The escape velocity for the earth is 11.2 km/s. The mass of another planet 100  times mass of earth and its radius is 4 times radius of the earth. The escape velocity for the planet is

• 280 km/s

• 56.0 km/s

• 112 km/s

• 56 km/s

Change in acceleration due to gravity is same upto a height h from each other the earth surface and below depth x then relation between x and his ( h and x<<<R_e )

• x = h

• x = 2h

• x = $\frac{\mathrm{h}}{2}$

• x = h²

 A uniform ring of mass m and radius a is placed directly above a uniform sphere of mass m and of equal to radius. The centre of the ring is at a distance $\sqrt{3}$a from the centre of the sphere. The gravitational force (Fl exerted by the sphere on the ring is

• $\frac{\sqrt{3} \mathrm{G} \mathrm{Mm}}{8 {\mathrm{a}}^2}$

• $\frac{2 \mathrm{G} \mathrm{M} \mathrm{m}}{3 {\mathrm{a}}^2}$

• $\frac{7 \mathrm{G} \mathrm{Mm}}{\sqrt{2} {\mathrm{a}}^2}$

• $\frac{3 \mathrm{G} \mathrm{Mm}}{{\mathrm{a}}^2}$

A body of mass 2m is placed on earth's surface. Calculate the change in gravitational potential energy, if this body is taken from earth's surface to a height of h, where h = 4R.

• $\frac{2 \mathrm{mgh}}{\mathrm{R}}$

• $\frac{2}{3} \mathrm{mgR}$

• $\frac{8}{5} \mathrm{mgR}$

• $\frac{\mathrm{mgR}}{2}$

A communication satellite of 500 kg revolves around the earth in a circular orbit of radius 4.0 x 10⁷ m in the equatorial plane of the earth from west to east. The magnitude of angular momentum of the satellite is

• ∼ 0.13  × 10¹⁴ kg m² s^-1

• ∼ 1.3  × 10¹⁴ kg m² s^-1

• ∼ 0.58 × 10¹⁴ kg m² s^-1

• ∼ 2.58  × 10¹⁴ kg m² s^-1

The change in the gravitational potential energy when a body of mass m is raised to a height nR above the surface of the earth is ( here R is the radius of the earth )

• nn + 1 mgR

• nn - 1 mgR

• nmgR

• mgRn

147.

Assertion:  Two bodies of masses M and m (M > m) are allowed to fall from the same height if the air resistance for each be the same then both the bodies will reach the earth simultaneously. 

Reason:  For same air resistance, acceleration of both the bodies will be same.

• If both assertion and reason are true and reason is the correct explanation of assertion.

• If both assertion and reason are true but reason is not the correct explanation of assertion.

• If assertion is true but reason is false.

• If both assertion and reason are false.

Assertion: Kepler's second law can be understood by conservation of angular momentum principle.

Reason:  Kepler's second law is related with areal velocity which can further be proved to be based on conservation of angular momentum as (dA/dt) = ( r²ω / 2 )

• If both assertion and reason are true and reason is the correct explanation of assertion.

• If both assertion and reason are true but reason is not the correct explanation of assertion.

• If assertion is true but reason is false.

• If both assertion and reason are false

Time period of pendulum, on a satellite orbiting the earth, is

• $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{\pi }$}\right.$

• zero

• $\mathrm{\pi }$

• infinity

The additional kinetic energy to be provided to a satellite of mass m revolving around a planet of mass M to transfer it from a circular orbit of radius R, to another of radius R₂ (R₂ > R₁ ) is

• GmM $\left(\frac{1}{{\mathrm{R}}_1^2} - \frac{1}{{\mathrm{R}}_2^2}\right)$

• GmM $\left(\frac{1}{{\mathrm{R}}_1} - \frac{1}{{\mathrm{R}}_2}\right)$

• 2 GmM $\left(\frac{1}{{\mathrm{R}}_1} - \frac{1}{{\mathrm{R}}_2}\right)$

• $\frac{1}{2}$ GmM $\left(\frac{1}{{\mathrm{R}}_1}- \frac{1}{{\mathrm{R}}_2}\right)$