JEE Physics

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Question
CBSEENPH11020342

The moment of inertia of a uniform cylinder of length

  • 1

  • fraction numerator 3 over denominator square root of 2 end fraction
  • square root of 3 over 2 end root
  • fraction numerator square root of 3 over denominator 2 end fraction

Solution

C.

square root of 3 over 2 end root straight I space equals space fraction numerator straight m calligraphic l squared over denominator straight I 2 end fraction space plus mR squared over 4
or space straight I space equals space straight m over 4 open parentheses calligraphic l squared over 3 plus straight R squared close parentheses space space.... space left parenthesis 1 right parenthesis
Also comma space straight m space equals space πR squared calligraphic l rho
rightwards double arrow space R squared space equals space fraction numerator m over denominator pi 																</p>
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Question
CBSEENPH11020345

A body of mass m = 10–2 kg is moving in a medium and experiences a frictional force F = –kv2. Its initial speed is v0 = 10 ms–1. If, after 10 s, its energy is 1/8 mv02,the value of k will be

  • 10-4 kg m-1

  • 10–1 kg m–1 s–1

  • 10-3 kg m-1

  • 10-3 kg s-1

Solution

A.

10-4 kg m-1

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Question
CBSEENPH11020346

A slender uniform rod of mass M and length

  • fraction numerator 3 straight g over denominator 2 calligraphic l end fraction space cos space straight theta
  • fraction numerator 2 space straight g over denominator 3 space calligraphic l end fraction space cos space straight theta
  • fraction numerator 3 straight g over denominator 2 space calligraphic l end fraction space sin space straight theta
  • fraction numerator 2 space straight g over denominator 3 space calligraphic l end fraction space sin space straight theta

Solution

C.

fraction numerator 3 straight g over denominator 2 space calligraphic l end fraction space sin space straight theta WiredFaculty
Taking torque about pivot τ = Iα
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Question
CBSEENPH11020349

The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by (R = Earth's radius)

  • WiredFaculty
  • WiredFaculty
  • WiredFaculty
  • WiredFaculty

Solution

B.

WiredFaculty WiredFaculty
Where M is mass of earth
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Question
CBSEENPH11020350

A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λA to λB after the collision is

  • straight lambda subscript straight A over straight lambda subscript straight B space equals space 2 over 3
  • straight lambda subscript straight A over straight lambda subscript straight B space equals space 1 half
  • straight lambda subscript straight A over straight lambda subscript straight B space equals space 1 third
  • straight lambda subscript straight A over straight lambda subscript straight B space equals space 2

Solution

D.

straight lambda subscript straight A over straight lambda subscript straight B space equals space 2 WiredFaculty
By conservation of linear momentum
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