If a clock based on oscillating pendulum is taken from the earth to moon, it will
- AOption (a)
Become slow
- BOption (b)
Become fast
- COption (c)
Give same time as on the earth
- DOption (d)
Stop working
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Correct answer
Become slow
Why this answer is correct
Concept:
- The time period of the simple pendulum is given by, \(T=2\pi √{\frac{l}{g}} \), Where, l = length of the string, T = time period, g = acceleration due to gravity.
- The acceleration on the moon is one-sixth of the acceleration due to gravity on the earth.
- The time period of the pendulum is inversely proportional to the acceleration due to gravity, \(T\propto \frac{1}{√{g}}\)
Explanation:
The time period of the pendulum is given by, \(T=2\pi √{\frac{l}{g}} \)
From the above expression, the time period of the pendulum is inversely proportional to the acceleration due to gravity,
\(T\propto \frac{1}{√{g}}\)
We know that the acceleration on the moon's surface is one-sixth of the acceleration due to gravity on the earth.
\(g'=\frac{g}{6} \)
\(\frac{T}{T'}= √\frac{\frac{g}{6}}{{g}} \)
\(\frac{T}{T'}= √\frac16\)
T' √6 T
So, when the pendulum is taken on the moon the value of g will decrease. Hence, the pendulum takes more time to complete one vibration. It will become slow.

