CPT-17 / XI-LJ /

1

NARAYANA I I T A C A D E M Y

CPT – 17 XI-LJ (Date: 2.01.17)

PHYSICS CHEMISTRY MATHEMATICS

1. (B)

2. (B)

3. (D)

4. (B)

5. (D)

6. (B)

7. (D)

8. (C)

9. (B)

10. (B)

11. (B)

12. (B)

13. (C)

14. (C)

15. (B)

16. (C)

17. (C)

18. (A)

19. (D)

20. (B)

21. (B)

22. (B)

23. (C)

24. (C)

25. (D)

26. (A)

27. (C)

28. (D)

29. (B)

30. (D)

31. (A)

32. (B)

33. (D)

34. (C)

35. (C)

36. (B)

37. (A)

38. (B)

39. (C)

40. (B)

41. (B)

42. (D)

43. (B)

44. (C)

45. (C)

46. (A)

47. (D)

48. (C)

49. (D)

50. (C)

51. (B)

52. (B)

53. (D)

54. (D)

55. (A)

56. (A)

57. (A)

58. (A)

59. (B)

60. (D)

61. (A)

62. (C)

63. (B)

64. (C)

65. (A)

66. (A)

67. (C)

68. (D)

69. (D)

70. (D)

71. (B)

72. (B)

73. (C)

74. (B)

75. (A)

76. (D)

77. (C)

78. (A)

79. (A) 80. (C)

81. (C)

82. (D)

83. (A)

84. (A)

85. (B)

86. (A)

87. (B)

88. (B)

89. (C)

90. (A)

CODE XI-LJ

CPT-17 / XI-LJ /

2

Hints & Solution PHYSICS

1. (B)

GMm

W U 0R

= ∆ = −

11 3

10

2

6.67 10 100 10 106.67 10 J

10 10

− −

−

−

× × × ×

= = ×

×

2. B

260

12 /5

f irad s

t

ω ω

α

−

= = =

3. At equator the value of g is minimum so it is profitable to purchase sugar at this position.

4. Acceleration due to gravity at height h is given by

2

+=′

hR

Rgg

2

100

+=⇒

hR

Rg

g

10

1=

+

⇒

hR

R Rh 9=⇒ .

5. If P is the point where net gravitational force is zero then PBPA

FF =

2

2

2

1

)( xd

mGm

x

mGm

−

=

By solving 21

1

mm

dmx

+

=

For the given problem Dd = , =1m earth, =

2m moon and

2181mm =

81

1

2

m

m =∴

So 21

1

mm

Dmx

+

= 10

9

9

11

81

1

1

1 DD

mm

Dm=

+

=

+

=

6.

−=′R

dgg 1 ⇒

−=R

dg

n

g1 R

n

nd

nR

d

−=⇒−=⇒

111

7. (D) Gravitational field inside the hollow sphere is equal to zero

8. (C)

At the point P, we have I2 – I2 = 0 (because the gravitational field inside a shell is zero).

Hence, I1 = I2

9. As dx

dVI −= , if 0=I then V = constant.

10. Potential increases by kgJ /10+ everywhere so it will be kgJ /5510 +=−+ at P

A

m1 m2

d

B

P

x d – x

FPA FPB

CPT-17 / XI-LJ /

3

11. Net potential at origin

+++−= .........

321r

Gm

r

Gm

r

GmV

+++−=8

1

4

1

2

1

1

1Gm GmGm 2

2

11

1−=

−

−=

12. Work done Rh

mgh

/1 += , If Rh = then work done

RR

mgR

/1 += mgR

2

1= .

13. 2

; 2 ;2 2

v v vx R

x x R Rω ω= = ⇒ =

+

14. ' 1d

g gR

= −

15. Distance of null point

2

1

1

dx

m

m

=

+

16. By the law of conservation of angular momentum ( )ωωω212211IIII +=+

Angular velocity of system 21

2211

II

II

+

+=

ωω

ω

Rotational kinetic energy ( ) 221

2

1ωII += ( )

2

21

2211

212

1

+

++=

II

IIII

ωω

)(2

)(

21

2

2211

II

II

+

+

=

ωω

.

