1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

At what depth from the surface of earth, the time period of a simple pendulum is 0.5% more than that on the surface of the earth? (Radius of earth is 6400 km)

A
32 km
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
B
64 km
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
96 km
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
D
128 km
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

The correct option is B 64 kmThe acceleration due to gravity at a depth d below the surface of the Earth is given by g′=g(1−dR) The time period of a simple pendulum is given by T=2π(lg)1/2 Hence, T is inversely proportional to g1/2 for the same pendulum. ∴T′T=(gg′)1/2 .....(1) From question, T′=T+0.5100T=1.005T putting all data in equation (1), we get 1.005TT=⎛⎜ ⎜ ⎜ ⎜⎝gg(1−dR)⎞⎟ ⎟ ⎟ ⎟⎠1/2 ⇒1.005=1(1−dR)1/2 Substituting, R=6400 km, we get d=63.5≈64 km Hence, option (b) is correct.

Suggest Corrections
3
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program