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 pendulum is 2 second. As the temperature increases time period increases. Clock
becomes slow and it loses the time. The change in time period is
` DeltaT = (1)/(2)T alpha Delta theta =(1/2) (2) (1.2xx10^(-5))(30^(@)-20^(@)) = 1.2 xx 10^(-4) s`
` therefore` New Time period is`T' = T +DeltaT =(2+1.2xx 10^(-4)) = 2.00012 s`
` therefore "Time lost in one week" Deltat = ((DeltaT)/(T'))t = (1.2xx10^(-4))/(2.00012) (7xx 24 xx 3600) = 36.28 s`