The balance wheel of a mechanical wriest watch has a frequency of oscillation given by `f = ((1)/2pi) sqrt(C//I)`, where `I` is the moment of inertia of the wheel and `C` is the torisonal rigidity of its spring. The wrist watch keeps accurate time at `25^(0)C`, How many seconds would it gain a day at `.-25^(0)C` if the balance wheel made of aluminium ?
(Given, `alpha_(Al) = 25.5 xx 10^(-6//0)C)`

Explanation

`f = (1)/(2pi)sqrt((C)/(1)) = (1)/(2pik)sqrt((C)/(M)) [:' = MK^(2)]`
`f prop (1)/(T), f prop (1)/(k) rArr (df)/(f) = (-dT)/(T) = (-dk)/(k)`
As `(dk)/(k) alphadt rArr (dt)/(T) = +alphadt`
Number of seconds gained/day
`dT = (8.64 xx 10^(4))(alphadT) = 110.2 s/day`