Subject

Physics

Class

JEE Class 12

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JEE Physics 2013 Exam Questions

Multiple Choice Questions

11.

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and cylinder have equal cross sectional area A. When the piston is in equilibrium, the volume of the gas is V0 and its pressure is P0. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency

  • fraction numerator 1 over denominator 2 straight pi end fraction fraction numerator straight A subscript straight gamma straight p subscript 0 over denominator straight V subscript 0 straight M end fraction
  • fraction numerator 1 over denominator 2 straight pi end fraction fraction numerator straight V subscript 0 Mp subscript 0 over denominator straight A squared straight gamma end fraction
  • fraction numerator 1 over denominator 2 straight pi end fraction square root of fraction numerator straight A squared γp subscript 0 over denominator MV subscript 0 end fraction end root
  • fraction numerator 1 over denominator 2 straight pi end fraction square root of fraction numerator MV subscript 0 over denominator straight A subscript straight gamma straight p subscript 0 end fraction end root

C.

fraction numerator 1 over denominator 2 straight pi end fraction square root of fraction numerator straight A squared γp subscript 0 over denominator MV subscript 0 end fraction end root


FBD of piston at equilibrium
⇒ Patm A + mg = P0A

FBD of piston when piston is pushed down a distance x
straight P subscript atm space plus mg space minus space left parenthesis straight P subscript 0 plus dP right parenthesis space straight S space equals space straight m fraction numerator straight d squared straight x over denominator dt squared end fraction space left parenthesis ii right parenthesis
process space is space adiabatic space
rightwards double arrow space PV to the power of straight gamma space equals space straight C
rightwards double arrow straight C space equals dp equals space γPdV over straight V space space.. space left parenthesis iii right parenthesis
from space left parenthesis straight i right parenthesis space comma space left parenthesis ii right parenthesis space and space left parenthesis iii right parenthesis
straight f space equals space fraction numerator 1 over denominator 2 straight pi end fraction square root of fraction numerator straight A squared γP subscript 0 over denominator MV subscript 0 end fraction end root

401 Views

12.

Two charges, each equal to q, are kept at x = −a and x = a on the x-axis. A particle of mass m and charge qo =-q/2 is placed at the origin. If charge qo is given a small displacement (y<< a) along the y-axis, the net force acting on the particle is proportional to

  • y

  • -y

  • 1/y

  • -1/y


A.

y




Net force in negative y -direction,
Fnet = 2F cos θ 
straight F subscript net space equals space fraction numerator 2 kq open parentheses begin display style straight q over 2 end style close parentheses over denominator left parenthesis square root of straight y squared plus straight a squared end root right parenthesis to the power of 3 divided by 2 end exponent end fraction
rightwards double arrow space straight F subscript net space equals space fraction numerator kq squared straight y over denominator straight a cubed end fraction
rightwards double arrow space straight F subscript net space proportional to space straight y 
635 Views

13.

Two capacitors C1 and C2 are charged to 120 V and 200 V respectively. It is found that by connecting them together the potential on each one can be made zero. Then

  • 5C1 = 3C2

  • 3C1 = 5C2

  • 3C1 = 5C2 = 0

  • 9C1 = 4C2


B.

3C1 = 5C2

q1 = q2
C1V1 = C2V2
120C1 = 200C2
⇒ 3C1 = 5C2

418 Views

14.

The supply voltage to a room is 120 V. The resistance of the lead wires is 6 Ω. A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?

  • zero

  • 2.9V

  • 13.3 V

  • 10.04 V


D.

10.04 V

Resistance of bulb =(120x 120)/60 = 240Ω
Resistance of Heater =(120x 120)/240 = 60Ω

Voltage across bulb before heater is switched on,straight V subscript 1 space equals space 120 over 246 space straight x space 240
Voltage across bulb after heater is switched on,straight V subscript 2 equals 120 over 54 straight x 48
A decrease in the voltage is V1 − V2 = 10.04 (approximately)

348 Views

15.

