The depth d at which the value of acceleration due to gravity becomes n 1/n times the value at the surface, is [R = radius of the earth]

The correct option is R(n-1)/n.

To find the depth at which the acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface, we use the formula for the acceleration due to gravity at a depth $d$ below the Earth's surface:

$g_d=g\left(1-\frac{d}{R}\right)$

where:
• $g_d$ is the acceleration due to gravity at depth $d$,
• $g$ is the acceleration due to gravity at the surface of the Earth, and
• $R$ is the radius of the Earth.

According to the problem statement, the acceleration due to gravity at depth $d$ is:

$g_d=\frac{g}{n}$

Substituting this relation into the depth formula, we obtain:

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

Dividing both sides of the equation by $g$:

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

Rearranging the terms to isolate the fraction with $d$:

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

Simplifying the right-hand side by finding a common denominator:

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

Multiplying both sides by $R$ to solve for the depth $d$:

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

This matches the option $\frac{R(n-1)}{n}$.