When a block of iron floats in mercury at `0^@C`, fraction `k_(1)` of its volume is submerged, while at the temperature `60^@C`, a fraction `k_(2)` is seen to be submerged. If the coefficient of volume expansion of iron is `gamma_(Fe)` and that of mercury is `Mg`, then the ratio `k_(1)//k_(2)` can be expressed as
(This problem is mixed concept of heat mechanics)

Options

A. `(1+60gammaFe)/(1+60gamma_(Hg))`
B. `(1-60gamma_(Fe))/(1+60gamma_(Hg))`
C. `(1+60gamma_(Fe))/(1-60gamma_(Hg))`
D. `(1+60gamma_(Hg))/(1+60gamma_(Fe))`

Answer

A

Explanation

`k_(1)=((rhoFe)/(rhoHg))andk_(2)=((rhoFe)/(rhoHg))_(60^(@)C`
Here `,rho`= Density
`therefore(k_(1))/(k_(2))=((rho_(Fe))_(0^(@)C))/((rho_(Hg))_(0^(@)C))xx((rho_(Hg))/(rho_(Fe)))_(60^(@)C)=((l+60gamma_(Fe)))/((1+60gamma_(Hg)))`