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A clock with an iron pendulum keeps correct time at `20^(@)C`. How much will it lose or gain per day if cubical expansion of iron `=36xx10^(-6) .^(@)C^(-1)` .

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Correct Answer - `[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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