A pendulum clock gives correct time at 20°C at a place where g = 9.800 m s–2. The pendulum consists of a light steel rod connected to a heavy ball. It is taken to a different place where g = 9.788 m s–2. At what temperature will it give correct time? Coefficient of linear expansion of steel = 12 × 10–6 °C–1.
Given:
Temperature at which the pendulum shows correct time:T1=20°C
Value of gravitational acceleration at a place where T1 is 20°C is: g1= 9.8 m s-2
Value of g at different place is: g2= 9.788 m s-2
Coefficient of linear expansion of steel: α = 12 x 10-6 °C–1
Let, T2 be the temperature at a place where value of g is 9.788 m s-2.
Formula used:
Now the formula for time period is:
Where, l is the length of the rod and g is the acceleration due to gravity.
For places where values of g is different ,say :
And
Where l1 is the original length of the rod at T1 and l2 is the changed length at T2.
Since, for obtaining correct time both the time periods should be same,
That is: t1=t2
Since length changes due to linear expansion, by formula of linear expansion : 
Where L’ is the changed length. 
Substituting this result in the equality above we get,
Hence, at T2 = -82.04 ° C the pendulum will give correct time when it is taken to a place where g is 9.788 m s-2.