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A second's pendulum clock has a steel wi...

A second's pendulum clock has a steel wire. The clock is calibrated at `20^@ C`. How much time does the clock lose or gain in one week when the temperature is increased to `30^@ C` ? `alpha_(steel = 1.2 xx 10^-5(^@ C)^-1`.

Text Solution

Verified by Experts

The time period of second's pendulum is 2 second. As the temperature increases length and hence, time period increases. Clock becomes slow and it loss the time. The change in time period is
`Delta T =(1)/(2) T alpha Delta theta =((1)/(2))(2) (1.2xx10^(-5))(30^(@)-20^(@))`
`=1.2xx10^(-4) s`
`therefore` New time period is,
`T'=T +Delta T =(2+1.2xx10^(-4))=2.0012 s`
`therefore` Time lost in one week
`Delta t =((Delta T)/(T'))t=((1.2xx10^(-4)))/((2.00012))(7xx24xx3600)`
`=36.28 s`
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