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. `60^@ C, alpha=1.85 xx 10^(-4)//""^@C`
B. `30^@ C, alpha=1.85 xx 10^(-3)//""^@C`
C. `55^@ C, alpha=1.85 xx 10^(-2)//""^@C`
D. `25^@ C, alpha=1.85 xx 10^(-5)//""^@C`

Answer

D

Explanation

`T=2pi sqrt(l/g) implies (DeltaT)/(T)=1/2(Delta l)/(l)`
`(12)/(T)=1/2 alpha (40-theta)` ….(i)
Putting the value of `theta` in eq. (i)
`(12)/(T)=1/2 alpha (40-25) implies (12)/(24 xx 3600)=1/2 alpha(15)`
`implies alpha =1.85 xx 10^(-5)//""^@C`
`4/T=1/2 alpha (theta-20)` ....(ii)
By (i) and (ii)
`3=(40-theta)/(theta-20) implies theta =25^@C`.