. (– 1) t = n

(1.5 – 1) t = 4 × 500 × 10–9 m

t = 4000 × 10–9 m

t = 4m

59. A force (3x2 + 2x – 5) N displaces a body from

x = 2 m to x = 4m. Work done by this force is

…………J.

Ans. (58)

Sol.

2

1

x

W Fdx

 

4

2

W 3x 2x – 5 dx



4

32

2

W x x 5x



  



 

W 60 2 J 58J  

60. At room temperature (27ºC), the resistance of a

heating element is 50. The temperature

coefficient of the material is 2.4 × 10–4 ºC–1. The

temperature of the element, when its resistance is

62 , is …………. ºC.

Ans. (1027)

Sol. R = R0(1 + T)

62 = 50 [1 + 2.4 × 10–4T]

T = 1000°C

C

T 1027 C



CHEMISTRY

TEST PAPER WITH SOLUTION

SECTION-A

61. The candela is the luminous intensity, in a given

direction, of a source that emits monochromatic

radiation of frequency 'A' × 1012 hertz and that has

a radiant intensity in that direction of

1

'B'

watt per

steradian. 'A' and 'B' are respectively

(1) 540 and

1

683

(2) 540 and 683

(3) 450 and

1

683

(4) 450 and 683

Ans. (2)

Sol. The candela is the luminous intensity of a source

that emits monochromatic radiation of frequency

radiation of frequency 540 × 1012 Hz and has a

radiant intensity in that direction of

1

683

w/sr. It is

unit of Candela.

62. The correct stability order of the following

resonance structures of CH3 –CH = CH–CHO is

CH3–CH–CH=C–H CH3–CH–CH=C–H



O



I



O



O

II

III

 CH3–CH=CH–C–H

(1) II > III > I

(2) III > II > I

(3) I > II > III

(4) II > I > III

Ans. (2)

Sol.

CH3–CH–CH=CH (II)



O



O



CH3–CH=CH–CH (III)

Non Polar R.S.

More No of covalent bond

CH3–CH–CH=CH (I)



O





Having –ve charge on more

electronegative atom

Having –ve charge on less

electronegative atom



Stability order III > II > I

63. Total number of stereo isomers possible for the

given structure:

Br

(1) 8 (2) 2

(3) 4 (4) 3

Ans. (1)

Sol.

Br

There are three stereo center

So No of stereoisomer = 23 = 8

64. The correct increasing order for bond angles

among BF3, PF3 and CF3 is :

(1) PF3 < BF3 < CF3 (2) BF3 < PF3 < CF3

(3) CF3 < PF3 < BF3 (4) BF3 = PF3 < CF3

Ans. (3)

Sol.

B

120º

F

Cl

87.5º

F

P

F

97º

Order of bond angle is

ClF3 < PF3 < BF3

65. Match List I with List II

LIST-I

(Test)

LIST-II

(Observation)

A.

Br2 water test

I.

Yellow orange or

orange red

precipitate

formed

B.

Ceric

ammonium

nitrate test

II.

Reddish orange

colour

disappears

C.

Ferric chloride

test

III.

Red colour

appears

D.

2, 4-DNP test

IV.

Blue, Green,

Violet or Red

colour appear

Choose the correct answer from the options given

below:

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

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

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

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

Ans. (2)

Sol. (A) Br2 water test is test of unsaturation in which

reddish orange colour of bromine water

disappears.

(B) Alcohols given Red colour with ceric

ammonium nitrate.

(C) Phenol gives Violet colour with natural ferric

chloride.

(D) Aldehyde & Ketone give Yellow/Orange/Red

Colour compounds with 2, 4-DNP i.e., 2,

4-Dinitrophenyl hydrazine.

66. Match List I with List II

LIST-I

(Cell)

LIST-II

(Use/Property/Reaction)

A.

Leclanche

cell

I.

Converts energy

of combustion into

electrical energy

B.

Ni-Cd cell

II.

Does not involve

any ion in solution

and is used in

hearing aids

C.

Fuel cell

III.

Rechargeable

D.

Mercury

cell

IV.

Reaction at anode

Zn  Zn2+ + 2e–

Choose the correct answer from the options given

below:

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

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

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

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

Ans. (3)

Sol. A-IV, B-III, C-I, D-II

67. Match List I with List II

LIST-I

LIST-II

A.

K2[Ni(CN)4]

I.

sp3

B.

[Ni(CO)4]

II.

sp3d2

C.

