We would like to make a vessel whose volume does not change with temperature . We can use brass and iron `(beta_(brass) = 6 xx10^(-5) // K and beta_(iron) = 3.55 xx 10^(-5)//K)` at create a volume of 100 c c. How do you think you can achieve this.

Explanation

In the previous problem the difference in the length was constant.
In this problem the difference in volume is constant.
The situation is shown in the diagram.

Let `V_(io), V_(bo)` be the volume of iron and brass vessel at `0^(@)C`
`V_(i), V_(b)` be the volume of iron and brass vessel at `Delta theta^(@)C`,
`gamma_(i), gamma_(b)` be the coefficient, of volume expansion of iron and brass.
As per question, `V_(io)-V_(bo) = 100 c c = V_(i) - V_(b)` ...(i)
Now, `V_(i) = V_(io) (1+gamma_(i)Delta theta)`
`V_(b) = V_(bo) (1+gamma_(b) Delta theta)`
`V_(i) - V_(b) = (V_(io)- V_(bo)) + Delta theta (V_(io) gamma_(i)-V_(bo)-gamma_(b))`
Since, `V_(i)-V_(b)` = constant.
So, `V_(io) gamma_(i) = V_(bo) gamma_(b)`
`rArr (V_(io))/(V_(bo)) = (gamma_(b))/(gamma_(i)) = ((3)/(2)beta_(B))/((3)/(2)beta_(i))= (beta_(b))/(beta_(i)) =(6 xx 10^(-5))/(3.55 xx 10^(-5)) = (6)/(3.55)`
`(V_(io))/(V_(bo)) = (6)/(3.55)` ...(ii)
Solving Eqs. (i) and (ii), we get
`V_(io) = 244.9 c c`
`V_(bo) = 144.9 c c`