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. `25^(@)C, alpha=1.85xx10^(-5) .^(@)C^(-1)`
B. `60^(@)C, alpha=1.85xx10^(-4) .^(@)C^(-1)`
C. `30^(@)C, alpha=1.85xx10^(-3) .^(@)C^(-1)`
D. `55^(@)C, alpha=1.85xx10^(-2) .^(@)C^(-1)`

Answer

A

Explanation

Time period of pendulum, `T=2pi sqrt((l)/(g))`
`Delta T =(2pi)/(sqrt(g))xx(1)/(2)l^(-1//2) Delta l=(pi Delta l)/(sqrt(g l))`
`:. (Delta T)/(T) =(pi Delta l//sqrt(gl))/(2pi sqrt(l//g)) =(1)/(2) (Delta l)/(l)`
When clock gains 12 s a day, then
`(12)/(T)=(1)/(2)alpha(40 - theta)` ....(i)
When clock gain loses 4 s a day, then
`(4)/(T) =(1)/(2)alpha(thata - 20)` ....(ii)
Dividing (i) by (ii) we get
`3=(40 - theta)/(theta - 20)`
or `3theta - 60=40 - theta` or `4theta =100` or `theta =25^(@)C`
Putting this value of `theta` in (i), we get
`(12)/(T) =(1)/(2)alpha(40 - 25)=(1)/(2)alphaxx15`
or `(12)/(24xx60xx60)=(1)/(2)alphaxx15`
or `alpha=(12xx2)/(24xx60xx60xx15) =1.85xx10^(-5) .^(@)C^(-1)`