Question

The length of the second's pendulum on the surface of the earth is $1 \, m$ . If the acceleration due to gravity on the surface of the moon is one-sixth of the acceleration due to gravity of the earth, then the length of the second's pendulum on the surface of the moon will be

• A. $\frac{1}{6 \, }m$
• B. $6 \, m$
• C. $\frac{1}{36 \, }m$
• D. $36 \, m$

Answer 1

Answer: (A)

Seconds pendulum is a special pendulum having a time period of $2 \, s$ .
Using the formula $T = 2 \pi \sqrt{\frac{L}{g}}$
If the time period remains the same and $g$ is reduced by a factor of $6$ .
L should be reduced by a factor of 6.
$\Rightarrow L=\frac{1}{6} \, m$ on moon.