The depth at which the effective value of acceleration due to gravity is $\frac{g}{4}$ is

$A$ pendulum is oscillating with frequency $n$ on the surface of the Earth. If it is taken to a depth $d = R/3$ below the surface of the Earth,what is the new frequency of oscillation? ($R$ is the radius of the Earth)

$A$ planet has the same density as that of the Earth,and the universal gravitational constant $G$ is twice that of the Earth. The ratio of the acceleration due to gravity on the planet to that on the Earth is:

At what distance above and below the surface of the earth will a body have the same weight? (Take the radius of the earth as $R$.)

Assume that the Earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the Earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is $100 \, g$. The time period of the motion of the particle will be (approximately) (take $g = 10 \, m/s^2$, radius of Earth $R = 6400 \, km$):

$A$ uniform solid sphere of radius $R$ produces a gravitational acceleration of $a_o$ on its surface. The distance of the point from the centre of the sphere where the gravitational acceleration becomes $\frac{a_o}{4}$ is,