Assuming earth to be a sphere of a uniform density, what is the value of gravitational acceleration in a mine 100 km below the earth's surface (Given R = 6400km)

The correct option is 9.66m / s².

Step-by-Step Explanation:

The acceleration due to gravity at a depth $d$ below the surface of the Earth is given by the formula:
$g_d=g\left(1-\frac{d}{R}\right)$
where:
- $g$ is the acceleration due to gravity at the surface of the Earth (approximately $9.8{\mathrm{m/s}}^2$ or more precisely $9.81{\mathrm{m/s}}^2$),
- $d$ is the depth below the Earth's surface,
- $R$ is the radius of the Earth.

From the given problem, we have the following values:
- Depth ($d$) = $100\mathrm{km}$
- Radius of the Earth ($R$) = $6400\mathrm{km}$
- Standard acceleration due to gravity ($g$) = $9.81{\mathrm{m/s}}^2$

Now, substituting these values into the formula:
$g_d=9.81\left(1-\frac{100}{6400}\right)$

Simplify the fraction:
$\frac{100}{6400}=\frac{1}{64}\approx 0.015625$

Substitute this value back into the expression:
$g_d=9.81\times \left(1-0.015625\right)$
$g_d=9.81\times 0.984375$
$g_d\approx 9.66{\mathrm{m/s}}^2$

Therefore, the value of gravitational acceleration in the mine is approximately 9.66m / s².