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» <html><body><p></p>Solution :We known `(<a href="https://interviewquestions.tuteehub.com/tag/delta-947703" style="font-weight:bold;" target="_blank" title="Click to know more about DELTA">DELTA</a> T)/(T) = (1)/(2) alpha Delta t` <br/> `rArr (T_(2) - T_(1))/(T_(1)) = (1)/(2) alpha (t_(2) - t_(1))` <br/> Which is the fractional <a href="https://interviewquestions.tuteehub.com/tag/loss-1079380" style="font-weight:bold;" target="_blank" title="Click to know more about LOSS">LOSS</a> of time . As the temperature period increases and the clock loses time or goes slow. <br/> The time lost in one day (or in <a href="https://interviewquestions.tuteehub.com/tag/24-295400" style="font-weight:bold;" target="_blank" title="Click to know more about 24">24</a> <a href="https://interviewquestions.tuteehub.com/tag/hours-1030090" style="font-weight:bold;" target="_blank" title="Click to know more about HOURS">HOURS</a>) <br/> `((1)/(2) alpha (t_(2) - t_(1)) ) = (86400 s) (1)/(2) alpha (t_(2) - t_(1))` <br/> = 43200 `alpha (t_(2) - t_(1)) s ` <br/> In this problem, `alpha = 12 xx 10^(-6)//^(0)C`, <br/> `t_(1) = 20^(0)C , t_(2) = 40^(0)C ` <br/> The time lost in one day = 43200 `xx 12 xx 10^(-6) (40 - 20)` = 10.368 s.</body></html>