If the length of a seconds pendulum is increased by `2%` then in a day the pendulum

Options

A. 3927 s
B. 392.7 s
C. 37.38 s
D. 864 s

Answer

D

Explanation

`(T_(2))/(T_(1))=sqrt(1.02)=(1+0.02)^(1//2)=1+0.01`
`(T_(2)-T_(1))/(T_(1))=0.01s`
For second's pendulum `T_(2)=2s`
`therefore T_(2)-T_(1)=0.01xx2=0.02s`
In one oscillation the pendulum lose 0.02 s
`therefore 43200` oscillations the pendulum loses is 864 s