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?

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Payal Gupta | Contributor-Level 10

7 months ago

This is a multiple choice answer as classified in NCERT Exemplar

Time period of simple pendulum T=2s

For simple pendulum T= $2 π \sqrt[]{\frac{l}{g}}$ where l is length and g = acceleration due to gravity.

T_e=2 $π \sqrt[]{\frac{l_{e}}{g_{e}}}$

On the surface of the moon T_m= 2 $π \sqrt[]{\frac{l_{m}}{g_{m}}}$

$\frac{T_{e}}{T_{m}}$ = $\frac{2 π}{2 π} \sqrt[]{\frac{l_{e}}{g_{e}}} * \sqrt[]{\frac{g_{m}}{l_{m}}}$

T_e=T_m to maintain the second's pendulum time period

1= $\sqrt[]{\frac{l_{e}}{g_{e}}} * \sqrt[]{\frac{g_{m}}{l_{m}}}$ …………….1

But the accel

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