A pendulum clock loses 12 s a day if be the temperature is `40^(@)C` and gains 4 s a day if the temperature is `20^(@)C`. The temperature at which the clock will show correct time and the coefficient of linear expansion `alpha` of the pendulum shaft are, respectively

Options

A. `25^(0)C,alpha = 1.85 xx 10^(-5) // ^(0)C`
B. `60^(0)C,alpha = 1.85 xx 10^(-4) // ^(0)C`
C. `30^(0)C,alpha = 1.85 xx 10^(-3) // ^(0)C`
D. `55^(0)C,alpha = 1.85 xx 10^(-2) // ^(0)C`

Answer

A

Explanation

Let at temperature `theta`, clock gives correc time
`DeltaT = ((1)/(2)alphaDeltatheta)T, T = 1day = 86400s`
`12 = (1)/(2)alpha(40-theta)T` ……(i)
`4 = (1)/(2)alpha(theta - 20)T` …….(ii)
`i//ii rArr theta = 25^(0)C`
substitung `theta` in equation `(ii)`, we get
`4 = (1)/(2)alpha(25 - 20) xx 86400`
`alpha = (1)/(5 xx 86400) = 1.85 xx 10^(-5 //0)C`