A clock with a brass pendulum keeps correct time at 30^@C . How many seconds will it loose or gain if the temperature falls by 10^@C ? Take, alpha_("brass") = 0.000019^@C^(-1)

A clock with a brass pendulum keeps correct time at 30^@C . How many s

1.	A clock with a brass pendulum keeps correct time at 30^@C . How many seconds will it loose or gain if the temperature falls by 10^@C ? Take, alpha_("brass") = 0.000019^@C^(-1)
Answer» Solution :Time period of pendulum ` T_30 = 2S` The new temperature will be `(30 - 10)^@C = 20^@C`and LET `T_20`be the time period at this temperature. Then, `l_30 = l_0 (1+ 30 alpha)` `l_20 = l_0 (1+20 alpha)` where `l_30, l_20` and `l_0`are the lengths of the pendulum at `30^@C, 20^@C` and `0^@C` , respectively. ` T_30 = 2pi sqrt( (l_30)/(g) ) = 2pi sqrt( (l_0 (1+ 30 alpha) )/(g) )` ` T_20 = 2pi sqrt( (l_20)/(g) ) = 2pi sqrt( (l_0 (1+ 20 alpha) )/(g) )` ` therefore (T_20)/(T_30) = sqrt( (1+ 20 alpha)/(1 + 30 alpha)) = (1+ 20 alpha)^(1//2) (1+ 30 alpha)^(-1//2)` `= (1+ 1/2 xx 20 alpha) (1- 1/2 xx 30 alpha)` `= (1 + 10 alpha )(1 - 15 alpha) = (1 -5 alpha)` ` therefore T_10 = T_30 (1 - 5alpha)` ` = 2 (1 - 5 xx 1.9 xx 10^(-5) )s ` ` = (2- 1.9 xx 10^(-4) ) s` SINCE `T_20 lt T_30` , the clock will gain time when the temperature falls to `20^@C` . Gain in time for 2 seconds = `T_30 - T_20 = 1.9 xx 10^(-4) s` Gain in time for 1 day ` = (1.9 xx 10^(-4) xx 24 xx 3600)/(2)` = 8.21 s