Question 733250
A metal cube of side l  is heated and each side increases by 1%
so the original side= l & after heating side=1.01l
so the original surface area=6l^2 & after heating surface area=6x(1.01)^2xl^2
increase in surface area=6x(1.01)^2xl^2-6l^2=6l^2x(1.0201-1)=0.0201*6l^2
% increase in surface area=6l^2x(0.0201)*100/6l^2=2.01 %

now the original volume=l^3 & after heating volume=(1.01l)^3=1.030301l^3
so the increase in volume=1.030301l^3-l^3=0.030301*l^3
% increase in volume=0.030301*l^3*100/l^3=3.0301 %
ans