Question

The value of g
$(a)$ Increases with increase in depth
$(b)$ Decreases with increase in depth
$(c)$ Remains same regardless of the depth
$(d)$ Depends upon the mass of object

Answer 1

Hint – In this question use the relationship between g, universal gravitational constant, mass of earth and radius of earth, to figure out the right option.

Complete step-by-step answer:
As we know (g) is known as acceleration of gravity and the value of (g) is constant at the earth's surface or at sea level.
 g = 9.8 m2/sec (at earth surface).
But as we know g is dependent on the radius of the earth which is given as
g = (GM/R2)……………….. (1), where G = universal gravitational constant.
                                                              M = mass of earth.
                                                               R = radius of the earth
Now as we know that the mass of the earth is the volume of the earth multiplied by density.
As we know earth is in the shape of a sphere so the volume (V) of the earth is $\dfrac{4}{3}\pi {r^3}$.
So the mass (M) of the earth is
M =$\dfrac{4}{3}\pi {r^3} \times \rho $
Now substitute this value in equation (1) we have,
g = (GM/R2) = $\dfrac{{\left[ {G \times \dfrac{4}{3}\pi {r^3} \times \rho } \right]}}{{{R^2}}} = G \times \dfrac{4}{3} \times \pi \times R \times \rho $
So as we see (g) is directly proportional to the mass and radius of the earth.
So the value of g is max at the surface of the earth because R is maximum at the surface of the earth.
Now as we increase the depth the mass and the radius of the earth decreases so the acceleration due to gravity decreases.
So this is the required answer.
Hence option (B) is correct.

Note – There is a very interesting fact about g, in case of freefall object objects fall freely with no forces acting upon it except gravity, a defined constant, $g = - 9.8m/{s^2}$, negative sign because it acts in downwards direction.