A pendulum clock ( fitted with a small heavy bob that is connected with a metal rod ) is 3 seconds fast each day at a temperature of `15^(@)C` and 2 seconds slow at a temperature of `30^(@)C`. Find the temperature ( in `""^(@)C` ) at which will show the correct time.

Answer

`24.00`

Explanation

Fractional loss of time per second = `= 1/2 alpha DeltaT`
Therefore `1/2 alpha (T_(0) - 15) xx (24 "hrs") = 3`............. Eq(1) and `1/2 alpha(30-T_(0)) xx (24 "hrs") = 2`.........Eq (2)
On solving, `T_(0) = 24^(@)` C