Q76 / 100 · BITSAT 2022 · Physics
What will be the acceleration due to gravity at a depth $d$, where $g$ is acceleration due to gravity on the surface of earth?
- A$\frac{g}{\left[1+\frac{d}{R}\right]^2}$
- B$g\left[1-\frac{2 d}{R}\right]$
- C$\frac{g}{\left[1-\frac{d}{R}\right]^2}$
- Correct answerD$g\left[1-\frac{d}{R}\right]$
Solution
Acceleration due to gravity at the surface of earth,
$$
g=\frac{G M}{R^2}=\frac{4}{3} \pi \rho G R
$$
$$
\left[\because M=\frac{4}{3} \pi R^3 \cdot \rho\right]
$$
Where, $\rho=$ density of earth.
Acceleration due to gravity at depth $d$ from the surface of earth,
$$
g^{\prime}=\frac{4}{3} \pi \rho G(R-d)
$$
From Eqs. (i) and (ii), we get
$$
g^{\prime}=g\left[1-\frac{d}{R}\right]
$$
Hence, the correct answer is option: 'D'.