Question

Assuming that the moon is a sphere of the same mean density as that of the earth and one quarter of its radius, the length of a seconds pendulum on the moon (its length on the earth's surface is 99.2 cm) is:

• A. 24.8 cm
• B. 49.6 cm
• C. 99.2
• D. $\frac{99.2}{\sqrt{2}}cm$

Answer 1

The correct option is A 24.8 cm

If radius of earth is $R$ then radius of moon will be $R/4$ thus volume of moon will be $(1/64{)}^{th}$ of the volume of earth

thus mass of moon will be $(1/64{)}^{th}$ of the mas of earth

because the mass is directly proportional to volume provided density is same.

As we know that $g=GM/R^2$ is gravity at earth so get the value of gravtiy at moon $\text{just replace M by M/64 and R by R/4}$ we get value of

gravity at moon as $g/4$

For second pendulum $length=l=g{T}^2/4{\pi }^2$ where $T=1second$

as we see that $l\propto g$ so at moon the length will be $\left(1/4\right)$ of length at earth so length at moon will be $99.2/4=24.8cm$