[Co(NH3)6]Cl3

III.

dsp2

D.

Na3[CoF6]

IV.

d2sp3

Choose the correct answer from the options given

below:

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

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

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

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

Ans. (4)

Sol. (A)

K2 [Ni(CN)4]

+2



Ni2+ [Ar]3d8 4sº ,

Pre hybridization state of Ni+2

4s

dsp2

3d

4p

(B) [Ni(CO)4]

Ni  [Ar] 3d8 4s2

CO isS.F.L , so pairing occur

Pre hybridization state of Ni

4s

sp3

3d

4p

(C)

 

3 6 3

Co(NH ) Cl

3 6 0

Co :[Ar]3d 4s



With Co3+, NH3 act as S.F.L

d2sp3

4s

4p

3d

(d) Na3 [CoF6]

Co3+ : [Ar] 3d6 (

F

W.F.L)

sp3d2

3d

4s

4p

4d

68. The coordination environment of Ca2+ ion in its

complex with EDTA4– is :

(1) tetrahedral

(2) octahedral

(3) square planar

(4) trigonal prismatic

Ans. (2)

Sol. EDTA4– Hexadentate ligand

[Ca(EDTA)]2–

So Coordination environment is octahedral

69. The incorrect statement about Glucose is :

(1) Glucose is soluble in water because of having

aldehyde functional group

(2) Glucose remains in multiple isomeric form in

its aqueous solution

(3) Glucose is an aldohexose

(4) Glucose is one of the monomer unit in sucrose

Ans. (1)

Sol. Glucose is soluble in water due to presence of

alcohol functional group and extensive hydrogen

bonding.

Glucose exist is open chain as well as cyclic forms

in its aqueous solution.

Glucose having 6C atoms so it is hexose and

having aldehyde functional group so it is aldose.

Thus, aldohexose.

Glucose is monomer unit in sucrose with fructose.

70.

OCH3

Br

KCN(alc)





Major Product 'P'

In the above reaction product 'P' is

(1)

OCH3

CN

(2)

OCH3

CN

(3)

OCH3

CN

(4)

OCH3

CN

Ans. (1)

Sol.

Br

OCH3

+

CN

OCH3

CN–

Due to NGP effect of phenyl ring Nucleophilic

substitution of Br will occurs.

(CN–is S.F.L)

71. Which of the following compound can give

positive iodoform test when treated with aqueous

KOH solution followed by potassium hypoiodite.

(1)

O

CH3CH2–C–CH2CH3

(2)

Cl

CH3CH2–C–CH3

Cl

(3) CH3CH2CH2CHO

(4)

O

CH3CH2–CH–CH2

Ans. (2)

Sol.

CH3 – CH2 – C – CH3

Cl

aq. KOH

CH3 – CH2 – C – CH3

OH

–H2O

CH3 – CH2 – C – CH3

O

KOI

CH3 – CH2 – COOK + CHI3↓

Yellow ppt

72. For a sparingly soluble salt AB2, the equilibrium

concentrations of A2+ ions and B– ions are

1.2 × 10–4 M and 0.24 × 10–3 M, respectively. The

solubility product of AB2 is :

(1) 0.069 × 10–12

(2) 6.91 × 10–12

(3) 0.276 × 10–12

(4) 27.65 × 10–12

Ans. (2)

Sol.

2

2(s) (aq) (aq)

AB A 2B





22

sp

K [A ][B ]





4 4 2

1.2 10 (2.4 10 )



   

12 3

6.91 10 M





73. Major product of the following reaction is

CO2CH3

CN

3

(i) CH MgBr(excess)

(ii) H O



(1)

CN

CH3

HO

CH3

(2)

C

CH3

HO

CH3

O

(3)

C

CO2CH3

CH3

O

(4)

CN

C

CH3

O

Ans. (2)

Sol.

C  N

C

OCH3

O

CH3MgBr (excess)

C

CH3

NMgBr

– +

BrMgO

–

CH3

H3O+

O = C – CH3

HO – C – CH3

CH3

74. Given below are two statements :

Statement I : The higher oxidation states are more

stable down the group among transition elements

unlike p-block elements.

Statement II : Copper can not liberate hydrogen

from weak acids.

In the light of the above statements, choose the

correct answer from the options given below :

(1) Both Statement I and Statement II are false

(2) Statement I is false but Statement II is true

(3) Both Statement I and Statement II are true

(4) Statement I is true but Statement II is false

Ans. (3)

Sol. On moving down the group in transition elements,

stability of higher oxidation state increases, due to

increase in effective nuclear charge.



