Key & Solut_new

NARAYANA IIT ACADEMY INDIA SR IIT-IZ3-SPARK

JEE MAINS MODEL)

TIME : 3 HRS

DATE :10-1-2015

CTM-2

MARKS : 360

KEY PHYSICS 01) 4

02) 1

03) 2

04) 1

05) 2

06) 2

07) 3

08) 1

09) 2

10) 4

11) 4

12) 3

13) 4

14) 4

15) 2

16) 3

17) 3

18) 1

19) 1

20) 2

21) 2

22) 2

23) 4

24) 2

25) 4

26) 3

27) 4

28) 2

29) 3

30) 4

CHEMISTRY 31) 3

32) 1

33) 2

34) 2

35) 2

36) 4

37) 1

38) 4

39) 3

40) 3

41) 3

42) 1

43) 3

44) 2

45) 3

46) 4

47) 1

48) 2

49) 4

50) 4

51) 2

52) 2

53) 2

54) 3

55) 3

56) 2

57) 1

58) 1

59) 2

60) 3

MATHS 61) 3

62) 4

63) 1

64) 2

65) 3

66) 4

67) 4

68) 1

69) 2

70) 4

71) 3

72) 1

73) 2

74) 1

75) 1

76) 3

77) 3

78) 1

79) 3

80) 2

81) 1

82) 1

83) 4

84) 2

85) 1

86) 3

87) 4

88) 3

89) 1

90) 4

PHYSICS anet  01.

m2 g 2  q 2 E 2 m

 T  2

 T  2

l anet l  qE g    m

2

2

Eq Mg

02. It is a balanced wheatstone bridge.

 C AB 

5 3 5 3  53 53

 C AB  2 

15 8

 C AB  3.75 F

03. Since capacitor is a dc blocking element, so no current flows through the branch containing the capacitor. However, a voltage drop will exist across this branch Rnet  4  I total 

V  1.5 A Rnet

Voltage drop across parallel combination of V2 = 6 – 4.2 = 1.8 V 2 If I2 is the current in resistor, then 1.8 I2   0.9 A 2 04. Since

Rl

2

and

3

is

 R1   R1 

R (a ) 2 a

R 2 R2 

R (2 a  a ) 2 a

Similarly R  R2  (2   ) 2 RAB 

Since, RAB 

R1 R2 R1  R2

R (2   ) 4 2

r | Fm | BIl 05. Magnetic force

acts in the direction shown in figure

Since the rod moves downwards with constant velocity, hence the net force on it is zero.

 mg  tan   Il 

B

sin   06. According to following figure 2mk 1 2mV r  qB B q

sin   Bd

d r

also

q 2mV

 0.51 0.1

1.6 1019 1  27 3 2 1.67  10  500  10 2

   30 07. Torque experienced by the coil is    BI ( r 2 ) This torque will be balanced by the torque produced by weight of coil (acting from the centre)  BI ( r 2 )  (mg ) r I

mg  rB

10 20 08. At t = 0 i.e. when the key is just pressed, no current exists inside the inductor. So and resistors (10  20)  30 are in series and a net resistance of exists across the circuit. 2 1 I1   A t  30 15 Hence As , the current in the inductor grows to attain a maximum value ie.e. the 10 entire current passes through the inductor and no current passes through resistor. 2 1 I2   A 20 10 Hence r r d Ñ  E. dr  (2 r ) E   dt 09.  dB  2 rE   a 2     dt E P

10.

1 a 2 dB 2 r dt

e2 d d ; e   ( BA)  A ( B0 e  t )  AB0 e  t R dt dt

P

A2 B02 e 2t 1 ( AB0e  t )2  R R

t0 At the time of starting , so 2 2 2 2 4 ( r ) B0 B0  r P  R R

P

A2 B02 R

11. The voltage VL and VC are equal and opposite so voltmeter reading will be zero.

R  30, X L  X C  25 Also V

i

R2  ( X L  X C )2



V 240   8A R 30

So, 12. Capacitor is a dc blocking element and hence no current flows in (1). An inductor offers a zero resistance path to flow of dc and hence maximum current flows through (2). Vr .m.s  T K . E T 13. and  New K .E  2 (6.211021 J )  12.42 10 21 J  2(484)  684ms 1

