A clock which keeps correct time at `20^(@)C`, is subjected to `40^(@)C`. If coefficient of linear expension of the pendulum is `12xx10^(-6//@)C`. How much will it gain or lose time?

Options

A. `10.3 s//day`
B. `20.6 s//day`
C. `5 s//day`
D. `20 min//day`

Answer

A

Explanation

Time period of oscillation , `T = 2 pi sqrt(l/g)`
`log T = log (l)`
On differentiating on both sides , we get,
`(Delta T) /(t) = (1)/(2) (Delta l)/(l)`
` implies (Delta T )/( T) = (alpha Delta t)/(2)`
`implies Delta T = ( 12 xx 10^(-6) xx(40 -20 ))/(2) xx T`
` implies Delta T = 120 xx 10^(-6) xx 24 xx 60 xx 60`
`therefore Delta T = 10.3 s|day`