A perpendicular clock (fitted with a small heavy bob that is connected with a metal rod) is `5` seconds fast each day at a temperature of `15^(@)C` and `10` seconds slow at a temperature of `30^(@)C`. The temperature at which it is designed to give correct time, is

Options

A. `18^(@)C`
B. `20^(@)C`
C. `24^(@)C`
D. `25^(@)C`

Answer

B

Explanation

Fractional loss of time per second =`1/2 alpha Delta T`
Therefore, `1/2 alpha(T_(0)-15) xx (24 hrs) = 5`
and `1/2 alpha (30-T_(0))xx(24 hrs) = 10`
On solving `T_(0)=20^(@)C`.