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 temperatures of both mercury and the metal are increased by an amount $\Delta T$, the fraction of the volume of the metal submerged in mercury changes by the factor ................

Step 1: Apply Archimedes' principle for floating.

For a floating body, the weight of the metal equals the weight of the displaced mercury.

$$\rho_m V g = \rho_{\text{Hg}} V_{\text{sub}} g$$

Here, $\rho_m$ is the density of the metal, $V$ is its total volume, $\rho_{\text{Hg}}$ is the density of mercury, and $V_{\text{sub}}$ is the submerged volume.

Step 2: Express the initial submerged fraction.

The fraction $f$ of the volume of the metal submerged in mercury is the ratio of the submerged volume to the total volume.

$$f = \frac{V_{\text{sub}}}{V}$$ $$f = \frac{\rho_m}{\rho_{\text{Hg}}}$$

Step 3: Determine the densities after the temperature increase.

When the temperature increases by $\Delta T$, the densities of the metal and mercury change due to volume expansion.

$$\rho_m' = \frac{\rho_m}{1 + \gamma_1 \Delta T}$$ $$\rho_{\text{Hg}}' = \frac{\rho_{\text{Hg}}}{1 + \gamma_2 \Delta T}$$

Step 4: Find the new submerged fraction.

The new fraction $f'$ of the metal submerged is determined by the ratio of the new densities.

$$f' = \frac{\rho_m'}{\rho_{\text{Hg}}'}$$

Substitute the expressions for the new densities.

$$f' = \frac{\rho_m}{\rho_{\text{Hg}}} \left( \frac{1 + \gamma_2 \Delta T}{1 + \gamma_1 \Delta T} \right)$$ $$f' = f \left( \frac{1 + \gamma_2 \Delta T}{1 + \gamma_1 \Delta T} \right)$$

Step 5: Calculate the factor of change.

The factor by which the submerged fraction changes is the ratio of the new fraction to the initial fraction.

$$\frac{f'}{f} = \frac{1 + \gamma_2 \Delta T}{1 + \gamma_1 \Delta T}$$

For small temperature changes, this can be approximated using the binomial expansion.

$$\frac{f'}{f} \approx 1 + (\gamma_2 - \gamma_1)\Delta T$$

Answer: $\dfrac{1 + \gamma_2 \Delta T}{1 + \gamma_1 \Delta T}$ (or approximately $1 + (\gamma_2 - \gamma_1)\Delta T$)