Physics, asked by shubhjoshi9457, 1 year ago

The length of a second’s pendulum at the surface of earthis 1 m. The length of second’s pendulum at the surface of moon where g is 1/6th that at earth’s surface is(a) 1/6 m(b) 6 m(c) 1/36 m(d) 36 m

Answers

Answered by avengerthorraknarok
0

does drop WP will amp will work

Answered by YTTechnicalGuy
0

of length L is given by :

T=2π∗sqrt(L/g)

Since a second’s pendulum advances one second, every time the pendulum reached an extreme position and the pendulum reaches its extreme position twice in one oscillation, the time period of a seconds pendulum is 2 seconds.

Put T=2 seconds and g= 1.62519 m/s^2 , we get

L=(1.62519/3.142)=0.164m

On earth surface, this length is approximately 1m.

Basically, time period depends on the ratio of L/g . Since the value of g on the surface of moon is about 16.6% of that on the earth, to get the same time period we have to reduce the length of the pendulum to approximately the same percentage to maintain the ratio.

Hope this helps :)

4.6k Views ·

Your feedback is private.

Is this answer still relevant and up to date?

as T is under root L so so T should be 6......

265 Views

Your feedback is private.

Is this answer still relevant and up to date?

If time period is kept constant, length of a seconds pendulum is directly proportional to acceleration due to gravity. As its length on Earth is approximately 1 m, it will be about 1/6th of a metre on the moon.

Similar questions