question_answer 53)

                  The length of a second's pendulum on the surface of Earth is 1m. What will be the length of a second's pendulum on the moon?

Answer:

                  \[T=2\pi \,\sqrt{l/g}\] on the surface of moon,                 \[T=2\pi \,\sqrt{\frac{lm}{gm}}\,=\,2\pi \,\sqrt{\frac{6lm}{g}}\] \[\,(\because \,{{g}_{m}}\,=\frac{g}{6})\]                 \[\therefore \] \[6\,l\,m=l\]                 or \[lm\,=\,\frac{1}{6}=\,\frac{1m}{6}\,=0.17\,m\]