A sinker of weight `w_0` has an apparent weight `w_1` when weighed in a liquid at a temperature `t_1 and w_2` when weight in the same liquid at temperature `t_2`. The coefficient of cubical expansion of the material of sinker is `beta`. What is the coefficient of volume expansion of the liquid.

Explanation

Here, weight of sinker in air `=W_(0)`
Weight of sinker in liquid at temperature `T_(1)=W_(1)`
Weight of sinker in liquid at temperature `T_(2)=W_(2)`
(i) At temperature `T_(1)`
Loss of weight of the sinker in liquid `=W_(0)-W_(1)`
Weight of the liquid displaced `=W_(0)-W_(1)`
Volume of liquid displaced
`V_(T_(1))=("weight of liquid displaced")/("density")`
`V_(T_(1))=((W_(0)-W_(1)))/(d_(1))` ....................1.51
Where `d_(1)` is the density of the liquid at temperature `T_(1)`
(ii) At temperature `T_(2)`
Similarly, we can calculate the volume of liquid displaced `V_(T_(2))` at temperature `T_(2)`
`V_(T_(2))=((W_(0)-W_(2)))/(d_(2))`.1.52
Where `d_(2)` is the density of the liquid at temperature `T_(2)`
We know that
`V_(T_(2))=V_(T_(1))[1+gamma(T_(2)-T_(1))]`
where `gamma` is the coefficient of cubical expansion of sinker
According to the given problem `gamma=beta` thus
`V_(T_(2))=V_(T_(1))[1+beta(T_(2)-T_(1))]`...........1.53
Substituting the value of `V_(T_(2))` and `V_(T_(1))` we have
`(((W_(0)-W_(2)))/(d_(2)))=(((W_(0)-W_(1)))/(d_(1)))[1+beta(T_(2)-T_(1))]`.1.54
But `(d_(1))/(d_(2))=(M//V_(T_(1)))/(M//V_(T_(2)))=(V_(T_(2)))/(V_(T_(1)))=(V_(T_(1))[1+gamma_(L)(T_(2)-T_(1))])/(V_(T_(1)))`
`(d_(1))/(d_(2))=[1+gamma_(L)(T_(2)-T_(1))]` ..................1.55
Where `gamma_(L)` is the coefficient of cubical expansion of the liquid,
From equations (1.54) and (1.55) we have
`[1+gamma_(L)(T_(2)-T_(1))]=((W_(0)-W_(1))/(W_(0)-W_(2)))[1+beta(T_(2)-T_(1))]`
Solving we have
`gamma_(L)=((W_(2)-W_(1))+beta(W_(0)-W_(1))(T_(2)-T_(1)))/((W_(0)-W_(2))(T_(2)-T_(1)))`