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

Options

A. 10.3 seconds/day
B. 20.6 seconds/day
C. 5 seconds/day
D. 20 minutes/day

Answer

A

Explanation

`Tpropsqrtl`
`implies (DeltaT)/(T)=(1)/(2)(Deltal)/(l) implies (1)/(2)(Deltal)/(l)=(1)/(2) alpha Delta theta`
Since, `l.=l(1+alpha Delta theta)`
`implies (Deltal)/(l)=alpha Delta theta`
`therefore (DeltaT)/(T)=(1)/(2) (Deltal)/(l)=(1)/(2) alpha Deltatheta=(1)/(2)xx12xx10^(-6)xx(40-20)`
`=12xx10^(-5)`
`implies DeltaT=12xx10^(-5)xx86400` sec/day
`implies DeltaT=10.3` sec/day.