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A clock which keeps correct time at 20^(...

A clock which keeps correct time at `20^(@)C` is subjected to `40^(@)C.` If coefficient of linear expansion of the pendulum is `12xx10^(-6)//"^(@)C.` How much will it gain or loss in time ?

A

10.3 seconds/day

B

20.6 seconds/day

C

5 seconds/day

D

20 minutes/day

Text Solution

Verified by Experts

The correct Answer is:
A

`Tpropsqrtl`
`implies (DeltaT)/(T)=(1)/(2)(Deltal)/(l) implies (1)/(2)(Deltal)/(l)=(1)/(2) alpha Delta theta`
Since, `l.=l(1+alpha Delta theta)`
`implies (Deltal)/(l)=alpha Delta theta`
`therefore (DeltaT)/(T)=(1)/(2) (Deltal)/(l)=(1)/(2) alpha Deltatheta=(1)/(2)xx12xx10^(-6)xx(40-20)`
`=12xx10^(-5)`
`implies DeltaT=12xx10^(-5)xx86400` sec/day
`implies DeltaT=10.3` sec/day.
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