Question 269: Gravitation

Question

The depth at which acceleration due to gravity becomes $\frac{g^{\prime}}{n}$ is ( $R=$ radius of earth, $\mathrm{g}=$ acceleration due to gravity) ( $\mathrm{n}=$ integer)

Accepted Answer

Step 1: Relation between gravity and depth
At depth $d$ below EarthÃ¢ÂÂs surface (assuming uniform density of Earth):
$ g' = g \left(1 - 	frac{d}{R}ight) $
where $R$ = EarthÃ¢ÂÂs radius, $g$ = acceleration due to gravity at surface.
Step 2: Condition given
We want:
$ g' = \frac{g}{n} $
Thus:
$ g \left(1 - \frac{d}{R}ight) = \frac{g}{n} $
Cancel $g$ (non-zero):
$ 1 - \frac{d}{R} = \frac{1}{n} $
Step 3: Solve for depth $d$
$ \frac{d}{R} = 1 - \frac{1}{n} = \frac{n-1}{n} $
So:
$ d = \frac{R(n-1)}{n} $
Final Answer:
$ \boxed{\frac{R(n-1)}{n}} $
Correct Option: A
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Answer

$\frac{R(n-1)}{n}$

Answer

$\frac{R(n+1)}{n}$

Answer

$\frac{R(n-1)^2}{n}$

Answer

$\frac{R(n+1)^2}{n}$

Gravitation

The depth at which acceleration due to gravity becomes \frac{g^{\prime}}{n} is ( R= radius of earth, \mathrm{g}= acceleration due to gravity) ( \mathrm{n}= integer)

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