NARAYANA IIT ACADEMYINDIASR IIT-IZ3-SPARK JEE MAINS MODEL) DATE :10-1-2015TIME : 3 HRS CTM-2MARKS : 360KEYPHYSICS01) 402) 103) 204) 105) 206) 207) 308) 109) 210) 411) 412) 313) 414) 415) 216) 317) 318) 119) 120) 221) 222) 223) 424) 225) 426) 327) 428) 229) 330) 4

CHEMISTRY31) 332) 1 33) 234) 2 35) 236) 437) 1 38) 439) 3 40) 341) 342) 1 43) 3 44) 245) 3 46) 4 47) 1 48) 2 49) 450) 451) 2 52) 253) 2 54) 3 55) 3 56) 2 57) 1 58) 1 59) 2 60) 3MATHS61) 362) 463) 164) 265) 366) 467) 468) 169) 270) 471) 372) 173) 274) 175) 176) 377) 378) 179) 380) 281) 182) 183) 484) 285) 186) 387) 488) 389) 190) 4

PHYSICS

01.

EqMg

02.It is a balanced wheatstone bridge.

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

Voltage drop across parallel combination of and isV2 = 6 4.2 = 1.8 V

If I2 is the current in resistor, then

04.Since

Similarly

Since,

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.

06.According to following figure also

07.Torque experienced by the coil is

This torque will be balanced by the torque produced by weight of coil (acting from the centre)

08.At t = 0 i.e. when the key is just pressed, no current exists inside the inductor. So and resistors are in series and a net resistance of exists across the circuit.

Hence As , the current in the inductor grows to attain a maximum value ie.e. the entire current passes through the inductor and no current passes through resistor.

Hence

09.

10.

At the time of starting , so

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

Also

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).

13. and

New velocity 14.Let h be the distance of the apex from the middle of the base. Further by Pythagoras theorem

15.16.For 1st reading of oscillator

or 512 HzFor 2nd reading of oscillator

or 504 Hz

A has a frequency of 516 Hz.

17.

Since

18.

For TIRi > C

19.

20.

21.

When either of the two slits is covered then

22.Shift,

Here

23.As N2 < N1Number of atoms disintegrated in time (t2-t1) is

24.According to Ritz combination principle

25.

26.

27.

(equation of a straight line)

28.

29.

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-diodePotential difference acrossRL = 12 0.3 = 11.7 VPotential of Y = 11.7VCHEMISTRY

31.n-factor =10

32.n-factor =0.85

33.

= 11.52 J/K

= -11 J/K

=0.52 J/K

34.

= - 6.2 cal / mole

35.

x = 0.1y = 0.05

Total pressure = = 3 (x + y)= 0.45 atm36.Conceptual37.Conceptual38.Conceptual39.Conceptual

40.

41.

42.Conceptual43.Conceptual44.Conceptual45.Conceptual

46.(i) For

n = 2

(ii)

(iii)

n = 0

(iv)

47.Conceptual48.Conceptual49.At, 710C

50.51.Conceptual52.Conceptual53.If I2 reacts with excess Cl2 - water it is oxidized to HIO3.54.XYApropynemethaneBN2ON2CNH3NH3DN2ONO55.Conceptual56.Conceptual57.Conceptual58.Conceptual59.Conceptual60.Conceptual

MATHEMATICS

61. A is set of all point lying with in the square fromed by x = 2, y =2, x = 4, y = 4

lying with in and on the circum

Circle

62.For any a , for

For and

a,b,c have same sign

(a c) equivalence

63.From the graph of y = [x] and y + x = 1

Domain is

-2-10123Y+x=1

64.If f is one-one f1(x) > 0 3x2+2(a+2)x+3a>0

65.using L1 hospital

x=1x=0

Required area =

66.

67.

68.

69.

70. is always an integer

also

f(x) is constant function

Differentiable

71.

At

72. is decreasing .(1)

For from .(1) and also sin(2x) > 0 h1(x) > 0Increasing

73.

Eq. of tangent

OB = minimum value of

74. Area

75. and

76.

77.since

78.Let such that

parallel to x 4y + 3z =1

79.

80.required vector is unit vector perpendicular to the vectors i-j+k and 2i+j-3k,

81.

82. and

for 2 district solutions b2-4ac>0

largest k value 1

83.

84.

85. solution x = 0

roots

86.

87.3x2 + 5y = 32 down word parabola with vertex at

B(3,1)y =x-2y=x+2(-2,4)A423-2solving, coordinates of A and B are (-2, 4), (3, 1) required area

88. symmetric about x and y axes.

-101Required area put

89.

90. for orthogonal trajectory