2

o

Cu /Cu

E

0.34 V



2

o

H /H

E

= 0

SRP : Cu2+ > H+

Cu can't liberate hydrogen gas from weak acid.

75. The incorrect statement regarding ethyne is

(1) The C–C bonds in ethyne is shorter than that

in ethene

(2) Both carbons are sp hybridised

(3) Ethyne is linear

(4) The carbon-carbon bonds in ethyne is weaker

than that in ethene

Ans. (4)

Sol. The carbon-carbon bonds in ethyne is stronger than

that in ethene.

(H–CC–H) Ethyne is linear and carbon atoms are

SP hybridised.

76. Match List I with List II

List-I

(Element)

List-II

(Electronic Configuration)

A.

N

I.

[Ar] 3d104s2 4p5

B.

S

II.

[Ne] 3s2 3p4

C.

Br

III.

[He] 2s2 2p3

D

Kr

IV.

[Ar] 3d10 4s2 4p6

Choose the correct answer from the options given

below :

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

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

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

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

Ans. (2)

Sol. (A)

23

7N :[He]2s 2p

(B)

24

16 S :[Ne]2s 3p

(C)

10 2 5

35 Br :[Ar]3d 4s 4p

(D)

10 2 6

36 Kr :[Ar]3d 4s 4p

77. Match List I with List II

List-I

List-II

A.

Melting

point [K]

I.

Tl > In > Ga > Al > B

B.

Ionic

Radius

[M+3/pm]

II.

B > Tl > Al  Ga > In

C.

iH1

[kJ mol–1]

III.

Tl > In > Al > Ga > B

D

Atomic

Radius

[pm]

IV.

B > Al > Tl > In > Ga

Choose the correct answer from the options given

below :

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

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

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

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

Ans. (3)

Sol. Melting point B > A > T> In > Ga

Ionic radius (M+3/pm) T> In > Ga > A > B

IE 1

kJ

()

ml

Ho









B > T > A Ga > In

Atomic radius (in pm)  T> In > A> Ga > B

78. Which of the following compounds will give silver

mirror with ammoniacal silver nitrate?

(A) Formic acid

(B) Formaldehyde

(C) Benzaldehyde

(D) Acetone

Choose the correct answer from the options given

below :

(1) C and D only

(2) A, B and C only

(3) A only

(4) B and C only

Ans. (2)

Sol. Apart from aldehyde, Formic acid

O

H—C—OH

also gives silver mirror test with ammonical silver

nitrate.

79. Which out of the following is a correct equation to

show change in molar conductivity with respect to

concentration for a weak electrolyte, if the symbols

carry their usual meaning :

(1)

2

2m a m a m m

C K K 0



      

(2)

1

2

mm

AC 0



    

(3)

1

2

mm

AC 0



    

(4)

2

2m a m a m m

C K K 0



      

Ans. (1)

Sol. HA(aq) H+ (aq) + A– (aq)

2

a

C

K1





2

aa

C K K 0    

2

mm

aa

mm

C K K 0







  







 

2

m a m m a m

C K K 0



      

80. The electronic configuration of Einsteinium is :

(Given atomic number of Einsteinium = 99)

(1) [Rn] 5f12 6d0 7s2 (2) [Rn] 5f11 6d0 7s2

(3) [Rn] 5f13 6d0 7s2 (4) [Rn] 5f10 6d0 7s2

Ans. (2)

Sol. Einsteinium (atomic No = 99) [Rn] 5f11 6d0 7s2

SECTION-B

81. Number of oxygen atoms present in chemical

formula of fuming sulphuric acid is _______.

Ans. (7)

Sol. Fuming sulphuric acid is a mixture of

conc. H2SO4 + SO3 Or H2S2O7

So, Number of Oxygen atoms = 7

82. A transition metal 'M' among Sc, Ti, V , Cr, Mn

and Fe has the highest second ionisation enthalpy.

