NARAYANA...NARAYANA IIT/PMT ACADEMY (2) Applying mirror formula, we get; 1 1 1 u f 1 1 1 1 2 1 or 2u...

NARAYANA IIT/PMT ACADEMY (1)

NARAYANA I I T / P M T A C A D E M Y

C o m m o n P r a c t i c e T e s t – 1 6 XII STD BATCHES [CF] Date: 08.08.16

AANNSSWWEERR

PHYSIS HEMISTRY MTHEMTIS 1. (D) 2. (A) 3. (D) 4. (D) 5. (B) 6. (A) 7. (A) 8. (A) 9. (B) 10. (A) 11. (D) 12. (B) 13. (B) 14. (C) 15. (D)

16. (A) 17. (C) 18. (B) 19. (B) 20. (C) 21. (C) 22. (C) 23. (A) 24. (A) 25. (C) 26. (C) 27. (C) 28. (B) 29. (A) 30. (B)

31. (C) 32. (C) 33. (B) 34. (C) 35. (B) 36. (C) 37. (A) 38. (B) 39. (C) 40. (D) 41. (C) 42. (B) 43. (C) 44. (A) 45. (B)

46. (D) 47. (A) 48. (A) 49. (A) 50. (B) 51. (C) 52. (A) 53. (A) 54. (A) 55. (C) 56. (B) 57. (B) 58. (C) 59. (A) 60. (A)

61. (C) 62. (B) 63. (C) 64. (B) 65. (B) 66. (C) 67. (A) 68. (A) 69. (A) 70. (A) 71. (C) 72. (A) 73. (D) 74. (D) 75. (D)

76. (A) 77. (A) 78. (D) 79. (A) 80. (A) 81. (A) 82. (C) 83. (B) 84. (D) 85. (C) 86. (B) 87. (A) 88. (A) 89. (D) 90. (C)

(Hint & Solution)

PART A : PHYSICS 3. Both V and I are in the same phase. So, let us calculate the time taken by the voltage to

change from peak value to rms value. Now, 220 = 220 sin 100 t1

or 11002

t

or 11

200t s

Again, 2220 220sin100

2t

or 21 sin1002

t or 210034

t

or 23

400t s

Required time 32 1

1 2.5 10400

t t s s

7. Since, image formed by mirror will be at distance OM on right side, so image will be formed

for convex mirror at OM MP. So, = OM – MP and object is at OP. u = OP 8. Since, image is virtual, v is +ve. f = 15, u = ?

m 2 or, 2uu


Applying mirror formula, we get;

1 1 1u f

1 1 1 1 2 1or2u u 15 2u 15

9. The impedance triangle for resistance (R) and inductor (L) connected in series is shown is the

fig.

R

X = LL R +

L2

2

Power factor cos = 2 2 2

RR L

10. Phase difference for R-L circuit lies between 0,2

11. According to the following ray diagram HI AB d and 2dDS CD

22 2

2dAH AD GH CD d

Similarly IJ d so 3GJ GH HI IJ d d d d 12. From the following ray diagram

0 0.2 2 30.2 tan 30 303 0.2 / 3

ldd

13. If end A of rod acts an object for mirror then it’s image will be A’ and if

523 3f fu f so by using 1 1 1

f v u


1 1 1 5

5 23

v fff v

Length of image 5 22 2

ff f

14. From fig. it is clear that relative velocity between object and it’s image = 2v cos

15.

200 1 12 200 200 2

I f I I mmO f u

16. Since 1 1 1 1 1f v u v u f

Putting the sign convention properly

1 1 1 1 1 1v u f v u f

Comparing this equation with y mx c

Slope 0 0tan 1 135 45m or and intercept 1Cf

17. i2 = 9t2

12

2 010 1

0

9t dti 3

dt

rmsi 3A


18. 4C 6

1 1X 10C 100 10

AC ammeter gives rms value of current.

