A body weighs $49 \mathrm{N}$ on a spring balance at the north pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator?

[Use $g=\frac{GM}{R^2}=9.8 \mathrm{m} {\mathrm{s}}^{-2}$ and radius of earth, $R=6400 \mathrm{km}.$]

A solid sphere of radius $R$ gravitationally attracts a particle placed at $3R$ from its centre with a force $F_1.$ Now a spherical cavity of radius $\left(\frac{R}{2}\right)$ is made in the sphere (as shown in figure) and the force becomes $F_2$. The value of $F_1:F_2$ is:

[ Given : radius of earth $=6400 \mathrm{km},$ mass of earth $=6\times {10}^{24} \mathrm{kg}$ ]

Given below are two statements : one is labelled as Assertion $A$ and the other is labelled as Reason $R$.

Assertion $A$ : The escape velocities of planet  $A$ and $B$ are same. But $A$ and $B$ are of unequal mass.

Reason $R$: The product of their mass and radius must be same. $M_1{R}_1=M_2{R}_2$ In the light of the above statements, choose the most appropriate answer from the options given below :

A planet revolving in elliptical orbit has :

A. a constant velocity of revolution.

B. has the least velocity when it is nearest to the sun.

C. its areal velocity is directly proportional to its velocity.

D. areal velocity is inversely proportional to its velocity.

E. to follow a trajectory such that the areal velocity is constant.

Choose the correct answer from the options given below:

Find the gravitational force of attraction between the ring and sphere as shown in the diagram, where the plane of the ring is perpendicular to the line joining the centres. If $\sqrt{8}R$ is the distance between the centres of a ring (of mass $m$) and a sphere (mass $M$) where both have equal radius $R$

In the reported figure of earth, the value of acceleration due to gravity is same at point $A$ and $C$ but it is smaller than that of its value at point $B$ (surface of the earth). The value of $OA:AB$ will be $x:5$. The value of $x$ is