If a pendulum clock keepds correct time at sea level is taken to a place 1 km above sea level the clock will approximately

Options

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

Answer

B

Explanation

`g_(h)=g(1-(2h)/(R))`
`therefore g_(h)=g(1-(2xx10^(3))/(6.4xx10^(6)))`
`(T_(2))/(T_(1))=sqrt((g)/(g_(h)))=sqrt((g)/g(1-(1)/(3.2xx10^(3))))`
`(T_(2))/(T_(1))=[1-(1)/(3.2xx10^(3))]^(-1//2)`
`=[1+(1)/(2)xx(1)/(3.2xx10^(3))]`
`therefore (T_(2)-T_(1))/(T_(1))=(10^(-3))/(6.4)=0.15623`
This is the time loses per oscillation
The time loses per day will be
`T_(2)-T_(1)=0.15625xx86400=13.5s`
Note : Time in one day is 86400 seconds
Alternative method :
`(DeltaT)/(T)=-(1)/(2)(Deltag)/(g)`