A pendulum 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 pendulum 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

Time lost/gained per day `=(1)/(2) prop Delta theta xx 86400` 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)`
`3theta-60=40-theta`
`4theta=100 rArr theta=25^(@)C`