Question

A piece of metal floats on mercury. The coefficients of volume expansion of the metal and mercury are ${{\gamma }_{1}}$ and ${{\gamma }_{2}}$ respectively. If the temperature of both mercury and metal are increased by an amount $\Delta T$, the fraction of the volume of the metal submerged in mercury changes by the factor.
$\text{A}\text{. }\dfrac{1}{\left( {{\gamma }_{2}}-{{\gamma }_{1}} \right)\Delta T}$
$\text{B}\text{. }\dfrac{1}{\left( {{\gamma }_{1}}-{{\gamma }_{2}} \right)\Delta T}$
$\text{C}\text{. }\left( {{\gamma }_{1}}-{{\gamma }_{2}} \right)\Delta T$
$\text{D}\text{. }\left( {{\gamma }_{2}}-{{\gamma }_{1}} \right)\Delta T$

Answer 1

Hint: The metal will float when the weight of the metal is balanced by the buoyant force of the mercury. Hence, they equate the weight and the buoyant force. Then calculate the new volume and density of the metal and mercury respectively. Since the weight of the metal will not change, both the buoyant forces will be the same. Hence, the equate the two buoyant forces.
Formula used:
F=mg
${{F}_{B}}=\rho g\Delta {{V}_{m}}$
$V{{'}_{m}}={{V}_{m}}\left( 1+{{\gamma }_{1}}\Delta T \right)$
$\rho '=\rho {{\left( 1+{{\gamma }_{2}}\Delta T \right)}^{-1}}$

Complete answer:
It is given that the piece of metal is floating on the mercury. That means that the buoyant force of mercury is balancing the weight (i.e. gravitational force) of the metal.
The weight of the metal will be F = mg, where m is the mass of the metal and g is acceleration due to gravity.
The buoyant force applied will be equal to ${{F}_{B}}=\rho g\Delta {{V}_{m}}$, where $\rho $ is the density of the liquid and $\Delta {{V}_{m}}$ is the submerged volume of the metal.
And $F={{F}_{B}}$.
$\Rightarrow mg=\rho g\Delta {{V}_{m}}$ ….. (i).
The fraction of the volume submerged of the metal in mercury will be $f=\dfrac{\Delta {{V}_{m}}}{{{V}_{m}}}$.
 $f{{V}_{m}}=\Delta {{V}_{m}}$
Substitute this value in equation (i).
$\Rightarrow mg=\rho gf{{V}_{m}}$ …. (ii).
Now, when the temperature is increased, both the metal and the mercury will expand. As a result, their volumes will increase.
The final volume of the metal is given by $V{{'}_{m}}={{V}_{m}}\left( 1+{{\gamma }_{1}}\Delta T \right)$.
Here, ${{\gamma }_{1}}$ is the coefficient of volume expansion of the metal and $\Delta t$ is the change in temperature.
Since the volume increases, its density will decrease. The final density of the liquid is given by $\rho '=\rho {{\left( 1+{{\gamma }_{2}}\Delta T \right)}^{-1}}$,
where ${{\gamma }_{2}}$ is the coefficient of volume of expansion of the mercury.
The metal is still floating on mercury but the volume of the metal and the density of mercury are changed.
Let the new fraction of the new volume of the metal submerged in mercury be f’.
Therefore, from equation (ii) we get,
$mg=\rho 'gf'V{{'}_{m}}$ ….. (iii).
Substitute the values of $V{{'}_{m}}$ and $\rho '$ in equation (iii).
$\Rightarrow mg=\rho {{\left( 1+{{\gamma }_{2}}\Delta T \right)}^{-1}}gf'{{V}_{m}}\left( 1+{{\gamma }_{1}}\Delta T \right)$
$\Rightarrow mg=\rho gf'{{V}_{m}}{{\left( 1+{{\gamma }_{2}}\Delta T \right)}^{-1}}\left( 1+{{\gamma }_{1}}\Delta T \right)$ …. (iv)
From (ii) and (iv), we get that
$\Rightarrow \rho gf{{V}_{m}}=\rho gf'{{V}_{m}}{{\left( 1+{{\gamma }_{2}}\Delta T \right)}^{-1}}\left( 1+{{\gamma }_{1}}\Delta T \right)$
$\Rightarrow f=f'{{\left( 1+{{\gamma }_{2}}\Delta T \right)}^{-1}}\left( 1+{{\gamma }_{1}}\Delta T \right)$
$\Rightarrow \dfrac{f}{f'}={{\left( 1+{{\gamma }_{2}}\Delta T \right)}^{-1}}\left( 1+{{\gamma }_{1}}\Delta T \right)$
$\Rightarrow \dfrac{f'}{f}=\left( 1+{{\gamma }_{2}}\Delta T \right){{\left( 1+{{\gamma }_{1}}\Delta T \right)}^{-1}}$
Generally, the coefficients of volume expansion are very small values. Hence,
${{\left( 1+{{\gamma }_{1}}\Delta T \right)}^{-1}}\approx \left( 1-{{\gamma }_{1}}\Delta T \right)$.
$\Rightarrow \dfrac{f'}{f}=\left( 1+{{\gamma }_{2}}\Delta T \right)\left( 1-{{\gamma }_{1}}\Delta T \right)$
$\Rightarrow \dfrac{f'}{f}=1-{{\gamma }_{1}}\Delta T+{{\gamma }_{2}}\Delta T-{{\gamma }_{1}}{{\gamma }_{2}}\Delta T$.
Since the term $-{{\gamma }_{1}}{{\gamma }_{2}}\Delta T$ is very small compared to other terms, it can be neglected.
$\Rightarrow \dfrac{f'}{f}\approx 1+\left( {{\gamma }_{2}}-{{\gamma }_{1}} \right)\Delta T$
$\Rightarrow \dfrac{f'}{f}-1=\left( {{\gamma }_{2}}-{{\gamma }_{1}} \right)\Delta T$
$\Rightarrow \dfrac{f'-f}{f}=\left( {{\gamma }_{2}}-{{\gamma }_{1}} \right)\Delta T$.

Hence, the correct option is D.

Note:
Note that the change in the fraction of the volume of the metal submerged can be positive or negative depending on the values of the coefficients of volume expansions of the metal and mercury.
If ${{\gamma }_{2}}$ > ${{\gamma }_{1}}$, then the change is positive. Meaning the fraction of volume submerged will be more when the temperature is increased.
If ${{\gamma }_{2}}$ < ${{\gamma }_{1}}$, then the change is negative. Meaning the fraction of volume submerged will be reduced when the temperature is increased.