New velocity 14. Let h be the distance of the apex from the middle of the base. Further by Pythagoras theorem L12 2 L2   h 2  cons tan t 4

 L12  ( h )    L       4 2

2 2

1  0  2 L2 L2  (2 L1L1 ) 4  L2 ( L2 2 T ) 



  1  



1 L1 ( L11T ) 4

L1 1 L 1  and  2  2  L1 T L2 T

L1  2 2 L2 1  v  v1 f  v  vs   1   f  v  v1  v  vs

f1  f 

15. 16. For 1st reading of oscillator f A  (514  2) Hz

 f A  516 Hz or 512 Hz For 2 reading of oscillator f A  (510  6) Hz nd

 f A  516 Hz  f A  516 Hz

or 504 Hz A has a frequency of 516 Hz.

VA 

V V  4l 0.60

VB 

V V  2l 0.61

17.

VA  VB  5 Since v v   5 0.60 0.61  v  183ms 1  VA 

183  305 Hz 0.6

VB  300 Hz sin C 

r 4 / 3 8   d 3 / 2 9

18.  8  C  sin 1    9

For TIR i>C   C

 sin   sin C  sin  

19.

 m  m0  me  m0  1    m0 

 m0  20.

8 9

m 

D  1  f 0  30 5 1 5

D  f 0

DBM  45  DM  h tan 45  h  32cm



sin i sin r

sin r 

sin i sin 45 3    4/3 4 2

 tan r 

sin r sin r  cos r 1  sin 2 r

3 3  tan r  4 2  9 23 1 32  CM  h tan r  32 

3  20cm 23

 CD  DM  CM  32  20  12cm I max  I  (a  a ) 2

21.  I  4a 2  4 I 0

When either of the two slits is covered then I I 1  ( a  0) 2  a 2  4 x  [(   1)t ] 22. Shift,

D d

x  5 Here D   D  5   [(   1)t ] d  d  5  (   1)t    (   1)  

t 5

(0.5)(6 106 ) 5

 6 10 7 m    6000Å

R1   N1 and R2   N 2 ( is same for a give sample) 23. As N2 < N1 Number of atoms disintegrated in time (t2-t1) is R R  R  R2 N1  N 2  1  2   1       N1  N 2 

( R1  R2 )T  ( R1  R2 )T 0.693

24. According to Ritz combination principle EC  A  EC B  EB A 

hc hc hc   3 1 2

 3 

12 1  2

N1  N 0 e 10 t

25. N 2  N 0 e  t



N1 1   e( 10    ) t N2 e

 9 t  1 t 

1 9

hv  Ek   26.  hv   

1 2 mv 2

v

2(hv   ) m

v

2(hc   ) m

EK  hv  hv0 27.

EK  y, v  x, hv0  cons tan t (k )  y  hx  k (equation of a straight line) h  2meV

28.

1 

h 2 Mev 

V  E   d 

Id   A

dE A dV  0 dt d dt

 Id 

8.86 10 12  3.14  (2  102 ) 2  5  1013 0.1103

29.

 I d  5.56  103 A

30. Ge conduct at 0.3 V and silicon at 0.7 V both Ge and Si diodes are connected in parallel. When current begins to flow, the potential difference remains at 0.3 V so no current flows through Si-diode Potential difference across RL = 12 – 0.3 = 11.7 V Potential of Y = 11.7V CHEMISTRY  5 2 31. n-factor =10 2     3   0.95 0.95  32. n-factor =0.85 M E 0.85  sys  nRln

33.

V2 V1

 2 R l n 2 = 11.52 J/K 3.41 1000 S sura  310 = -11 J/K

Stotal  11.52  11 =0.52 J/K I 2( s )  I 2( g ) 34.

H 2  H1  C p (T2  T1 ) C p  C p ( I 2( g ) )  C p ( I 2( s ) )

= - 6.2 cal / mole H 2  U  1 2  523

5786  U  1 2  523 U  4740 x( s ) ƒ

A( g )  B( g ) x (2 x  2 y )

35. Kp1  PA .PB2

y( s ) ƒ

C( g )  2 B( g ) y (2 y  2 x)

K P2  Pc .PB2 K P1 x   x  2y KP2 y K P1  x(2 x  2 y ) 2

x = 0.1 y = 0.05

36. 37. 38. 39. 40.

