Properties Of Matter Ques 26
- When a block of iron floats in mercury at $0^{\circ} \mathrm{C}$, fraction $k _ 1$ of its volume is submerged, while at the temperature $60^{\circ} \mathrm{C}$, a fraction $k _ 2$ is seen to be submerged. If the coefficient of volume expansion of iron is $\gamma _ {\mathrm{Fe}}$ and that of mercury is $\gamma _ {\mathrm{Hg}}$, then the ratio $k _ 1 / k _ 2$ can be expressed as
(2001, 2M)
(a) $\frac{1+60 \gamma _ {\mathrm{Fc}}}{1+60 \gamma _ {\mathrm{Hg}}}$
(b) $\frac{1-60 \gamma _ {\mathrm{Fe}}}{1+60 \gamma _ {\mathrm{Hg}}}$
(c) $\frac{1+60 \gamma _ {\mathrm{Fe}}}{1-60 \gamma _ {\mathrm{Hg}}}$
(d) $\frac{1+60 \gamma _ {\mathrm{Hg}}}{1+60 \gamma _ {\mathrm{Fc}}}$
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Answer:
Correct Answer: 26.( a )
Solution:
- $k _ 1=\left(\frac{\rho _ {\mathrm{Fc}}}{\rho _ {\mathrm{Hg}}}\right) _ {0^{\circ} \mathrm{C}}$ and $k _ 2=\left(\frac{\rho _ {\mathrm{Fc}}}{\rho _ {\mathrm{Hg} _ g}}\right) _ {60^{\circ} \mathrm{C}}$.
Here, $\rho=$ Density
$ \therefore \frac{k _ 1}{k _ 2}=\frac{\left(\rho _ {\mathrm{Fc}}\right) _ {0 \circ} \mathrm{C}}{\left(\rho _ {\mathrm{Hg}}\right) _ {0^{\circ} \mathrm{C}}} \times\left(\frac{\rho _ {\mathrm{Hg}}}{\rho _ {\mathrm{Fe}}}\right) _ {60^{\circ} \mathrm{C}}=\frac{\left(1+60 \gamma _ {\mathrm{Fc}}\right)}{\left(1+60 \gamma _ {\mathrm{Hg}}\right)} $
NOTE In this problem two concepts are used (i) When a solid floats in a liquid, then
Fraction of volume submerged $(K)=\frac{\rho _ {\text {sdid }}}{\rho _ {\text {liquid }}}$ This result comes from the fact that
$ \begin{aligned} \text { Weight } & =\text { Upthrust } \\ V _ {\rho _ {\text {selid }} g} & =V _ {\text {salmergad }} \rho _ {\text {liquia }} g \\ \therefore \quad \frac{V _ {\text {submeged }}}{V} & =\frac{\rho _ {\text {cold }}}{\rho _ {\text {dquid }}} \end{aligned} $
(ii) $\frac{\rho _ {\theta C C}}{\rho _ {D C C}}=\frac{1}{1+\gamma \cdot \theta}$
This is because $\rho \propto \frac{1}{\text { Volume }}$ (mass remaining constant)
$ \therefore \frac{\rho _ {\operatorname{OCC}}}{\rho _ {\sigma C C}}=\frac{V _ {O C}}{V _ {\theta C C}}=\frac{V _ {0 C C}}{V _ {0 \cdot C}+\Delta V}=\frac{V _ {0 C C}}{V _ {0^{\circ} C}+V _ {0 C C} \gamma \theta}=\frac{1}{1+\gamma \theta} $
BETA
