Subject

Physics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

21.

The momentum of a body is increased by 25%. The kinetic energy is increased by about 

  • 25 %

  • 5 %

  • 56 %

  • 38 %


22.

A particle of mass m = 5 units is moving with a uniform speed v = 32 m in the XOY plane along the line Y = X + 4. The magnitude of the angular momentum about origin is

  • zero

  • 60 unit

  • 7.5 unit

  • 402 unit


23.

A string of density 7.5 g cm3 and area of cross-section 0.2 mm2 is stretched under a tension of 20 N. When it is plucked at the mid-point, the speed of the transverse wave on the wire is

  • 116 ms-1

  • 40 ms-1

  • 200 ms-1

  • 80 ms-1


24.

A work of 2 x 10-2 J is done on a wire of length 50 cm and area of cross-section 0.5 mm2. If the Young's modulus of the matenal of the wire is 2 × 1010 Nm-2, then the wire must be

  • elongated to 50.1414 cm

  • contracted by 2.0 mm

  • stretched by 0.707 mm

  • of length changed to 49.293 cm


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25.

Water rises in a capillary tube to a height h. Choose false statement regarding capillary rise from the following

  • On the surface of Jupiter, height will be less than h

  • In a lift moving up with constant acceleration height is less than h

  • On the surface of moon the height is more than h

  • In a lift moving down with constant acceleration height is less than h


26.

The mstantaneous displacement of a simple harmonic oscillator is given by y = Acos ωt + π4. Its speed will be maximum at the time 

  • 2πω

  • ω2π

  • ωπ

  • π4ω


27.

A particle of mass 5 g is executing simple harmonic motion with amplitude of 0.3 m and time period π/5s. The maximumvalue of the force acting on the particle is

  • 5 N

  • 4 N

  • 0.15 N

  • 0.3 N


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28.

In a gas, two waves of wavelengths 1 m and 1.01 m are superposed and produce 10 beats in 3 s. The velocity of sound in the medium is

  • 300 m/s

  • 336.7 m/s

  • 360.2 m/s

  • 270 m/s


B.

336.7 m/s

The velocity (v) is given by

     v =  f = vλ

Number of beats = difference in frequencies

 f1 - f2 = v 1λ1 - 1λ2           103 = v 11 - 11.01           103 = v 0.011.01          v = 10 × 1.013 × 0.01 = 336.7 m/s


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29.

The pressure inside two soap bubbles is 1.01 and 1.02 atmosphere respectively. The ratio of their respective volumes is

  • 2

  • 4

  • 6

  • 8


30.

In a certain region of space there are only 5 molecules per cm3 on an average. The temperature there is 3 K.The pressure of this dilute gas is  (k = 1.38 x 10-23 J/K)

  • 20.7 × 10-17 N/m2

  • 15.3 × 10-15 N/m2

  • 2.3 × 10-10 N/m2

  • 5.3 × 10-5 N/m2


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