The spin only magnetic moment value of M+ ion is

______ BM (Near integer)

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

24, Mn : 25, Fe : 26)

Ans. (6)

Sol. Among given metals, Cr has maximum IE2

because Second electron is removed from stable

configuration 3d5

Cr+ [Ar] 3d5 4s0

No of unpaired e– in Cr+ is 5, n = 5

So, Magnetic moment =

n(n 2) B.M

=

5(5 2)

= 5.92 BM  6

83. The vapour pressure of pure benzene and methyl

benzene at 27°C is given as 80 Torr and 24 Torr,

respectively. The mole fraction of methyl benzene

in vapour phase, in equilibrium with an equimolar

mixture of those two liquids (ideal solution) at the

same temperature is____× 10–2 (nearest integer)

Ans. (23)

Sol. Xmethylbenzene = 0.5

methylbenzene

total

P

YP



methylbenzene

0.5 24

Y0.5 80 0.5 24



  

2

12 0.23 23 10

40 12



   



84. Consider the following test for a group-IV cation.

M2+ + H2S  A (Black precipitate) + byproduct

A + aqua regia  B + NOCl + S + H2O

B + KNO2 + CH3COOH  C + byproduct

The spin only magnetic moment value of the metal

complex C is ______BM.

(Nearest integer)

Ans. (0)

Sol. Co2+ + H2S  CoS  (Black)

(A)

CoS + Aqua-regia Co2+ (aq) + NOCl + S +H2O

(A) (B)

Co2+ (aq) + KNO2 + CH3COOH



K3[Co(NO2)6] + NO + S + H2O

In K3[Co(NO2)6]  Co+3d6 4s0

Co3+ : d2sp3 Hybridisation

Number of unpaired e–= 0

Magnetic moment =

n(n 2)

= 0 B.M

85. Consider the following first order gas phase

reaction at constant temperature

A(g)  2B(g) + C(g)

If the total pressure of the gases is found to be

200 torr after 23 sec. and 300 torr upon the

complete decomposition of A after a very long

time, then the rate constant of the given reaction

is____× 10–2 s–1 (nearest integer)

[Given : log10(2) = 0.301]

Ans. (3)

Sol. A(g)  2B(g) + C(g)

23 0

P P 2x 200  

0

P 3P 300



P0 = 100

0

t

PP

1

K ln

t P P







2.3 300 100

K log

23 300 200





2 –1

2.3 0.301 0.0301 3.01 10 sec

23





   

86.

1cm

 B

 A

10cm

Bottom

Solvent

front

Top

In the given TLC, the distance of spot A & B are

5 cm & 7 cm, from the bottom of TLC plate,

respectively.

Rf value of B is x × 10–1 times more than A. The

value of x is_____.

Ans. (15)

Sol.

f

Distance moved bysubstance from base line

R= Distance moved by solvent from base line

1cm

 B

 A

10cm

Bottom

Solvent

front

Top

Base line

6cm

4cm

8cm

 

fA

4

R8



 

fB

6

R8



 

fB

fA

R68

R 8 4



(Rf)B = 1.5 (Rf)A

x = 15

87. Based on Heisenberg's uncertainty principle, the

uncertainty in the velocity of the electron to be

found within an atomic nucleus of diameter

10–15 m is ______× 109 ms–1 (nearest integer)

[Given : mass of electron = 9.1 × 10–31 kg,

Plank's constant (h) = 6.626 × 10–34 Js]

(Value of  = 3.14)

Ans. (58)

Sol.

h

m V. x 4

   

34

31 15

6.626 10

V9.1 10 10 4 3.14







    

9

57.97 10 m/sec





88. Number of compounds from the following which

cannot undergo Friedel-Crafts reactions is :____

toluene, nitrobenzene, xylene, cumene, aniline,

chlorobenzene, m-nitroaniline, m-dinitrobenzene

Ans. (4)

Sol. Compounds which can not undergo Friedel Crafts

reaction are

NO2

NH2

NO2

Nitrobenzene

Aniline

m-nitroaniline

m-dinitrobenzene

89. Total number of electron present in (*) molecular

orbitals of O2 ,

2

O

and

2

O

is_______.

Ans. (6)

Sol. O2(16e)

 

22

**

1s 1s 2s 2s

   

     

   

11

2 2 2 **

2p 2 p 2 p 2p 2 p

,



      





Number of e– present in

 

*

2

of O 2

Number of e– present in

 

*

2

of O 1





Number of e– present in

 

*–

2

of O 3

So total e– in

 

*



= 2 + 1 + 3 = 6

90. When Hvap = 30 kJ/mol and Svap=75 J mol–1 K–1,

then the temperature of vapour, at one atmosphere

is ______K.

Ans. (400)

Sol. At equilibrium GPT = 0

vap vap

H T S  

30 1000 T 75  

T 400K