Δl/l = 2/100
αΔT = 2/100
ΔT = 2/(100α)
ΔA/A = βΔT
= 2α 2/100α
= 4/100
Thus area of base will inc. 4% [1]
Hope I'm correct cos I don't want answer wrongly in the very few eureka threads to which I reply[4]
If length of cylinder on heating inc. 2%,then are of base will increase ___%
Δl/l = 2/100
αΔT = 2/100
ΔT = 2/(100α)
ΔA/A = βΔT
= 2α 2/100α
= 4/100
Thus area of base will inc. 4% [1]
Hope I'm correct cos I don't want answer wrongly in the very few eureka threads to which I reply[4]