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.

We would like to make a vessel whose volume does not change with tempe

1.	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.
Answer» Solution :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"`