When a block of iron floats in mercury at , fraction of its volume is submerged, while at the temperature a fraction is seen to be submerged. If the coefficient of volume expansion of iron is and that of mercury is then the ratio can be expressed a

Question

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 $γ_{Fe}$ and that of mercury is $γ_{Hg},$ then the ratio $\frac{k_{1}}{k_{2}}$ can be expressed as [2001]

Smart Achievers · Accepted Answer

(1)

For equilibrium when temperature $0 ° C$ fig (a)

$Upthrust = Wt. of body$

$k_{2}$ $K_{1} V d_{2} g = V d_{1} g$

$\Rightarrow K_{1} = \frac{d_{1}}{d_{2}} \dots (i)$

For equilibrium when temperature increases to $60 °$ fig. (b)

When the temperature is increased the density will decrease.

$∴$ $d_{1}' = d_{1} (1 + γ_{Fe} \times 60)$

and $d_{2}' = d_{2} (1 + γ_{Hg} \times 60)$

Again upthrust = Wt. of body

$∴$ $K_{2} V' d_{2}' g = V' d_{1}' g$

$∴$ $K_{2} [\frac{d_{2}}{1 + γ_{Hg} \times 60}] = \frac{d_{1}}{1 + γ_{Fe} \times 60}$

$∴$ $K_{2} [\frac{1 + γ_{Fe} \times 60}{1 + γ_{Hg} \times 60}] = \frac{d_{1}}{d_{2}}$

$\Rightarrow \frac{K_{1}}{K_{2}} = \frac{1 + γ_{Fe} \times 60}{1 + γ_{Hg} \times 60}$

Solution

1 $\frac{1 + 60 γ_{Fe}}{1 + 60 γ_{Hg}}$

2 $\frac{1 - 60 γ_{Fe}}{1 + 60 γ_{Hg}}$

3 $\frac{1 + 60 γ_{Fe}}{1 - 60 γ_{Hg}}$

4 $\frac{1 + 60 γ_{Hg}}{1 + 60 γ_{Fe}}$

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Solution

Q. When a block of iron floats in mercury at 0°C, fraction k1 of its volume is submerged, while at the temperature 60°C, a fraction k2 is seen to be submerged. If the coefficient of volume expansion of iron is γFe and that of mercury is γHg, then the ratio k1k2 can be expressed as [2001]

1 1+60γFe1+60γHg

2 1-60γFe1+60γHg

3 1+60γFe1-60γHg

4 1+60γHg1+60γFe

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1 $\frac{1 + 60 γ_{Fe}}{1 + 60 γ_{Hg}}$

2 $\frac{1 - 60 γ_{Fe}}{1 + 60 γ_{Hg}}$

3 $\frac{1 + 60 γ_{Fe}}{1 - 60 γ_{Hg}}$

4 $\frac{1 + 60 γ_{Hg}}{1 + 60 γ_{Fe}}$