JEE Physics

Question 1

A particle located at x = 0 at time t = 0, starts moving along the positive x-direction with a velocity ‘v’ that varies as v= α√x . The displacement of the particle varies with time as

  • t3

  • t2

  • t

  • t1/2

Solution

B.

t2

straight v equals space straight alpha square root of straight x
dx over dt space equals space straight alpha square root of straight x space space space space open parentheses therefore space straight v space equals space dv over dt close parentheses
fraction numerator dx over denominator square root of straight x end fraction space equals space straight alpha space dt
Perform space integration
integral subscript 0 superscript straight x fraction numerator dx over denominator square root of straight x end fraction space equals space integral subscript 0 superscript straight t straight alpha space dt
because space at space straight t space equals space 0 comma space straight x space equals space 0 space and space let space at space any space time space straight t comma space particle space is space at space straight x right square bracket
rightwards double arrow right enclose space fraction numerator straight x to the power of 1 divided by 2 end exponent over denominator 1 divided by 2 end fraction end enclose subscript 0 superscript straight x space equals space αt
rightwards double arrow space straight x to the power of 1 divided by 2 end exponent space equals space straight alpha over 2 straight t
rightwards double arrow space straight x space equals space straight alpha squared over 4 straight x space straight t squared space rightwards double arrow space straight x proportional to space straight t squared
Question 2

A bomb of mass 16 kg at rest explodes into two pieces of masses of 4 kg and 12 kg. The velocity of the 12 kg mass is 4 ms−1 . The kinetic energy of the other mass is

  • 144 J

  • 96 J

  • 288 J

  • 192 J

Solution

C.

288 J

m1v1 = m2v2
KE space equals space 1 half straight m subscript 2 straight v subscript 2 superscript 2 space
space equals space 1 half space straight x space 4 space straight x space 144 space equals space 288 space straight J

Question 3

A mass of M kg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of 45° with the initial vertical direction is

  • Mg left parenthesis square root of 2 minus 1 right parenthesis
  • Mg left parenthesis square root of 2 plus 1 right parenthesis
  • Mg square root of 2
  • fraction numerator Mg over denominator square root of 2 end fraction

Solution

A.

Mg left parenthesis square root of 2 minus 1 right parenthesis straight F calligraphic l space sin space 45 space equals space M g left parenthesis calligraphic l minus calligraphic l space cos space 45 right parenthesis
F space equals space M g left parenthesis square root of 2 minus 1 right parenthesis
Question 4

A particle of mass 100 g is thrown vertically upwards with a speed of 5 m/s. the work done by the force of gravity during the time the particle goes up is

  • 0.5 J

  • -0.5 J

  • −1.25 J

  • 1.25 J

Solution

C.

−1.25 J

negative mgh space equals negative space mg open parentheses fraction numerator straight v squared over denominator 2 straight g end fraction close parentheses space equals space minus space 1.25 space straight J
Question 5

The potential energy of a 1 kg particle free move along the x-axis is given by
straight V left parenthesis straight x right parenthesis space equals space open parentheses straight x to the power of 4 over 4 minus straight x squared over 2 close parentheses space straight J

The total mechanical energy of the particle 2 J. Then, the maximum speed (in m/s) is

  • 2

  • 3 divided by square root of 2
  • square root of 2
  • 1 divided by square root of 2

Solution

B.

3 divided by square root of 2 kE subscript max space equals space straight E subscript straight T minus straight U subscript min
straight U subscript min space left parenthesis plus-or-minus 1 right parenthesis space equals negative 1 fourth straight J
KE subscript max space equals space 9 divided by 4 space straight J thin space
rightwards double arrow space straight U space equals space fraction numerator 3 over denominator square root of 2 end fraction space straight J