JEE physics

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Question
CBSEENPH11020389

The velocity of a particle is v = v0 + gt + ft2. If its position is x = 0 at t = 0, then its displacement after unit time (t = 1) is

  • v0 + 2g + 3f

  • v0 + g/2 + f/3

  • v0 + g + f

  • v0 + g/2 + f

Solution

B.

v0 + g/2 + f/3

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Question
CBSEENPH11020392

A block of mass ‘m’ is connected to another block of mass ‘M’ by a spring (mass less) of spring constant ‘k’. The blocks are kept on a smooth horizontal plane. Initially the blocks are at rest and the spring is stretched. Then a constant force ‘F’ starts acting on the block of mass ‘M’ to pull it. Find the force on the block of mass ‘m’.

  • mF/M

  • (M+m)F/m

  • mF/(m+ M)

  • MF (m+M)

Solution

C.

mF/(m+ M)

Question
CBSEENPH11020396

If gE and gm are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan’s oil drop experiment could be performed on the two surfaces, one will find the ratio (electronic charge on the moon/ electronic charge on the earth) to be

  • 1

  • 0

  • gE/gM

  • gM/gE

Solution

A.

1

Question
CBSEENPH11020397

A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the circumferences of the discs coincide. The centre of mass of the new disc is α/R from the centre of the bigger disc.The value of α is

  • 1/3

  • 1/2

  • 1/6

  • 1/4

Solution

A.

1/3

In this question distance of centre of mass of new disc is αR not α/R.

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Question
CBSEENPH11020398

A round uniform body of radius R, mass M and moment of inertia ‘I’, rolls down (without slipping) an inclined plane making an angle θ with the horizontal. Then its acceleration is

  • fraction numerator straight g space sin space straight theta over denominator 1 space plus space straight I divided by MR squared end fraction
  • fraction numerator straight g space sin space straight theta over denominator 1 plus space MR squared divided by straight I end fraction
  • fraction numerator straight g space sin space straight theta over denominator 1 minus straight I divided by MR squared end fraction
  • fraction numerator straight g space sin space straight theta over denominator 1 minus MR squared divided by straight I end fraction

Solution

A.

fraction numerator straight g space sin space straight theta over denominator 1 space plus space straight I divided by MR squared end fraction WiredFaculty
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