3rmsrms 4

C

E 200 2I 20 10 A 20mAX 10 100

19. As L CV V 300V,

and 22R L CV V V V

RV V 220V

Also V 220I 2.2AR 100

20. I 4sin 100 t Amp6

Initial value of current t 0I 4sin 100 06

14sin 4 2Amp

6 2

21. Power loss = 20 watt

Output power = VI = 220 × 4 = 880watt Power generated by dynamo = 900 watt

volt900 900El 900 E 225

I 4

22. Compare with p p pi i sin t i sin t cos i cos t sin

Thus p pi cos 10, i sin 8. Hence 4tan5

23. A solenoid consists of inductance and resistance.

When 100 V dc is applied, = 0 Z = R

rms

rms

V 100Z R 100I 1

When 100 V, 50 Hz ac is applied rms

rms

V 100Z 200I 0.5

22 2 2 2 2L CZ R X 200 100 X

L100 3X 100 3 2 fL 100 3 L 0.55 H

2 50


24. Since, VL = VC = VR XL = XC = R and V = VR = 10V

when capacitor is short circuited

L2 2L

10 10i as X R2RR X

L L10 10V iX R V2R 2

26. 2 / 2 2 2 2T 200

Trms0

3 sin t 4 cos t 2 3 4sin t cos .dtI dtIdt 2 /

rmsI 5 / 2 Ampere .

27. 2

2 1Z R LC

28. 2 2120 160 200vE V 29. Here, VL = VC. They are in opposite phase. Hence,. They will cannot each other. Now the

resultant potential difference is equal to the applied potential difference = 100 V Z = R ( XL = XC)

100 250

rms rmsrmsI

V V AZ R

30. Given that E0 = 10 V, 1600

t s

0 cos 2E E ft

10cos12 50

600

10 cos / 6 10 3 / 2 5 3V

PART B : CHEMISTRY 31. 2A 3B 4C

1 dA 1 dB 1 dC2 dt 3 dt 4 dt

32. xA yB C

dA dB dC1.5dt dt dt

1 dA 1 dB 1 dC3 dt 3 dt 2 dt

33. 22NO Cl 2NOCl

22Rate k NO Cl


34. Fraction of B =4

14 5

1 2

K 1.26 10 76.83%K K (1.26 10 ) (3.8 10 )

Fraction of C = 5

24 5

1 2

K 3.8 10 23.17%K K (1.26 10 ) (3.8 10 )

35. aE / RTe showing fraction of molecules having energy greater than Ea

aE / RTx e aEln x

RT

38. r = K[O3]2 [O]

Also 2

3c

O OK

O

3

2

cK OO

O

r = K [O3]2 . [O2]-1

39. r = k[X] . . . .(i) where [X] = 0.01 M

K =

0

t

X2.303 logt X

. . . . . (ii) where t = 40 min, [X]0 = 0.1, [X]t = 0.025

from (i) and (ii)

2.303 0.1r log (0.01)40 0.025

= 3.47 10-4 M min-1 40. k = Ea / RTAe [Arrhenius Equation]

log K = log A - EaRT

or log k = Ea 1 log AR T

y = mx + c means negative slope and non-zero intercept 45. E is activation energy. Thus order with respect to NO is 2. If Ea + ER = EThreshold energy then

only molecule may react. 46. Rate equation is derived from slowest step, thus first step in mechanism A is rate determining

step. Rate 2 2K Cl H S 48. 0 6 12.75 10G J mol 0G nE F

6

0 2.75 1026 96500

E

=1.096 V

49. Cathode: 22 2H e H

Anode: 2 212 22

OH H O O e

50. 0 0

/ /0.44 0.33 0.11cell OPM M RPX X

E E E V For .M X M X Thus reaction is non-spontaneous. The spontaneous reaction in

0; 0.11M X M X E V


51. The salt bridge possesses the electrolyte having nearly same ionic mobilities of its cation and anion.

52. Equivalent conductivity of Fe2(SO4)3 is given by 3 2

4eq eq eqFe SO 53. More negative potential will constitute negative terminal (i.e. anode) of the galvanic cell. 54. A positive potential implies that the given ion has a large tendency to reduce as compared to

H+. Thus, Cu2+ ion can be reduced by H2(g).

55. For the reaction 21 ,2

H aq e H g the standard potential is given by

2

1/ 2

3 0

/ln

/ /Hp barRTE

nF H mol dm c

E will be reactive if the numerical value of

2

1/ 2 / 1.Hp H

56. For the reaction 2 2 ,Cu e Cu Nernst equation is 02 0

1ln2 /RTE E

F Cu c

2

2 1 2

2 0.059ln ln 0.01 0.0592 2 20.01

CuRT RT VE E VF FCu

60. The S.R.P. of 0

B / BE 0.80V is higher

It acts as cathode.

