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Mercury of volume 0.2m^(3) at 10^(@)C ...

Mercury of volume `0.2m^(3)` at `10^(@)C` is heated to `80^(@)C` . What will be the increase in volume of mercury. Given `alpha_(V)=2xx10^(-3)C^(-1)`.

Text Solution

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We are given : `V_(o)=0.2m^(3)`
`alpha_(v)=2xx10^(-3)""^(@)C^(-1)`
`DeltaV=80-10=70^(@)C`
We are to calculate : increase in volume , `DeltaV=alpha_(v)V_(o)DeltaT`
Using relation , `DeltaV=alpha_(V)V_(o)DeltaT=2xx10^(-3)xx0.2xx80=32xx10^(-3)m^(3)`.
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