A second's pendulum clock has a steel wire. The clock is calibrated at `20^@ C`. How much time does the clock lose or gain in one week when the temperature is increased to `30^@ C` ? `alpha_(steel = 1.2 xx 10^-5(^@ C)^-1`.

Explanation

The time period of second's prnfulum id 2 d. As the temperature increases length and hence, time period increases. Clock becomes show and it loses the time. The change in time period is
`DeltaT=(1)/(2)TalphaDeltatheta=((1)/(2))(2)(1.2xx10^(-5))(30"^(@)C-20^(@))`
`=1.2xx10^(-4)s`
`therefore` New time period is, `T=T+DeltaT=(2+1.2xx10^(-4))=2.00012 s`
`therefore` Time lost in one week `Deltat=((DeltaT)/(T))t=((1.2xx10^(-4)))/((2.00012))(7xx24 xx3600)=36.28s`