Answer and Solution

1.(C)

Sol. where r1 and r2 are radii of spheres A and B.

Now, when B is earthed, potential of B becomes zero. Thus,

Now,

2.(D)

Sol.

Solving

3.(B)4.(B)Sol.It is independent of the temperature.5.(B)

Sol.

The internal resistance must be equal to external resistance for maximum power transfer.

The Req for circuit =

Thus, . Thus 6.(B)

Sol.

4y = 2 ; y = 0.5 m7.(C)Sol.Above equivalent circuit can be drawn as follow

Where

Time constant ; Correct answer (C)8.(C)Sol.Since voltmeter is ideal and reading of voltmeter is zero. So, either fault is in box or 5 resistance.So, sum of potential difference across box and 5 resistance will be 12 volt9.(C)

Sol.

; ; 10.(D)Sol.Drawing F.B.D. of rod PQ

Fx = 0Fy = 0 N = mg .. (1) about centre of mass = 0

Hence

Thus,

11.(D)

Sol.

For V to be positive

12.(B)Sol.Magnitude of force on AB (F1) and magnitude of force on CD (F2) will be equal in magnitude and opposite in direction.

Hence resultant of will be away from the long straight wire.13.(A) Sol.COM

.. (1)

(2)From (1) and (2)

;

Angle rotated by radian and angle rotated by radian14.(D)Sol.Since P1V1 > P2V2 TA > TB

Total work done is positive and efficiency is positive quantity . Therefore there must be some heat absorbed in the process.During adiabatic process Q = 0 therefore during linear process heat will enter and leaves system at different times.15.(A)Sol.Velocity of image of particle B

velocity of image of particle A

Relative velocity of image

16.(B) Sol.Backlash error occurs only in instruments using screws

17.(a,b,d) Sol. wavelength end correction

or

18.(c,d)Sol.In the three experiments, the jockey is moved according to the direction of deflection of the galvanometer.19.(a,b)20.(a,c,d)Sol.Charge Q must be induced on inner surface.

So + 2Q charge appears an outer surface.

21.(a,c)

Sol.

for x to minimum

22.(B)23.(a,b,d)Sol.Option a and b are of type y = f(t) g(x) and option d is the superposition of two waves traveling in opposite direction 24.(a,b)25.(c, d)

26.(C)

Sol.The disc will stop (Translation) at time Since, = I

and Also at the same time, linear speed also ceases. V = v0 gtO = v0 gtT = v0/gThus it will not regain .Since linear speed and angular speed becomes zero at the same time.27.(D)Sol.At time t = 4v0/5g the angular speed will be zero.At this instant v = v0/5.

Now when pure rolling starts, it will take a time from t = 0

Thus finally . Hence first will decrease from to zero and then increase from zero to . Hence (D) is correct.28.(C)Sol.COAM : about point of contact

; 29.(D)Sol.The cube will become spherical because of surface tension.30.(C)

Sol.

31.(A)

Sol. ; Where 32.(B)

Sol.After a long time inductor will short the two resistances. Thus, current will be only.33.(D)

;

Now for maximum rate of power generation

Thus, 34.(B)

Sol.At , the power is

On solving, we get

35.(C)Sol.As pressure is constant, thus it is a isobaric process.36.(D)Sol.For isobaric process

T = 500 k 37.(B)38.A (p,q) ; B (s) ; C (p,q) ; D (r)

39.

= = 6 106 N mAns is 6

40.

Also,

Now,

Ans is 14

41.. After connecting XC = 400 . Power factor becomes 1. box contains inductor of XL = 400 ;

Power factor of Box = .Ans is 6

42. Nmax = Mg + KR [sec 1] cos

= Mg +

= = Mg + 2Mg = 3Mg; Nmin = MgThen the ratio = 3Ans is 3

43.a1 = R a22 = 2 4 = 3 rad/s2.Ans is 3

44. Velocity of bullet + ice block, V = m/sV =4 m/s

Loss of K.E. =

= [0.01(400)21(4)2] = [1600 16]= (1584/2) J

Heat generated = = 95 Cal

Mass of ice melted = = 1.2 gm.Ans is 12

45. Ans is 7

46. Let us mark the capacitors as 1, 2, 3 and 4 for identifications. As is clear, 3 and 4 are in series, and they are in parallel with 2. Then 2, 3, 4 combine is in series with 1.

The `q on 1 is 48 C, thus V1=q/c=6v[v1 = ]VPQ = 10 6 = 4VBy symmetry of 3 and 4, we say, VAB = 2V.Ans is 2

47. P.d.across the points = V2 V1 = = 2 6 + 12 = 4 voltsAns is 4

48. As internal resistance of an ideal voltmeter is infinite, the resistance of the battery across which it is connected will not change by its presence as r' = rNow as the given 8 batteries are discharging in series i.e. Eeq = 8 5 = 40 V and req = 8 0.2 = 1.6 so current in the circuit I = = 25 AHence potential difference across the required battery v = E Ir = 5 25 0.2 = 0V.Ans is 049. ans is 15

50. x = 3t2

Force acting on particle = ma = 1 6 = 6NDisplacement = 2

work = ans is 12

51 Gentle Placing. No external force. Apply conservation servation of momentum.

Ans is 5

52 Let Pulley P moves up with velocity VP thenVP + 12 = 2 4 VP = 4 m/s (up)So VB = 4 + 3 = 7 m/s (up)Ans is 7

53. Power of reactor Where 'n' is number of fissions 't' is time 'E' is energy released per fission

Ans is 6

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Answer and Solution