Correct Answer - A
(a) `R = rho (l)/(A) implies (R_(1))/(R_(2)) = (A_(2))/(A_(1)) (rho, L "constant") implies (A_(1))/(A_(2)) = (R_(2))/(R_(1)) = 2`
Now, when a body dipped in water, loss of weigh `V sigma_(L) g Al sigma_(L) g`
So, `(("Loss of weigth")_(1))/(("Loss of wight")_(2)) = (A_(1))/(A_(2)) = 2`, so `A` has more loss of weight.