17. Initial angular momentum of the system = Angular momentum of bullet before collision

=

2

LMv .....(i)

let the rod rotates with angular velocity .ω

Final angular momentum of the system ωω22

212

+

=

LM

ML ......(ii)

By equation (i) and (ii) ω

+=

4122

22MLMLL

Mv or Lv 2/3=ω

18. =×=→→→

prL

243

121

ˆˆˆ

−

−

kji

kjkji ˆ2ˆˆ2ˆˆ0 −−=−−= and the X- axis is given by kji ˆ0ˆ0 ++

Dot product of these two vectors is zero i.e. angular momentum is perpendicular to X-

axis.

19. Rotational kinetic energy 2222

1

2

1

2

1ωω

= MRI ( ) J25020)5.0(10

2

1

2

1 22 =

××=

m

1m

2m

4m

8m

O

m m m

M

v

CPT-17 / XI-LJ /

4

20. ( )m

Used v A tv

= = ×

21

2I mR=

21. As, I

LE

2

2

= ⇒

2

1

2

1

2

=

L

L

E

E2

1

13

=

L

L [As

112.%200 LLL += = 3L1]

⇒ 12

9EE = %8001+= E of

1E

22. Velocity at the bottom (v) ghgh

R

K

gh

3

4

2

11

2

1

2

2

2=

+

=

+

= .

23.

2

2

1

sin

R

k

ga

+

=θ

11

30sin

+

=

o

g

4

g= [As 1

2

2

=

R

k and o30=θ ]

24. (C)

Gravitational potential at the surface of the earth GM

R= −

Gravitational potential at 0∞ =

Potential difference GM

0R

= − −

GMV

R∆ =

Workdone in shifting a body of m kg from the surface of the earth to infinity.

GMm

W m VR

= ∆ = Workdone per GM

kg gRR

= =

25. (C) Gravitational field inside a shell is zero, but for points outside it, the shell behaves as if

whole of its mass is concentrated at its centre.

26. Initial angular momentum of ring ωω 2MRI ==

If four object each of mass m, and kept gently to the opposite ends of two perpendicular

diameters of the ring then final angular momentum = ')4( 22 ωmRMR +

By the conservation of angular momentum

Initial angular momentum = Final angular momentum

')4( 222 ωω mRMRMR += ⇒ ωω

+=

mM

M

4' .

28. ( ) 1 2 1 2 312

, ,

net

GmmGPE U use U U U U

r= = + +

29.

2

2

1

1 3

n kn

n R

= =

+

30.

2 0

2

sin sin30 5

21 141

5

g k ga n g

n R

θ = = = =

+ +

CHEMISTRY

CPT-17 / XI-LJ /

5

31. A

Half neutralization

[ ] [ ]

[ ]

[ ]

5.4

log

log

log

10 10H

H

a

H

a

H

a

H

a

P

acid salt

saltP Pk

acid

P Pk

P k

k anti P

− −

=

= +

=

= −

= −

= =

32. B

Equilibrium shift in backward direction

33. D Isoelectric point is the condition, when zwitterions or sol particles do not move under the

influence of electric field, i.e., they lose their charge.

34. C

Maximum buffer capacity when [salt] = [acid] P

H=Pka

35. C

36. B

A buffer of H2CO3 and 3HCO− is formed.

37. A

pH = –logKa + log[Salt]

[Acid]

= 41

log10 log1

−

− + = 4

[Since Ka × Kb = 10–14

Given Kb = 10–10

∴ Ka = 10–4]

38. B Mixing of NH4OH and (NH4)2SO4 gives a basic buffer.

pOH = pKb + log4

4

[NH ]

[NH OH]

+

;NH

4

apK

+

= 9.26

∴ NH OH4

bpK = 4.74

5.74 = 4.74 + loga

5; where a is millimole of [NH4]

+ obtained on mixing.

∴ a = 50 ∴ millimole of (NH4)2SO4 = 25

or, mole of (NH4)2SO4 = 0.025

CPT-17 / XI-LJ /

6

39. C

pH = pKa + log1

1

(300 V ) 1

1 V

− ×

×

; let V1 be volume of benzoic acid and thus (300 – V1) is

volume of benzoate.

