Question

3. Suppose you take a pendulum with length L and mass m having a period T to a

Answer 1

(C)

Explanation:

$t = 2\pi \sqrt{ \frac{l}{g} }$

If g is only 1/6 on another planet, then

$t = 2\pi \sqrt{ \frac{l}{ \frac{g}{6} } } = 2\pi \sqrt{ \frac{6l}{g} }$

$= \sqrt{6} \: (2\pi \sqrt{ \frac{l}{g} } ) = 2.4 \times t(on \: earth)$