A pendulum clock loses 12 s a day if the temperature is `40^(@)` and gains 4 s 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. `25^(@)C,alpha=1.85xx10^(-5)//^(@)C`
B. `60^(@)C,alpha=1.85xx10^(-4)//^(@)C`
C. `30^(@)C,alpha=1.85xx10^(-3)//^(@)C`
D. `55^(@)C,alpha=1.85xx10^(-2)//^(@)C`

Answer

D

Explanation

Time period of a pendulum, `T=2pisqrt((l)/(g))`
where l is length of pendulum and g is acceleration due to gravity.
Change in time period of a pendulum, `(DeltaT)/(T)=(1)/(2)(Deltal)/(l)`
When clock gains 12 s, we get `(12)/(T)=(1)/(2)alpha(40-theta)`
When clock loses 4s, we get `(4)/(T)=(1)/(2)alpha(theta-20)`
Comparing Eqs. (i) and (ii), we get `3=(40-theta)/(theta-20)`
`rArr3theta-60=40-thetarArr4theta=100rArrtheta=25^(@)C`
Substituting the value of `theta` in Eq . (i), we have `(12)/(T)=(1)/(2)alpha(40-25)rArr(12)/(24xx3600)=(1)/(2)alpha(15)`
`alpha=(24)/(24xx3600xx15),alpha=1.85xx10^(-5)//^(@)C`
Thus, the coefficient of linear expansion in a pendulum clock `=1.85xx10^(-5)//^(@)C`