Question

On the surface of the earth acceleration due to gravity will be $g$ and gravitational potential will be $V$.

	COLUMN 1		COLUMN 2
a	At height$h=R$, value of $g$	p	Decreases by a factor \[\dfrac{1}{4}\]
b	At depth\[h=\dfrac{R}{2}\], value of $g$	q	Decreases by a factor \[\dfrac{1}{2}\]
c	At height$h=R$, value of $V$	r	Increases by a factor \[\dfrac{11}{8}\]
d	At depth\[h=\dfrac{R}{2}\], value of $V$	s	Increases by a factor \[2\]
		t	None

$\begin{align}
  & A.\left( a-p \right),\left( b-q \right),\left( c-s \right),\left( d-t \right) \\
 & B.\left( a-q \right),\left( b-p \right),\left( c-t \right),\left( d-s \right) \\
 & C.\left( a-t \right),\left( b-s \right),\left( c-p \right),\left( d-q \right) \\
 & D.\text{none of these} \\
\end{align}$

Answer 1

Hint: The value of acceleration due to gravity at the surface can be found by taking the ratio of the product of the gravitational constant and the mass of the earth to the square of the radius of the earth. The acceleration due to gravity at a specific height from the surface of earth will be found by taking the ratio of the product of the gravitational constant and the mass of the earth to the square of the height on the surface.

Complete step by step answer:
Value of $g$ at a distance $r$ from the centre of the earth $=\dfrac{GM}{{{r}^{2}}}$
The value of $g$ at the surface \[=\dfrac{GM}{{{R}^{2}}}\]
The value of $g$ at a height \[R\] from the surface \[=\dfrac{GM}{{{\left( 2R \right)}^{2}}}\]
Therefore the acceleration due to gravity decreased by a factor of \[\dfrac{1}{4}\].
Dependence of the acceleration of gravity $g$ on the depth \[h\] be,
\[{{g}_{h}}=\dfrac{4\pi }{3}G\rho h\]
Hence it will decrease by a factor of \[\dfrac{1}{2}\] ​ at \[h=\dfrac{R}{2}\].
 The gravitational potential varies with distance \[r\] from the centre of the earth as \[\dfrac{-GM}{r}\].
Hence at the surface will be having a value be \[\dfrac{-GM}{R}\]​
At a height \[R\], it will be having a value as, \[\dfrac{-GM}{2R}\]​
Therefore it will be increased by a factor \[2\], as the potential is negative.

So, the correct answer is “Option A”.

Note: The gravitational potential at a point will be equivalent to the work per unit mass that is required to move a body to that point from a fixed reference point. It will be analogous to the electric potential where the mass is playing the role of the charge. In short the gravitational potential at a point will be the gravitational potential energy per unit time.