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We would like to make a vessel whose vol...

We would like to make a vessel whose volume does not change with temperature (take a hint from the problem above). We can use brass and iron `(beta_("vbrass")=6xx10^(-5)//K" and "beta_("viron")=3.55xx10^(-5)//K)` to create a volume of 100 cc. How do you think you can achieve this.

Text Solution

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Let, `V_(io)` and `V_(bo)` are volume of iron and brass at `0^(@)C` respectively.

`V_(i),V_(b)` are volume of iron and brass respectively of `DeltaT^(@)C`. `gamma_(i)` and `gamma_(b)` are coefficient of volume expansion of them respectively.
According to question,
`V_(io)-V_(bo)=V_(i)-V_(b)=100" cc"` . . .(1)
`:.V_(i)=V_(io)(1+gamma_(i)DeltaT)` and `V_(b)=V_(bo)(1+gamma_(b)DeltaT)`
`:.` From equation (1),
`V_(i)-V_(b)=V_(io)(1+gamma_(i)DeltaT)-V_(ib)(1+gamma_(b)DeltaT)`
`V_(i)-V_(b)=V_(io)-V_(ib)+V_(io)gamma_(i)DeltaT-V_(ib)gamma_(b)DeltaT`
But `V_(i)-V_(b)=V_(io)-V_(ib)`
`:.V_(io)gamma_(i)DeltaT=V_(ib)gamma_(b)DeltaT`
`:.(V_(io))/(V_(bo))=(gamma_(b))/(gamma_(i))` for midpoint `=(6xx10^(-5))/(3.55xx10^(-5))=(6)/(3.55)`
`:.V_(io)=(6)/(3.55)V_(bo)` . . . (2)
`:.` From equation `V_(io)-V_(ib)=100`,
`(6)/(3.55)V_(bo)-V_(bo)=100`
`:.6V_(bo)-3.55V_(bo)=355`
`:.2.45V_(bo)=355`
`:.V_(io)=144.89" cc"`
`:.V_(io)~~144.9" cc"`
Now, again `V_(io)-V_(bo)=100`
`:.144.89-V_(bo)=100`
`:.144.89-100=V_(bo)`
`:.44.89=V_(b)`
`:.V_(bo)~~44.9" cc"`
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