Gravitation

Sponsor Area

Question
CBSEENPH11020710

(1) Centre a gravity (CG) of a body is the point at which the weight of the body acts.

(2) Centre  of mass coincides with the centre of gravity if the earth is assumed to have infinitely large radius

(3) To evaluate the gravitational field intensity due to anybody can be considered to be concentrated at its CG.

(4) The radius of gyration of anybody rotating about an axis is the length of the perpendicular dropped from the CG of the body to the axis.

Which one of the following pairs of statements is correct?

  • (4) and (1)

  • (1) and (2)

  • (2) and (3)

  • (3) and (4)

Solution

A.

(4) and (1)

Centre of gravity of a body is the point at which the weight of the body acts and the radius of gyration of any body rotating about an axis is the length of the perpendicular dropped from the CG of the body to the axis.

Sponsor Area

Question
CBSEENPH12039771

A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would earth (mass = 5.98 x 1024 kg) have to be compressed to be a black hole?

  • 10-9 m

  • 10-6 m

  • 10-2 m

  • 100 m

Solution

C.

10-2 m

For a black hole, escape speed is more than the speed of light.
Escape velocity is given by,
straight v subscript straight e space equals space square root of fraction numerator 2 GM over denominator straight R apostrophe end fraction end root space semicolon
where space straight R apostrophe space is space the space new space radius
Approximating the escape velocity to the speed of light, we have
space space space space space space space space straight c space equals space square root of fraction numerator 2 GM over denominator straight R apostrophe end fraction end root

rightwards double arrow space straight c squared space equals space fraction numerator 2 GM over denominator straight R apostrophe end fraction
straight R apostrophe space equals space fraction numerator 2 GM over denominator straight c squared end fraction
space space space space space equals space fraction numerator 2 space straight x space 6.67 space straight x space 10 to the power of negative 11 end exponent space straight x space 6 space straight x space 10 to the power of 24 over denominator 9 space straight x space 10 to the power of 16 end fraction
space space space space space equals space fraction numerator 4 space straight x space 6.67 over denominator 3 end fraction straight x 10 to the power of negative 3 end exponent
space space space space space equals space 8.89 space straight x space 10 to the power of negative 3 end exponent
space space space space space equals space 0.889 space straight x space 10 to the power of negative 2 end exponent
space space space space space almost equal to space 10 to the power of negative 2 space end exponent straight m

Question
CBSEENPH11020721

A body mass 1 kg is thrown upwards with velocity 20 ms-1. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction?  (g = 10 ms-2)

  • 20 J

  • 30 J

  • 40 J

  • 10 J

Solution

A.

20 J

The energy lost due to air friction is equal to the difference of initial kinetic energy and final potential energy.
Initially, body possess only kinetic energy and after attaining a height the kinetic energy is zero.
Therefore, loss of energy = KE - PE
=1 half mv squared minus mgh

equals space 1 half space straight x 1 space straight x space 1400 space minus 1 space straight x space 18 space straight x space 10

space equals space 200 minus 80 space straight J space equals space 20 space straight J

1 half mv squared minus mgh

equals space 1 half space straight x 1 space straight x space 1400 space minus 1 space straight x space 18 space straight x space 10

space equals space 200 minus 80 space straight J space equals space 20 space straight J

Question
CBSEENPH11020634

A body projected vertically from the earth reaches a height equal to earth's radius before returning to the earth. The power exerted by the gravitational force is greatest

  • at the instant just before the body hits the earth

  • it remains constant all through

  • at the instant just after the body is projected

  • at the highest position of the body

Solution

A.

at the instant just before the body hits the earth

We know
P = F. v = Fv cos θ
so just before hitting θ is zero and both F and v are maximum

Question
CBSEENPH11020635

A boy standing at the t op of a tower of 20 m height drops a stone. Assuming g = 10 ms-2, the velocity with which it hits the ground is 

  • 20 m/s

  • 40 m/s

  • 5 m/s

  • 10 m/s

Solution

A.

20 m/s

Given, g = 10 ms-2 and h = 20 m
We have
straight v space equals space square root of 2 gh end root

space equals space square root of 2 space straight x space 10 space straight x 20 end root space equals space square root of 400 space equals space 20 space ms to the power of negative 1 end exponent

20