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A clock with an Iron Pendulum keeps corr...

A clock with an Iron Pendulum keeps correct time at `20^(@)C` How much will it lose or gain if temperature changes to `40^(@)C`? [Given cubical expansion of iron `=36xx10^(-6)""^(@)C^(-1)`]

Text Solution

Verified by Experts

The correct Answer is:
`[10.368 sec. loss]`
`Delta L=L alpha Delta T`
`Delta T=(Delta L)/(Lalpha)=(5xx10^(-4))/(1xx1.32xx10^(-5))=37.8^(@)C`
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