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BITSAT 2022 Paper

Section: Physics

Question : 6 of 130

Marks: +1, -0

d

g

Solution: 👈: Video Solution

Acceleration due to gravity at the surface of earth,

g=\frac{GM}{R^2} = \frac{4}{3}\pi\rho G R \text{.................(i)}

\left[\because M=\frac{4}{3}\pi R^3\rho\right]

Where,

\rho =

density of earth.

Acceleration due to gravity at depth d from the surface of earth,

g'=\frac{4}{3}\pi\rho G(R-d) \text{...........(ii)}

From Eqs. (i) and (ii), we get

g' = g\left[1 - \frac{d}{R}\right]

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