Question 271: Gravitation

Question

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

Accepted Answer

Step 1: Gravity inside the Earth
For a uniform Earth of radius $R$,
$ g(h) = g\left(1 - \frac{h}{R}ight) $
where:
$g$ is surface gravity,
$h$ is depth below the surface.
Equivalently, at radial distance $r = R - h$:
$ g(r) = g \cdot \frac{r}{R} $
Step 2: Set condition
We want:
$ g(h) = \frac{1}{n} g $
So:
$ g\left(1 - \frac{h}{R}ight) = \frac{1}{n} g $
Cancel $g$:
$ 1 - \frac{h}{R} = \frac{1}{n} $
Step 3: Solve for depth
$ \frac{h}{R} = 1 - \frac{1}{n} = \frac{n-1}{n} $
$ h = \frac{R(n-1)}{n} $
Ã¢ÂÂ Final Answer:
$ \boxed{	frac{R(n-1)}{n}} $
Correct Option: A

Answer

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

Answer

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

Answer

$\frac{\mathrm{Rn}}{(\mathrm{n}-1)}$

Answer

$\frac{R}{n}$

Gravitation

The depth at which the value of acceleration due to gravity becomes \left(\frac{1}{n}\right) times the value at the surface of the earth is ( \mathrm{R}= radius of the earth)

Choose the correct answer: