Question

On the surface of earth acceleration due to gravity is g and gravitational potential is V.

Column-I	Column-II
a) At height h = R, value of g	(p) decreases by a factor 1/4
(b) At depth h = R/2 value of g	(q) decreases by a factor 1/2
(c) At height h = R, of V	(r) Increases by a factor 11/8
(d) At depth h = R/2 of V	(s) Increases by a factor 2 (t) None

• A. (a - p), (b - q), (c - s), (d - t)
• B. (a - q), (b - p), (c - t), (d - s)
• C. (a - t), (b - s), (c - p), (d - q)
• D. None of these

Answer 1

Answer: (A)

(a - p), (b - q), (c - s), (d - t)

$\text{Value of g at a distance 'r' from center of earth}=\frac{GM}{r^2}$

$\text{'g' at surface}=\frac{GM}{R^2}$

${}^{\prime }{g}^{\prime }\text{at height}{}^{\prime }{R}^{\prime }\text{from surface}=\frac{GM}{(2R{)}^2}$

$\text{Hence decreased by a factor of}\frac{1}{4}.$

$\text{Dependence of 'g' on the depth h is}{g}_h=\frac{4\pi }{3}G\rho h$

$\text{Hence decreases by a factor of}\frac{1}{2}\text{at}h=\frac{R}{2}.$

$\text{Gravitational potential varies with distance 'r' from center of earth as}-\frac{GM}{r}$

$\text{Hence at surface it has value}-\frac{GM}{R}$

$\text{At a height 'R' it has value}-\frac{GM}{2R}$

$\text{Hence it has 'increased' by a factor 2, since potential is negative.}$