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

Answer

B

Explanation

`T = 2pisqrt((l)/(g))`
`(DeltaT)/(T) = (1)/(2)(Deltal)/(l)`
When clock gain `12` sec
`(12)/(T) = (1)/(2)alpha(40-theta) …. (1)`
When clock lose `4` sec.
`(4)/(T) = (1)/(2)alpha (theta-20) ….. (2)`
From equation (1) & (2)
`3 = (40-theta)/(theta-20)`
`3theta-60 = 40 - theta`
`4theta = 100`
`theta= 25^(@)C` from equation `(1)` `(12)/(T) = (1)/(2)alpha(40-25)`
`(12)/(24 xx 3600) = (1)/(2) alpha xx 15`
`alpha = (24)/(24 xx 3600 xx 15) alpha = 1.85 xx 10^(-15)//^(@)C`