PART C : MATHEMATICS

61. 2

2 2

01

xF x f t dt x x

Differentiating both the sides, we get 2 2.2 2 3f x x x x

2 312

f x x

34 1 22

f or 31 22

4 2or

62. 2

1

loge ee

xI dxx

21

1/ 1

log logee ee

x xdx dxx x

212 2

1/ 1

log log2 2

e

e e

e

x x

1 522 2


63. 3 2 3cos

2 2 2sin 2xf x dx e x dx dx

0 2 2 64.

2

02g f t dt

1 2

0 11f t dt f dt

Now 1

0

1 1 12

f dt and 2

1

102

f t dt

1 2

0 1

1 12 2

f t dt f t dt

1 32 0 2 22 2

g g

65. 2

2 21

1limn

x r

rn n r

2

21

/1lim1 /

n

x r

r nn r n

2

2 2

0 20

1 5 11

x dx xx

66. From the adjacent of y = 2sinx, we find that 2sinx attains integral value for

5 7 3, , , ,2 6 6 2

x

3 /2

/ 22sin x dx

5 /6 7 /6 3 /2

/2 5 /6 7 /61 0 1 2dx dx dx dx

5 7 3 70 26 2 6 2 6

23 6 3 2

67. 3 /4

/4 1 cosdxI

x

3 /4

/4 31 cos4 4

dx

x

3 /4

/4 1 cosdx

x


3 /4

/4

1 121 cos 1 cos

I dxx x

3 / 4 3 /4

2 /4/42 2 cot 4

sindx x

x

2I 68.

1

0

x

xf t dt x tf t dt

Differentiating both the sides with respect to x, we have 1f x xf x

11

f xx

112

f

69. If 1 0x then 1 1x f x x If 0 1x then 0x f x x

' 0 1

1 1 01

If x dx x dx x dx

0 12 2

1 0

12 2x xx

70. 4

0cos

xg x t dt

4 4

0cos cos

x x

xt dt t dt

4 4

0 0cos cos

xt dt t dt

( cos4 t is periodic with period ) g x g 71. 1 1

1k

kI xf x x dx

1

1 1 1 1k

kx f x x dx

b b

a af x dx f a b x dx

2 111 1

k

kx f x x dx I I

11 2

2

122

II II

72. For < x < 2, we have


72sin 1,6

x x

7 112sin 2,6 6

x x

112sin 1, 26

x x

2

2sinI x dx

7 /6 11 /6 2

7 /6 11 /61 2 1dx dx dx

7 11 7 112 1 26 6 6 6

46 3 6

53

73. sin2xf x A B

' cos2 2

A xf x

1' 22

f

cos 22 4

A

4A

Now 1

0

2Af x dx

1

0

2sin2x AA B dx

1

0

2 2cos2

A x ABxx

2 2 0A AB B

So, 4A

and 0B

74. /2

30 1 tandxI

x

3/2

3 30

coscos sin

x dxx x

3

/2

0 3 3

cos2

cos sin2 2

xdx

x x

3/2

3 30

sinsin cos

x dxx x

/2

02 1

2I dx

4

I


75. Let h x f x f x g x g x

h x f x f x g x g x h x h x is odd function.

/2

/20h x dx

76. Given 5x2 – y = 0, and 2x2 – y + 9 = 0 Eliminating y, we get 5x2 – (2x2 + 9) = 0 3x2 = 9 x = 3, 3

Therefore, required area

3 2 2

02 2 9 5x x dx

3 2

02 9 3x dx

33

02 9x x

2 9 3 3 3 12 3 sq. units

77. Curve tracing: y = x logex Clearly, x > 0 For 0 < x < 1, a logex < 0 and for x > 1, x logex > 0 Also, x logex = 0 x = 1

Further, 0 1 log 0 1/ ,edy x x edx

which is a point of minima.

