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.