PA  PB  PC

Total pressure = = 3 (x + y) = 0.45 atm Conceptual Conceptual Conceptual Conceptual A  N 6.023  3.7 1010  3.7 104 N

N  6.023  106 atoms  1015 gm Atoms of Sr  0.05 Atoms of Rb

41.

Atoms of ( Sr  Rb)  1.05 Atoms of Rb

t

n 2.303 log 0  n

 3.28 109 year 42. 43. 44. 45.

Conceptual Conceptual Conceptual Conceptual [ Ni (CN )3 CH 2O)3 ]1 46. (i) For   2(2  2) n=2  8 B.M [Co( H 2O )3 F3 ]o

(ii) n  4    4(4  2)

 24BM [ Fe(CN )3 ( H 2O)3 ]1

(iii)

  0 n=0 [Cr (CN )3 ( H 2O)3 ]o (iv) n  3    3(3  2)

 15BM 47. Conceptual 48. Conceptual 49. At, 710°C CO2  C ƒ 2CO

G  0 Cr 3  6 NH 3  [Cr ( NH 3 ) 6 ]2 (excess ) 50. 51. Conceptual 52. Conceptual 53. If I2 reacts with excess Cl2 - water it is oxidized to HIO3. 54. X Y A propyne methane B N2O N2 C NH3 NH3 D N2O NO

55. 56. 57. 58. 59. 60.

Conceptual Conceptual Conceptual Conceptual Conceptual Conceptual

MATHEMATICS | x  3 | 1  2  x  4| y  3 | 1  2  y  4 61.

A is set of all point lying with in the square fromed by x = 2, y =2, x = 4, y = 4 ( x  3) 2  ( y  32 )  2 B  {( x, y) lying with in and on the circum  A  B A  B   Circle

(ab)  S  ab  0  ba  0  (ba)  S  R aa  a  0 (a, a )  S 62. For any ‘a’ , for ( a, b)  S  ab  0 (b, c)  S  bc  0 For and  a.c  0 a,b,c have same sign S   (a c) equivalence

f ( x )  [ x]  1  x  0  [ x]  1  x 63. From the graph of y = [x] and y + x = 1 x  [1, )[ x]  1  x [1, ) Domain is

-2

-1

0

1

2 Y+x=1

3

64. If f is one-one f1(x) > 0 3x2+2(a+2)x+3a>0 x  R, b 2  4ac  0  a  [1, 4]

65.

 x tan(log(sec   ))  n f ( x)  lim n  1 n2 f ( x) 

x 2



1

0

Required area = 1  6

x2 dx 2

x=0

 e x 1  1 if  1  x  0   x 1 f ( x)   1 if x  0  ex 1 if 0  x  1  x

66.

e x 1  1 1  1 x 0 x  1 e

lim f ( x)  lim

x0

ex 1 lim f ( x)  lim 1 x 0 x  0 x

e

67.

e

  x   x  1  2 a    a

lim tan 

xa

x   1  a lim  x a   x cot    2 a

x  2  1  . . a 2  2/  lim e xa  x tan  1  2 a 

e

using L1 hospital

2

lim

x a

e

1

x a

   x tan    2 2 a

x=1

e f ( x )  log ex  f ( x )  log e log ex  y (let )

68. y

log ex  e y  x  e e f 1 ( x)  ee x

x

x

g ( x)  ee  g 1 ( x)  ee .e x  e ( e

x

 x)



69.

x  1  x if x  0  f ( x )   0 if x  0  x if x  0  1  x f 1(x ) 

(1  x)  x  f 1 (0 )  1 2 (1  x )

f 1 ( x ) 

(1  x)  x  f 1 (0  )  1 2 (1  x)

sin[ x   ]  0

[x   ] 70.

is always an integer 1  [ x]2  0 x  R also f ( x)  0x  R f(x) is constant function x  R Differentiable f 11 ( x)  6 x  4  f 1 ( x)  3 x 2  4 x  c

71.

x  1, f 1 ( x)  0  C  1, f 1 ( x)  3 x 2  4 x  1 At f ( x)  x3  2 x 2  x  k , f (1)  5

 5  k ,  f ( x)  x 3  2 x 2  x  5 f (0)  5 f 11 ( x )  0  f 1 ( x ) 72.

is decreasing ……….(1) h ( x)  sin(2 x)[ f (sin 2 x)  f 1 (cos 2 x)] 1

1

  x   0,  sin 2 x  cos 2 x  f 1 (sin 2 x)  f 1 (cos 2 x) 4 

73.

