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    We know that for an ideal pendulum time period ##T## is given by the expression ##T=2pisqrt(L/g)## where ##L## is the length of the pendulum and ##g## due to gravity. Given that the pendulum keeps correct time at ##20^@C## ##:.T_”correct”=2pisqrt((L_”20″)/g)## …….(1)

    Now we need to find time period at ##40^@C##

    We know that with increase in temperature length ##L_0## of metallic rod expands and the expression is ##L_T = L_0 ( 1 + ? ?T)## …….(2) where ##?## is the coefficient of linear expansion and ##DeltaT## is the change in temperature.

    It can be shown that the coefficient of volumetric expansion ##gamma~~3alpha## As such from (2) we get ##L_40= L_20 ( 1 + (36xx10^-6)/3xx (40-20))## ##L_40=1.00024 L_20 ## …..(3)

    We see that at ##40^@C## length of the pendulum is greater than at ##20^@C##. As the time period is directly proportional to the square root of length, increase in length implies increase in time period. Which amounts to clock loosing time. From (1) ##T_40=2pisqrt((L_40)/g)## Using (3) ##T_40=2pisqrt((1.00024 L_20 )/g)## ##=>T_40=sqrt1.00024 xxT_”correct”## To calculate loss per day we insert seconds in ##1## day (##24## hours) as the correct time and deduct seconds equal to one day. We get ##DeltaTime=sqrt1.00024 xx86400-86400## ##=10.4s##, rounded to one decimal place.

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