The depth at which the effective value of acceleration due to gravity is g/4 g is (R = radius of the earth)

The correct option is 3R/4.

To find the depth at which the effective value of acceleration due to gravity becomes:
$\frac{g}{4}$
we can use the formula for acceleration due to gravity at a depth 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 Earth's surface,
• $d$ is the depth below the surface, and
• $R$ is the radius of the Earth.

Given that the effective value of acceleration due to gravity at depth $d$ is:
$g_d=\frac{g}{4}$
we can substitute this value into the formula:

$\frac{g}{4}=g\left(1-\frac{d}{R}\right)$

Divide both sides of the equation by $g$:
$\frac{1}{4}=1-\frac{d}{R}$

Rearranging the equation to solve for $\frac{d}{R}$:
$\frac{d}{R}=1-\frac{1}{4}$

This gives:
$\frac{d}{R}=\frac{3}{4}$

Solving for the depth $d$:
$d=\frac{3R}{4}$

Therefore, the acceleration due to gravity becomes $\frac{g}{4}$ at a depth of $\frac{3R}{4}$ below the surface of the Earth.