∴ 4.5 = 4.2 + log 1

1

300 V

V

−

∴ 1

1

300 V

V

−

= 2

∴ V1 = 100 mL

40. B

HIn ��⇀↽�� H+ + In–

Ka = [H ][In ]

[HIn]

+ −

∴ Ka = w

b

K

K =

14

11

10

10

−

−

= [H+]

∵ [In–] = [HIn]

or [H+] = 10–3

or pH = 3

41. B NH3 + HCl in 2 : 1 will give NH3 + NH4Cl in 1 : 1 ratio.

42. D

Buffer solution 43. B

Find solubility for each separately by S2 = KSP for MnS and ZnS.

108S5 = KSP for Bi2S

3 and 4S

3 = KSP for Ag2S.

44. C

Let S is the solubility of BaF2 in a solution of BaNO3, then

KSP = [Ba2+][F–]2; then [F–] = 2S; then

1[F ]

2

− = S

45. C Presence of common ion decreases the solubility of salt.

46. A Presence of common ion decreases the solubility of salt.

47. D

KSP of PbCl2 = 4S3 = 4 × (0.01)

3 = 4 × 10

–6

In NaCl solution for PbCl2;

KSP = [Pb2+][Cl

–]2

or, 4 × 10–6 = [Pb

2+][0.1]

2

∴ [Pb2+] = 4 × 10

–4 M

CPT-17 / XI-LJ /

7

48. C

AgS

+= SAgCl = SPK =

101.8 10

−

× = 1.34 × 10–5 M

AgS

+= SAgBr = SPK =

135.0 10

−

× = 7.07 × 10–7 M

2 4

Ag CrOS = SP3

K

4 =

12

32.4 10

4

−

×

= 0.8 × 10–4 M

Also, AgS

+= 2 ×

2 4Ag CrOS = 0.8 × 10

–4 × 2 = 1.6 × 10

–4 M

This will give maximum value.

49. D

4

9 2

4

7

4

1.5 10 10

1.5 10

Ksp Ba SO

SO

SO

−−++

−−− −

−− −

=

× =

= ×

50. C

2 2

mm before 10 20 0 0

reaction 0 0 10 20

MgCl 2NaOH Mg(OH) 2NaCl+ → +

Thus, 10 milli-mole of Mg(OH)2 are formed.

The product of [Mg2+][OH

–]2 is therefore

2

10 20

200 200

×

= 5 × 10–4 which is more than

KSP of Mg(OH)2. Now solubility (S) of Mg(OH)2 can be derived by KSP = 4S

3

∴ S = SP3K

4=

11

31.2 10

4

−

×

= 1.4 × 10–4

∴ [OH–] = 2S = 2.8 × 10

–4

51. B

Due to common ion effect

52. B

2 2[Mg ][OH ]+ − = 1 × 10–12;

∴ [OH ]− =

1210

0.01

−

= 10–5

or pOH = 5 and thus pH = 9

53. D

[OH–] = Cα = b

KC

C

= b

K .C = 121.0 10 0.01−× × = 1.0 × 10–7 M

54. D

CPT-17 / XI-LJ /

8

MX M X+ −+��⇀↽�� ; KSP = x2 ⇒ x = (KSP)

1/2

M2X ��⇀↽�� 2M+ + X

–; KSP = 4x

3 ⇒ x =

1/3

SPK

4

MX3 ��⇀↽�� M3+ + 3X

–; KSP = 27x

4 ⇒ x =

1/4

SPK

27

55. A 56. A 57. A

58. A 59. B

60. D

MATHEMATICS

61. Given a,b,c are in G.P, then

Let these are a, ar, ar2

Now, given

2

12,ab

a b=

+

and 2

36,bc

b c=

+

( )6 ..... 2ab

a b=

+

( )18 ..... 3bc

b c=

+

From (2) 6a ar

a ar

×=

+

61

ar

r

⇒ =

+

…. (4)

From (3)

2

218

ar ar

ar ar

×=

+

2

181

ar

r

⇒ =

+

… (5)

From (1) and (2) r = 3 and a = 8 Then a + b + c = 104

62. Let ,A m h nλ λ= =

2 2

G AH mnλ∴ = =

∴ a, b, are the roots of ( )2 0t a b t ab− + + = or

2 2

2 0t m t mnλ λ− + =

( )

