A pendulum clock of an iron rod is connected to a small, heavy bob. If it shows correct time at 20^(0)C, how fast or slow will itgo in one day at 40^(0)C ? (Coefficient of linear expansion for steel is 12 xx 10^(-6)//^(0)C . )

A pendulum clock of an iron rod is connected to a small, heavy bob. If

1.	A pendulum clock of an iron rod is connected to a small, heavy bob. If it shows correct time at 20^(0)C, how fast or slow will itgo in one day at 40^(0)C ? (Coefficient of linear expansion for steel is 12 xx 10^(-6)//^(0)C . )
Answer» Solution :We known `(DELTA T)/(T) = (1)/(2) alpha Delta t` `rArr (T_(2) - T_(1))/(T_(1)) = (1)/(2) alpha (t_(2) - t_(1))` Which is the fractional LOSS of time . As the temperature period increases and the clock loses time or goes slow. The time lost in one day (or in 24 HOURS) `((1)/(2) alpha (t_(2) - t_(1)) ) = (86400 s) (1)/(2) alpha (t_(2) - t_(1))` = 43200 `alpha (t_(2) - t_(1)) s ` In this problem, `alpha = 12 xx 10^(-6)//^(0)C`, `t_(1) = 20^(0)C , t_(2) = 40^(0)C ` The time lost in one day = 43200 `xx 12 xx 10^(-6) (40 - 20)` = 10.368 s.