A body weighs W_(0) in air. Its apparent weights in a liquid at t_(1)""^(@)C " and " t_(2)""^(@)C " are " W_(1) " and " W_(2) respectively. If the coefficient of volume expansion of the material of the body is gamma,find the coefficient of real expansion of the liquid.

A body weighs W_(0) in air. Its apparent weights in a liquid at t_(1)"

1.	A body weighs W_(0) in air. Its apparent weights in a liquid at t_(1)""^(@)C " and " t_(2)""^(@)C " are " W_(1) " and " W_(2) respectively. If the coefficient of volume expansion of the material of the body is gamma,find the coefficient of real expansion of the liquid.
Answer» Solution :Weight of displaced liquid at `t_(1)""^(@)C=W_(0)-W_(1)` Let `V_(1) " and " V_(2)` be the volumes of the BODY and `d_(1) " and " d_(2)` be the densities of the liquid at `t_(1)""^(@)C "and "t_(2)""^(@)C` RESPECTIVELY. We have, `W_(0)-W_(1)=V_(1)d_(1)g "...(1)"` and `" " W_(0)-W_(2)=V_(2)d_(2)g "...(2)"` DIVIDING equation (1) by equation (2) we get, `""(W_(0)-W_(1))/(W_(0)-W_(2))=(V_(1)d_(1))/(V_(2)d_(2))` Now, `V_(2)=V_(1)[1+gamma(t_(2)-t_(1))] " and " d_(1)=d_(2)[1+gamma_(l)(t_(2)-t_(1))]`, where `gamma_(l)` is the coefficient of real EXPANSION of the liquid. `therefore ""(W_(0)-W_(1))/(W_(0)-W_(2))=(1+gamma_(l) times t)/(1+gamma times t) " "["where "t=(t_(2)-t_(1))]` or, `" " 1+gamma_(l)t=(W_(0)-W_(1))/(W_(0)-W_(2)) times (1+gammat)=(W_(0)-W_(1))/(W_(0)-W_(2))+(gammat(W_(0)-W_(1)))/(W_(0)-W_(2))` or, `""gamma_(l)t=(W_(2)-W_(1))/(W_(0)-W_(2))+(gammat(W_(0)-W_(1)))/(W_(0)-W_(2))` or, `""(W_(0)-W_(1))(1+gammat)=(W_(0)-W_(2))(1+gamma_(l) times t)` or, `""gamma_(l)=(W_(2)-W_(1))/((W_(0)-W_(2))t)+(gamma(W_(0)-W_(1)))/(W_(0)-W_(2)).`