A clock pendulum made of invar has a period of 0.5s, at `20^(@)C`. If the clock is used in a climate where the temperature averages to `30^(@)C`, how much time does the clock lose in each oscillation? (For invar `alpha = 9xx10^(-7)//^(@)C,` g= Constant)

Options

A. `2.25 xx 10^(-6)s`
B. `2.5 xx 10^(-7)s`
C. `5 xx 10^(-7)s`
D. `1.125 xx 10^(-6)s`

Answer

A

Explanation

Time period of oscillation, `T = 2pi sqrt (l/g)`
`logT=log [ 2pi((l)/(g))^(1//2)]`
On differentiating the above equation, we get,
`(dT)/(T)=(1)/(2)(dl)/(l)`
`dl=lalpha dt`, (`alpha'` is the coefficient of linear expansion)
`implies (dT)/(T)=(1)/(2)alphadt=(1)/(2)xx9xx10^(-7)xx(30-20)=4.5xx10^(-6)`
Therefore, Loss in time in each oscillation`(dT)=4.5xx10^(-6)xx0.5`
`=2.25xx10^(-6)s`