Question

(a) A pendulum clock consists of an iron rod connected to a small, heavy bob. If it is designed to keep correct time at `20^(@)C` , how fast or slow will it go in 24 hours at `50^(@)C` ? `alpha_(iron)=1.2xx10^(-5)//^(@)C` .
(b) A pendulum clock having copper rod keeps correct time at `20^(@)C` . It gains 15 seconds per day if cooled to `0^(@)C` . Find the coefficient the of linear expansion of copper.

Answer 1

(a) `If theta gt theta_(0)` , loss of time
`theta lt theta_(0)` , gain of time
`theta_(0)` : temperature at which clock gives correct time
`theta_(0)20^(@)C` , `theta=50^(@)C`
Loss of time per day
`=(1)/(2)alpha_(i)(theta-theta_(0))xx86400`
`=(1)/(2)xx1.2xx10^(-5)(50-20)xx86400`
`=15.6s`
(b) Gain of time per day `(1)/(2)alpha(theta_(0)-theta)xx86400=15`
`(1)/(2)alpha(20-0)xx86400=15`
`alpha=(15)/86400=1.7xx10^(-5)//^(@)C` .