( ){ }2 2 2

4 4

2

m m mn

t m m m n

λ λ λ

λ

2 ± −

∴ = = ± −

( ) ( ): :a b m m n m m n⇒ = + − − − or ( ) ( ):m m n m m n− − + −

63. 2 2 2 2 2 21 2.2 3 2.4 5 ... n+ + + + + + as n is odd

( ){ }22 2 2 2 21 2.2 3 2.4 ... 2 1n n= + + + + + − + ( ) ( )2

2 21 1

2 2

n n n

n n

− + = + =

64. Multiply and divide by (1 – a) in the given series

CPT-17 / XI-LJ /

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65. 100

2 2 4 200

1

...

n

n

a a a a α

=

= + + =∑

100

2 1 1 3 199

1

2

1

...

n

n

a a a a

a

r

a

α

α

β

−

=

= + + =

= =

∑

66. 2PQ

HP Q

=

+

67. If x, y, z are in G.P, then log , log , logn n nx y z are in A.P.

68. Given a1 = 2 a10 = a1 + 9d = 3

=2+9d = 3 d = 1/9

then a4 = a1 + 3d = 2+1/3 = 7/3

10 1

1 1 19 ' ,

3d

h h= + =

1'

54d = − (given h10 = 3 and h1 = 2)

Now, 7 1

1 1 76 '

18d

h h= + =

Then 7

18

7h = therefore

4 7

7 186

3 7a h = × =

69. If a1, a2,….. an are in H.P, then 1 2

1 1 1, ,...,

na a a

are in A.P.

Now, let 2 1 3 2 1

1 1 1 1 1 1.....

n n

da a a a a a

−

− = − = = − =

2 3 11 2

1 2 2 3 1

......

n n

n n

a a a aa ad

a a a a a a

−

−

− −−

= = = =

70.

( )( )( )

( )

3

1 log

1 2 log

1 3 15log

1 1 1 2 1 3 15

2

1 3 15log

5 15log

2

3log 1

y

z

x

x

x

x

d x

d y

d z

d d d

d

d z

z

d

z

z x

+ =

+ =

+ = −

+ + + = −

= −

+ = −

− = −

= −

=

=

∵

CPT-17 / XI-LJ /

10

71. Use A.M. > G.M for (a, b, c)

Then a + b + c > 3 (abc)1/3 …..(1)

Now ab2c3, a

2 b

3c4, a

3 b

4 c

5 are in A.P. (a, b, c, > 0)

Then 1, abc, a2b2c2 are also in A.P.

2abc = 1 + a2b2c2, (abc – 1)

2 = 0, abc = 1, put abc = 1 in equation (1)

72.

( )

( )

( )

( ) ( )

1.3.5............... 2 1

2 1 11

2 1.3.5............. 2 1

1 1 1

2 1.3.5........ 2 1 1.3.5......... 2 1

n

nt

n

n

n

n n

=+

+ −=

+

= −

− +

Given series

( )

( )

1

1 1 1

2 1 1.3.5........ 2 1

1

2

n

n

t

n

as n

∞

=

= −

+

=

→∞

∑

73. Let G = x, S = 5 And r is common ratio

Then 1

aS

r=

−

, 1 / 51

xS r x

r= = −

−

1 1,r− <

CPT-17 / XI-LJ /

11

( )

( )

3

1 2 1 4

1

11 4 1 4 1 2

1 1

r r

r

a a a a

a a

a a a a a a+

=

−

∴ = ×

−∑ 4 1

2 1

1 1

33

1 1

a a

a a

λ

λ

−

= = =

−

Hence, 3 is a root of 2

2 5 0x x+ − =

75. 1 2 2 2 1 1

...

n n n na b a a a a a a

− ++ = + = + = = + and

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

n n n nab g g g g g g

− += = = =

and 2

.ab

ha b

=

+

1 2 2 2 1 1

1 2 2 2 1 1

...

n n n n

n n n n

a a a a a a

g g g g g g

− +

− +

+ + +∴ + + + =

22

2

a b a b nn n

ab ab h

+ + = = =

76. Given ( ) ( ) ( )2log log log 2c a a c a b c− = + + − + ( ) ( )log 2a c a b c= + × − +

( ) ( ) ( )2

log log 2c a a c a b c− = + × − + ⇒ ( ) ( ) ( )2

2c a a c a b c− = + − +

After solving,

2ac

ba c

=

+

Hence, a, b, c are in H.P

77. ( )( )( )

