A second-pendulum reads correct time on the earth surface. What will be the time period of it if it is placed on the moon's surface? The radius and mass of the earth is 4 times and 81 times that of the moon 

respectively.

(b); $g_{m}=\frac{G_{m}}{R_{m}^{2}}=\frac{G_{M}}{R_{e}^{2}} \times \frac{4^{2}}{81}=\frac{16}{81} \times 9.8=1.9358 \mathrm{~ms}^{-2} \Rightarrow T=2 \pi \sqrt{\frac{L}{g}}$

$\therefore \mathrm{T}_{\mathrm{m}}=\mathrm{T}_{\mathrm{e}} \sqrt{\frac{\mathrm{g}_{\mathrm{e}}}{\mathrm{g}_{\mathrm{m}}}}=2 \sqrt{\frac{81}{16}}=2 \times \frac{9}{4}=4.5 \mathrm{~s}$