JEE Physics

Question 1

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

integral subscript 0 superscript straight x space dx space equals space integral subscript 0 superscript 1 space left parenthesis straight V subscript 0 space plus space gt space plus ft squared right parenthesis space dt
straight x space equals straight v subscript 0 space plus space straight g space open parentheses 1 half close parentheses space plus straight f open parentheses 1 third close parentheses
Question 2

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 3

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 4

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.

negative fraction numerator 3 straight M over denominator 4 end fraction space straight alpha space straight R space plus space straight M over 4 space straight R space equals 0
rightwards double arrow space straight alpha space equals space 1 third
Question 5

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
Mg space sin space straight theta space minus space straight f space equals space Ma
fR space equals space straight I straight a over straight R
rightwards double arrow space straight a space equals space fraction numerator straight g space sin space straight theta over denominator open parentheses 1 plus begin display style straight I over MR squared end style close parentheses end fraction