A clock is calibrated at a temperature of `20^@C` Assume that the penduum is a thin brass rod of negligible mass with a heavy bob attached to the end `(alpha_("brass")=19xx10^(-6)//K)`

Options

A. On a hot day at `30^@C` the clock gains 8.2 s
B. On a hot day at `30^@C` the clock loses 8.2 s
C. On a cold day at `10^@C` the clock gains 8.2 s
D. On a cold day at `10^@C` the clock loses 8.2 s

Answer

B, C

Explanation

`T=2pisqrt((l)/(g))=2pisqrt((l_0+alphal_0Deltatheta_0)/(g))`
`=T_0sqrt(1+(1)/(2)alphaDeltatheta)`
At `30^@C` fraction loss of time `=(T_(30^@C)-T_(20^@C))/(T_(20^@C))`
`=5alpha=5xx19xx10^-6`
time lost in 24 h
`=86400xx95xx10^-6=8.2s`
On a cold day at `10^@C`, fraction gain of time
`=(T_(10^@)-T_(20^@))/(T_(20^@C))=-5alpha`
Time graph in `24h=8.2s`