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When a block of iron floats in mercury at $\;0^{0}C\;$ , a fraction $\;K_{1}\;$ of its volume is submerged. While at the temperature $\;60^{0}C\;$ , a fraction $\;K_{2}\;$ is seen submerged . If the coefficient of volume expansion of iron is $\;r_{Fe}\;$ and that of mercury is $\;r_{Hg}\;$ , then the ratio $\;\large\frac{K_{1}}{K_{2}}\;$ can be expanded as

$(a)\;\large\frac{1+60 r_{Fe}}{1+60 r_{Hg}}\qquad(b)\;\large\frac{1-60 r_{Fe}}{1+60 r_{Hg}}\qquad(c)\;\large\frac{1+60 r_{Fe}}{1-60 r_{Hg}}\qquad(d)\;\large\frac{1+60 r_{Hg}}{1+60 r_{Fe}}$

1 Answer

Answer : $\;\large\frac{1+60 r_{Fe}}{1+60 r_{Hg}}$
Explanation :
Since iron block in floating weight of iron = weight of mercury displaced
Let $\;V_{Fe}\;$ be Volume of iron at $\;0^{0}C\;$
$V_{e}\;$ Volume submerged in mercury at $\;0^{0}C$
$\rho_{Fe}\;$ - density at $\;0^{0}C$
$\rho_{Hg}\;$ - density of mercury at $\;0^{0}C$
$V_{Fe}\; \rho_{Fe} = V_{i} \; \rho_{Hg}$
$\large\frac{V_{i}}{V_{Fe}} = \large\frac{\rho_{Fe}}{\rho_{Hg}} = K$
$V_{Fe}^{'}\; \rho_{Fe}^{'} = V_{i}^{'} \; \rho_{Hg}^{'}\;\; [at\;60^{0}C]$
$\large\frac{V_{i}^{'}}{V_{Fe}^{'}} = \large\frac{\rho_{Fe}^{'}}{\rho_{Hg}^{'}} = \large\frac{\rho_{Fe}}{(1+r_{Fe})} \;(1+r_{Hg})$
$K_{2} = K_{1} \;(\large\frac{1+r_{Hg}}{1+ r_{Fe}})$
$\large\frac{K_{1}}{K_{2}} = \large\frac{1+ r_{Fe}}{1+ r_{Hg}}$
answered Apr 8, 2014 by yamini.v

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