Question

Calculate the length of a second's pendulum on the surface of moon,where g is 1/6th the value of 'g' on the surface of earth.[length of second's pendulum on the surface of earth is 1m]

Answer 1

$$\begin{gathered} \mathrm{Time}\mathrm{period}: \\ \mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g}}} \\ \mathrm{Length}\mathrm{of}\mathrm{seconda}\mathrm{pendulum}\mathrm{on}\mathrm{moon}: \\ \mathrm{L}=\frac{{\mathrm{gT}}^2}{4{\mathrm{\pi }}^2} \\ =\frac{{\displaystyle \frac{1}{6}}\times 9.8\times 2^2}{4{\mathrm{\pi }}^2} \\ =0.166\mathrm{m} \end{gathered}$$