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A thin copper wire of length L increase in length by 2 % when heated from `T_(1)` to `T_(2)`. If a copper cube having side 10 L is heated from `T_(1)` to `T_(2)` when will be the percentage change in (i) area of one face of the cube (ii) volume of the cube |
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Answer» (i) Area A=10Lxx10L=100`L^(2)` percentage change in area `=(DeltaA)/(A)xx100=2xx(DeltaL)/(L)xx100=2xx=4%` `(therefore(DeltaA)/(A)=(Delta100)/(100)+2(DeltaL)/(L)=0+2(DeltaL)/(L)=2(DeltaL)/(L))` Note:- Constants do not have any error in them. (ii) Volume `V=10Lxx10Lxx10L=1000 L^(3)` percentage change in volume `=(DeltaV)/(V)xx100=3(DeltaL)/(L)=3xx2%=6%` |
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