The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is :

  • 3 V/m

  • 6 V/m

  • 9 V/m

  • 12 V/m


B.

6 V/m

Ε = cB
Where E = electric field,
B = magnetic field
c = speed of EM waves
= 3 × 108 × 20 × 10–9 = 6 V/m

216 Views

16.

Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T, the density of liquid is ρ and L is its latent heat of vaporisation. 

  • ρL/T

  • square root of straight T divided by ρL end root
  • T/ρL

  • 2T/ρL


D.

2T/ρL

Decrease in surface energy = Heat required in vapourisation.... (i)
Decrease in surface energy = T x ΔA

 = 2T x 4πrdr  [ For two surface]
heat required in vapourisation = Latent heat
= ML = V x ρL
= (4πr2dr) ρL
Now from eq (i)
2T x 4πrdr = 4πr2ρL
r = 2T/ρL

400 Views

17.

This question has statement I and statement II. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement- I: Higher the range, greater is the resistance of ammeter.
Statement- II: To increase the range of ammeter, additional shunt needs to be used across it.

  • Statement – I is true, Statement – II is true, Statement – II is the correct explanation of statement- I.

  • Statement – I is true, Statement – II is true, Statement – II is not the correct explanation of Statement–I.

  • Statement – I is true, statement – II is false.

  • Statement – I is false, Statement – II is true


D.

Statement – I is false, Statement – II is true

For Ammeter, S = fraction numerator straight I subscript straight g straight G over denominator straight I minus straight I subscript straight g end fraction
So for I to increase, S should decrease, so additional S can be connected across it.

234 Views

18.

In an LCR circuit as shown below both switches are open initially. Now switch S1 is closed, S2 kept open. (q is charge on the capacitor and τ = RC is capacitive time constant). Which of the following statement is correct?

  • Work done by the battery is half of the energy dissipated in the resistor

  • At t = τ, q = CV/2

  • At t = 2τ, q = CV (1-e-2)

  • At t = τ/2, q = CV (1-e-1)


C.

At t = 2τ, q = CV (1-e-2)

For charging of capacitor
q = CV (1-e1/τ)
At t = 2τ, q =CV (1-e-2)

248 Views

19.

If a piece of metal is heated to temperature θ and then allowed to cool in a room which is at temperature θ0 the graph between the temperature T of the metal and time t will be closest to


C.

305 Views

20.

Two short bar magnets of length 1 cm each have magnetic moments 1.20 Am2 and 1.00 Am2 respectively. They are placed on a horizontal table parallel to each other with their N poles pointing towards the South.
They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the mid - point O of the line joining their centres is close to(Horizontal component of earth’s magnetic induction is 3.6 × 10–5 Wb/m2)

  • 3.6 x10-5 Wb/m2

  • 2.56 x10-4 Wb/m2

  • 3.50  x10-4 Wb/m2

  • 5.80 x  x10-4 Wb/m2


B.

2.56 x10-4 Wb/m2

straight B subscript net space equals space straight B subscript straight M subscript 1 end subscript space plus straight B subscript straight M subscript 2 end subscript plus straight B subscript straight H
space equals fraction numerator straight mu subscript 0 straight M subscript 1 over denominator 4 πx cubed end fraction space plus fraction numerator straight mu subscript 0 straight M subscript 2 over denominator 4 πx cubed end fraction space plus straight B subscript straight H
space equals space fraction numerator straight mu subscript straight o over denominator 4 πx cubed end fraction left parenthesis straight M subscript 1 plus straight M subscript 2 right parenthesis space plus space straight B subscript straight H
space equals 10 to the power of negative 7 end exponent over 10 to the power of negative 3 end exponent straight x 2.2 space plus 3.6 space straight X 10 to the power of negative 5 end exponent
equals 2.56 space straight X 10 to the power of negative 4 end exponent space Wb divided by straight m squared
251 Views

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