If a clock based on oscillating pendulum is taken from the earth to moon, it will
- ((a))
Become slow
- ((b))
Become fast
- ((c))
Give same time as on the earth
- ((d))
Stop working
Show Answer
Become slow
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.



































