A pendulum clock gives correct time at `20^@C`at a place where `g=9.800 m s^(-2)`.The pendulum consists of a light steel rod connected to a heavy ball. It is taken to a different place where `g=9.788 m s^(-2)`. At what temperature will it give correct time ? coefficient of linear expansion of steel `= 12 xx 10^(-6)C^(-1)`.

Explanation

`g^1=9.8 m/s^2`
` g^2=9.788 m/s^2`
`T_1= (2pi sqrt (l_1))/(sqrt(q_1))`
` T_2 =(2pi sqrt(l_2))/ ( sqrt(g_2))`
` = (2 pi sqrt(l_2(1+alphaT)))/(sqrt (g_2))`
` alpha_steel= 12 xx 10^-6/^0C`
` Q_1=20^0C`
` Q_2=?`
` T_1=T_2`
`rArr (2pi sqrt (l_1))/(sqrt(g_1)) = (2pi sqrt(l_1(1+alpha DeltaT)))/(sqrt (g_1))`
` rArr (sqrt(l_1/g_1))= sqrt(l_1(1+alpha Delta T))/ sqrt (g_2)`
`rArr 1/9.8= (1+12 xx 10^-6xx Delta T)/ 9.788`
` rArr 9.788/9.8= 1+12 xx 10^-6 xx Delta T`
` rArr 9.788/9.8 -1 =12 xx 10^6 xx Delta T `
`rArr Delta T= -0.00122/(12 xx 10^-6)`
`rArr (T_2-20)= -102.4`
`rArr T_2 =-102.4+20`
`= -82.4 ~~82^0C`