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At 30^(@)C the diameter of a brass disc ...

At `30^(@)C` the diameter of a brass disc is 8 cm. What will be the increase in surface area if it is heated to `80^(@)C`? `alpha` of brass `=18 times 10^(-6@)C^(-1)`.

Text Solution

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Increase in surface area `=S_(2)-S_(1)=beta S_(1)(t_(2)-t_(1))`
Here, `beta=2 alpha=2 times 18 times 10^(-6@)C^(-1) " and " S_(1)=pi times (8/2)^(2)cm^(2)`.
Increase in temperature `=t_(2)-t_(1)=80-30=50^(@)C`
`therefore` Increase in surface area,
`S_(2)-S_(1)=2 times 18 times 10^(-6) times pi times (8/2)^(2) times 50`
`" " =36 times 10^(-6) times 16pi times 50=0.0905 cm^(2)`.
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