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

Explanation

(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%`