( )1

1 2 3

8

n

r n

r

n n n nt S say

=

+ + +

∴ = =∑

( ) ( ) ( )1

1

1

1 1 2

8

n

r n

r

n n n nt S

−

−

=

− + +

∴ = =∑

( )( )

1

1 2

2n n n

n n nt S S

−

+ +

∴ = − =

( )1 1

1 2lim lim

1 ) 2

n n

n n

r rrt n n n→∞ →∞

= =

∴ =

+ +∑ ∑ =

( ) ( )( )1

1 1lim

1 1 2

n

n

rn n n n→∞

=

− + + +

∑

( )( ) ( )1

1 1lim

1 2 1

n

n

rn n n n→∞

=

= − − + + + ∑

( )( )

1 1lim

1 2 2n n n→∞

= − − + +

1 1

02 2

∴− − =

78. d,e,f are in G.P. 2e df⇒ =

2

2 0dx ex f+ + = ( )2

22 0 0⇒ + + = ⇒ + =dx df x f d x f

f

xd

∴ = −

Now, substituting the value of x in 2

2 0ax bx c+ + =

i.e., 2

0b faf

cd d

− + = 2

0 0a b c a b c

ord f d e fdf

2⇒ − + = − + =

, ,a b c

d e f∴ are in HP

CPT-17 / XI-LJ /

12

79. 1 1 1 1 1

1

jn i n n

i j k i j

j= = = = =

=∑∑∑ ∑∑ ( )

1

1

2

n

i

i i

=

+

=∑2

1 1

1

2

n n

i i

i i

= =

= +

∑ ∑

( )( ) ( ) ( ) ( )1 2 1 1 1 21

2 6 2 6

+ + + + + = + =

n n n n n n n n

80. ∵ H = 2ab

,a b+

∴ H

a =

2b

a b+,

⇒ H a 3b a

H a b a

+ +=

− −

…(i)

Similarly,

H b 3a b

H b a b

+ +=

− −

…(ii)

⇒ H a H b

2H a H b

+ ++ =

− −

81.

[ ]

[ ]

1 24 1 24

24

1 24

1

3 225 75

2412 75 900

2i

i

a a a a

a a a

=

+ = ⇒ + =

= + = × =∑

82. AM of positive numbers ≥ G.M. of positive numbers

( )

( ) ( ) ( )

12 2 2 2 2 2

6 6 6 6

2 2 2 2 2 2

6

6.

ab ac bc ba ca cba b c

a b c b c a c a b abc

+ + + + +⇒ ≥

⇒ + + + + + ≥

6l∴ = 83. In first 10 minutes, the number of notes counted is

150 × 10 = 1500

⇒ a11 + a12 + a13 + …. Upto n

terms = 3000

⇒ n2 – 149 n + 3000 = 0

84.

( )

2

2 2

2 2

2

2 2

a cb b b b

ab bc x yx aband y bc

+ = × =

+ += = = =∵

85.

CPT-17 / XI-LJ /

13

( ) ( ) ( )

( ) ( ) ( )

2 2 2

2 2 2

0

0

, , , . .

ap b bp c cp d

ap b bp c cp d

b c dp a b c d areinG P

a b c

− + − + − ≤

⇒ − + − + − =

⇒ = = = ⇒

86. 4 = 2ab

a b+

2A + G2 = 27

⇒ a + b + ab = 27

87. ⇒ 2 2y z b y z

a and ba y z z y= = ⇒ = =

⇒ 2 2y z

a b 2xz y

+ = + =

88. Given (y – x), 2(y – a), (y – z) are in H.P.

⇒ 1 1 1

, , are in A.P.y x 2(y a) y z− − −

⇒ x a y a

y a z a

− −

=

− −

⇒ x – a, y – a, z – a are in G.P.

89.

( )

( )

11

11

a m dn

a n dm

+ − =

+ − =

Subtracting :

( )

( )

1

1

1 11 1

mn

m nm n d d

mn mn

amn

T mnmn mn

−− = ⇒ =

=

= + − =

90. 2 2 2

1 1 1, ,

a b c are in A.P.

⇒ 2 2

1 1

b a− =

2 2

1 1

c b−

��