Subject

Physics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

31.

Two moles of exygen is mixed with eight moles of helium. The effective specific heat of the mixture at constant volume is

  • 1.3 R

  • 1.4 R

  • 1.7 R

  • 1.9 R


32.

On heating, the temperature at which water has minimum volume is

  • 0° C

  • 4° C

  • 4 K

  • 100° C


33.

In damped oscillations, the amplitude of oscillations is reduced to one-third of its initial value a0 at the end of 100 oscillations. When the oscillator completes 200 oscillations, its amplitude must be

  • a02

  • a06

  • a09

  • a04


34.

A particle executes simple harmonic motion with a time period of 16 s. At time t = 2 s, the particle crosses the mean position while at t = 4s, its velocity is 4 ms-1. The amplitude of motion in metre is

  • 2 π

  • 162 π

  • 322π

  • 4π


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35.

For a simple pendulum, the graph between T2 L and is

  • a straight line passing through the origin

  • parabola

  • circle 

  • ellipse


36.

Charges + 2q, + q and + q are placed at the comers A, B and C of an equilateral triangle ABC. If E is the electric field at the circumcentre 0 of the triangle, due to the charge + q, then the magnitude and direction of the resultant electric field at O is

  • E along AO

  • 2E along AO

  • E along BO

  • E along CO


37.

N identical drops of mercury are charged sumultaneously to 10 V. When combined to form one large drop, the potential is found to be 40 V, the value of N is

  • 4

  • 6

  • 8

  • 10


38.

The electrostatic potential energy between proton and electron separated by a distance 1 A is

  • 13.6 V

  • 27.2 eV

  • 14.4 eV

  • 1.44 eV


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39.

The plates of a parallel plate capacitor with air as medium are separated by a distance of 8 mm. A medium of dielectric constant 2 and thickness 4 mm having the same area is introduced between the plates. For the capacitance to remain the same, the distance between the plates is

  • 8 mm

  • 6 mm

  • 10 mm

  • 12 mm


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40.

The resistance of a wire at room temperature 30˜C is found to be 10 Ω. Now to increase the resistance by 10%, the temperature of the wire must be [The  temperature coefficient of resistance of the material of the wire is 0.002/°C]

  • 36°C

  • 83°C

  • 63°C

  • 33°C


B.

83°C

R = R0 1 + αt       R0 1 + 30 α = 10 Ωand       R0 1 + αt   = 11 Ω                        1110 = 1 + αt1 + 30αor           11 + 330 α  = 10 + 10 αtor 11 + 330 × 0.002 = 10 + 10 × 0.002 tor                      11.66 = 0.02 t + 10or                      0.02 t = 1.66or                              t = 83°C


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