A pendulam clock loses 12s a day if the temperature is `40^@C` and gains 4s a day if the temperature is `20^@C`, The temperature at which the clock will show correct time, and the co-efficient of linear expansion `(alpha)` of the metal of the pendulam shaft are respectively:

Options

A. `30^@C: alpha=1.85xx10^-3//^@C`
B. `55^@C: alpha=1.85xx10^-2//^@C`
C. `25^@C: alpha=1.85xx10^-5//^@C`
D. `60^@C: alpha=1.85xx10^-4//^@C`

Answer

C

Explanation

(c) Time lost//gained per day =1/2 prop Delta thetaxx86400 second`
`12=1/2 alpha(40- theta)xx86400….(i)`
`4=1/2 alpha( theta-20)xx86400….(ii)`
on dividing we get, `3=(40- theta)/( theta-20)`
`3 theta-60=40- theta`
`4 theta=100implies theta=25^@C`