A clock with an iron pendulum keeps correct time at 20C.How much will it lose or gain per day if temperature changes to 40C?Coefficient of cubical expansion of iron =36×10-6C-1

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.