Required area

1 1

2

0 0

2 2 logx x dx x x dx


1 13 2 2

2

0 0

2 log3 2 4x x xx x

2

0

2 1 11 0 lim log3 4 2 x

x x

1 1 73 4 12

78. 2

11 1

yx

Area

12 1

1

12 2 tan 11 1

dx xx

sq. units

79.

Integrating along the x-axis, we get

2

1 1

1

secA cosec x x dx

Integrating along the y-axis, we get

/4

0

2 sec 1A y dy

/4

02 log sec tany y y

2 log 2 14

log 3 2 22

sq. units


80.

1 2

2

1 1

2 2 1A x dx x dx

13 22

11

23x x x x

163

sq. units

81.

2

1

lnA x dx

2

1logx x x

2log 2 1 Required area = 4 – 2(2 ln 2 – 1) 6 – 4 ln 2 sq. units. 82.

First consider y = 3 - |3 – x|


For x < 3; y = 3 – (3 – x) = x For x 3; y = 3 – (x – 3) = 6 – x

Consider 6| 1

yx

For 61;1

x yx

1 6x y

For 61;1

x yx

Required area 3 5

2 3

6 661 1

x dx x dxx x

3 52 2

5

22 3

6 6 log 12 2x xx x

5 4 6log 22

13 6 ln 22

sq. units

83. Given curves are y = logex and y = (logex)2 Solving logex = (logex)2 logex = 0, 1 x = 1 and x = e Also, for 1 < x < e, 0 < loge x < 1 logex > (logex)2 For x > e, logex < (logex)2 Y = (logex)2 > 0 for all x > 0 and when x 0, (logex)2 . From these information, we can plot the graph of the functions.

Then the required area 2

1

log loge

ex x dx

2

1 1

log loge e

exdx x dx

2

1 11

2 loglog log

eee ee e

xx x x x x xdxx

11 2 log eee x x x

3 e sq. units


84. Curve tracing: y = x + sin x

1 cos 0dy x xdx

Also 2

2 sin 0d y xdx

when x = n, nZ

Hence, x = n are points of inflection, where curve changes its concavity Also for x(0, ), sin x > 0 x + sin x > x. And for x(, 2), sinx < 0 x > sinx < x From these information, we can plot the graph of y = f(x) and its inverse.

Required area = 4A, where

0 0

sinA x x dx xdx

2 2

cos cos 0 22 2

sq. units

85. /4

sin cos 24

f x dx

Differentiating both sides w.r.t. , we get

cos sin sin 24

f

' sin cos cos cos4

f

'2 2

f

86.

1.5 1 2 1.52 2 2 2

0 0 1 2x dx x dx x dx x dx

2 1.5

1 20 1dx 2dx 2 1 3 2 2 2 2.

87. Required area is 1/ a

2 2

0

x 1 1ax dx 1 a aa 3 3

88. Required area is 1/ 2

2

0

1 12 (x 1) dx sq. units4 3


89. Required area 1

2

0

2 4 x 3x dx

11/ 2

2 1

0

x 4 x 3.2x2 4 x sin2 2 2 3

2 33 3

90. y = 2x4 x2 y = 8x3 2x = 0 2x(4x2 1) = 0

x = 0, 1 1x ,2 2

y = 24x2 2

min at 1 1x ,2 2

12

4 2

0

A 2 (2x x )dx

1

5 3 2

0

x x4. 2.5 3

7120

TOPIC/SUB TOPIC Q. NOS. PHYSICS

Reflection of light 1, 2, 4, 5, 6, 7, 8, 12, 13, 14, 15 LR Circuit 9, 10, 28 Reflection 11, 25 Mirror equation 16 AC Circuit 3, 18, 19, 20, 22, 23, 26, 30 LCR Circuit 24, 27, 29

CHEMISTRY Chemical Kinetics 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42,

43, 44, 45, 46, 47 Electrochemistry 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59,

60 MATHEMATICS

Area 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 87, 88, 89, 90

Definite Integration 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 86

NARAYANA...NARAYANA IIT/PMT ACADEMY (2) Applying mirror formula, we get; 1 1 1 u f 1 1 1 1 2 1 or 2u...

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Transcript of NARAYANA...NARAYANA IIT/PMT ACADEMY (2) Applying mirror formula, we get; 1 1 1 u f 1 1 1 1 2 1 or 2u...