The time period of a second's pendulum on the surface of moon will be:

\frac{2}{\sqrt{6}} s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6 \sqrt{2} s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 \sqrt{6} s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2 s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $2 \sqrt{6} s$
On surface of moon, acceleration due to gravity is g/6. Time period of seconds pendulum on earth is 2 seconds.
On Earth, $2 = 2 π \sqrt{\frac{l}{g}}$
On moon, $T_{m o o n} = 2 π \sqrt{\frac{l}{\frac{g}{6}}}$
Dividing the above two equations, we get, $T_{m o o n} = 2 \sqrt{6} s$

Suggest Corrections

Similar questions

Q. The time period of a second's pendulum on the surface of moon is found to be

5 s

. The acceleration due to gravity on the surface of moon is:

Q. The time period of a loaded spring on earth is 6s. On the surface of moon, the time period of the same loaded spring will be :

Q. A second's pendulum has a time period of 2 seconds on the earth. In order to have the same time period of the pendulum on the moon, one has to

What is the effective length of a second's pendulum for a place where g = 9.8 $m s^{- 2}$ ? [Given: The time period of a seconds' pendulum = 2 s]

The time period of a simple pendulum on earth is 2 s. What will be the time period of same pendulum when it is brought on moon? (Value of g at moon is 1/6 of value of g at earth’s surface:

Join BYJU'S Learning Program

The time period of a second's pendulum on the surface of moon will be:

The correct option is A 2√6sOn surface of moon, acceleration due to gravity is g/6. Time period of seconds pendulum on earth is 2 seconds. On Earth, 2=2π√lgOn moon,Tmoon=2π√lg6Dividing the above two equations, we get, Tmoon=2√6s