For and also sin(2x) > 0 h1(x) > 0 Increasing dy  sin  y  8sin 3   3 x  8cos  dx cos 

x y   8, OA  8cos  cos  sin 

Eq. of tangent 8sin  OA  OB  8 2 OB = minimum value of

from …….(1)

74.

1 b 4x2  b2 A 2 2

dx  3cm / sec. dt

Area  dA 1  1 dx  b 8 x.  dt 4  2 4 x 2  b 2 dt

dA b2 ( when x  b)  (3)   3b dt 3b f (1)  f (1) 75.

and

 1 f1   0  2

2 1  x  9  16  (4 x  2) 2  4  3 x 2  4 x  4  0 2

76.  x  2,

2 3

a .r 1 |a |

77.

| a | 3  a .r  3

since r  a  b  a  ( r  a )  a  b  4i  2 j  5k  (a .a )r  (a .r )a  4i  2 j  5k 9r  3(2i  j  2k )  4i  2 j  5k 9r  2i  j  11k

Q  ( x, y , z ) OQ  r 78. Let such that PQ  r  (3i  2 j  6k ) sub : r PQ  (2  3s )i  ( s  3) j  (5s  4)k PQ

parallel to x – 4y + 3z =1  2  3s  4 s  12  15s  12  0  S 

1 4

3a 2  AB.BC   | AB || BC | cos( AB,BC )   a .cos(120)   2 2

79.

nˆ   80. required vector is unit vector perpendicular to the vectors i-j+k and 2i+j-3k, 222  cos(90   )  3 5



81.



a b | a b |

1 2  5    3 3 5 3





x2  6x  82. x2  6 x  k 

83.

17 k 2

1  k  1

and

17 k 0 2

for 2 district solutions b2-4ac>0

1 1 k  ( ,1] 2 2

largest k value 1  x 1 1 1     e  log x  x  x  x 2 dx dx 

1 e x log x  e x   c1 ]dx x

 

e x log x  c1 x  c2 3

3 r  (log x) dx  x Ar (log x)  B r 0

84.

log x  t  x  et  dx  et dt

 t .e dt  t e 3

t

3 t

 3t 2et  6tet  6e t

 x  (log x)3  3(log x) 2  6(log x)  6 3

A

 1  3  6  6  2

r

r 0

x 2  ax  b 

85.

4(b)  a 2 4

solution x = 0  4b  4b  a 2  a  0

f ( x)  x 2  b roots 

 b

b

 x dx   3



x 3dx  0

 b

a

a

dx dx I  1  f ( x) 1  f (a  x) 0 0

I  86.

a

a

1  f (a  x )  1  f ( x) 2  f ( x )  f (a  x)  dx [1  f ( x ][1  f ( a  x )] 1  f ( x )  f ( a  x )  f ( x ) f ( a  x ) 0 0

2I  

a

2 I   1 dx  I  0

a 2 (0,

87. 3x2 + 5y = 32 down word parabola with vertex at solving, coordinates of A and B are (-2, 4), (3, 1) required area 3 32  3 x 2 1 1 2 5 dx  2  4  4  2 11 y=x+2 (-2,4)

32 ); y | x  2 | 5

y =x-2

A

B(3,1)

4 -2

3

2

y  0  x  0,1, 1 88.

symmetric about x and y axes. 1

4  x 1  x 2 dx 0

Required area

put

x  sin 

-1

89.

0

1

t 2 f ( x )  x 2 f (t ) 2tf ( x )  x 2 f 1 (t ) lim  1  lim 1 tx tx tx 1

 2 xf ( x)  x 2 f 1 ( x)  1  x 2 f 1 ( x )  2 xf ( x )  1 dy 2 1  y  2 L.D.E dx x x x2  y 2  a2  2x  2 y 90.

dy dy 0 x y dx dx

 xy  c  y (1)  1  c  1  xy  1 y (2) 

1 2

x  y

for orthogonal trajectory

dx  x dy  ydx  0 dy