Q131 / 150 · MHT CET 2017 · Physics

The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the earth's surface is $(R$ = radius of the earth $)$

A
$d = R (\frac{n}{n - 1})$
B
$d = R (\frac{n - 1}{2 n})$
Correct answer
C
$d = R (\frac{n - 1}{n})$
D
$d = R^{2} (\frac{n - 1}{n})$

Solution

At depth d we have

g^{'} = g (1 - \frac{d}{R})

g^{'} = \frac{g}{n}

\frac{g}{n} = g (1 - \frac{d}{R})

∴ \frac{1}{n} = 1 - \frac{d}{R}

∴ \frac{d}{R} = 1 - \frac{1}{n} = \frac{n - 1}{n}

d = R (\frac{n - 1}{n})

Hence, the correct answer is option: 'C'.

Relevant Questions

The percentage decrease in the weight of a body when taken to a height of km above the surface of the earth is (Radius of the earth is km)20 Apr, 2026 (M)Acceleration Due to Gravity
Select the correct statement out of the following20 Apr, 2026 (E)Acceleration Due to Gravity
Time period of a simple pendulum is when on the earth's surface and when taken to a height '2R' above the earth's surface, where 'R' is the 18 Apr, 2026 (M)Acceleration Due to Gravity
The lengths of seconds pendulums on the surface of the earth and at an altitude ' ' from the surface of the earth are and respectively. The 18 Apr, 2026 (M)Acceleration Due to Gravity
The acceleration due to gravity at a height 'h' above the surface of earth is ' '. At the depth 90 km below the earth's surface the accelera18 Apr, 2026 (E)Acceleration Due to Gravity
The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh th as much as at present is (17 Apr, 2026 (M)Acceleration Due to Gravity