A pendulum clock, keeps correct time at sea level, when taken to a place 1km below sea level, the clock loses or gain per day will be

Options

A. gains 13.5 s
B. loses 13.5 s
C. loses 7 s
D. gains 7 s

Answer

C

Explanation

The period of clock pendulum `T_(1)=1s`
`T_(1)-2pisqrt((L)/(g_(1)))andT_(2)=2pisqrt((L)/(g_(2)))`
Where `T_(1)` period on the earth and `g_(1)` acceleration due to gravity on the earth. `g_(2)` acceleration due to gravity at depth 1km below the earth and `T_(2)` corresponding period.
`(T_(2))/(T_(1))=sqrt((g_(1))/(g_(2)))=(1-(d)/(R))^(1//2)` `(T_(2)-T_(1))/(T_(1))=-(d)/(2R)`
`therefore (T_(1)-T_(2))/(T_(1))=(d)/(2R)=(1xx10^(3))/(2xx6.4)xx10^(-6)=(10^(-3))/(12.8)`
In one sec. the pen. clock loses time `(10^(-3))/(12.8)s`
`therefore` In one day (86400 osci.) the clock lose time is `(86400xx10^(-3))/(12.8)=6.750`