A bakelite beaker has volume capacity of 500 cc at 30°C. When it is partially filled with $$V_m$$ volume (at 30°C) of mercury, it is found that the unfilled volume of the beaker remains constant as temperature is varied. If $$\gamma_{beaker} = 6 \times 10^{-6}$$ °C$$^{-1}$$, where $$\gamma$$ is the coefficient of volume expansion, then $$V_m$$ (in cc) is close to...

We have a bakelite beaker whose capacity at the reference temperature $$30^{\circ}\text{C}$$ is

$$V_{b0}=500\ \text{cc}.$$

It is partly filled with mercury. Let the volume of mercury taken at the same reference temperature be

$$V_m\ \text{cc}.$$

Therefore the empty (unfilled) space inside the beaker at $$30^{\circ}\text{C}$$ is

$$V_{\text{empty,0}} \;=\; V_{b0}-V_m.$$

The condition given in the statement is that this empty space remains the same when the temperature changes. Mathematically, for an infinitesimal rise $$dT$$ in temperature, the change in empty space must be zero:

$$dV_{\text{empty}} \;=\; 0.$$

But

$$dV_{\text{empty}} \;=\; dV_{\text{beaker}} \;-\; dV_{\text{mercury}}.$$

For any substance the change in volume caused by a small rise $$dT$$ is given by the volume-expansion formula

$$dV \;=\; V_0\,\gamma\,dT,$$

where $$\gamma$$ is the coefficient of volume expansion and $$V_0$$ is the initial volume at the reference temperature.

Applying this formula to the bakelite beaker, we get

$$dV_{\text{beaker}} \;=\; V_{b0}\,\gamma_{\text{beaker}}\,dT.$$

Similarly, applying it to the mercury, we have

$$dV_{\text{mercury}} \;=\; V_m\,\gamma_{\text{mercury}}\,dT.$$

Now we impose the zero-change condition:

$$0 \;=\; dV_{\text{empty}} \;=\; V_{b0}\,\gamma_{\text{beaker}}\,dT \;-\; V_m\,\gamma_{\text{mercury}}\,dT.$$

The factor $$dT$$ is common to both terms, so it can be cancelled. Rearranging, we obtain

$$V_m\,\gamma_{\text{mercury}} \;=\; V_{b0}\,\gamma_{\text{beaker}}.$$

Hence the required initial volume of mercury is

$$V_m \;=\; \frac{V_{b0}\,\gamma_{\text{beaker}}}{\gamma_{\text{mercury}}}.$$

We are already given

$$\gamma_{\text{beaker}} = 6 \times 10^{-6}\,^{\circ}\text{C}^{-1}.$$

The standard value for the coefficient of volume expansion of mercury around ordinary temperatures is approximately

$$\gamma_{\text{mercury}} \approx 1.5 \times 10^{-4}\,^{\circ}\text{C}^{-1}.$$

Substituting the numerical data, we get

$$V_m = \frac{500\ \text{cc}\;\bigl(6 \times 10^{-6}\bigr)}{1.5 \times 10^{-4}}.$$

We simplify step by step. First evaluate the ratio of the coefficients:

$$\frac{6 \times 10^{-6}}{1.5 \times 10^{-4}} = \frac{6}{1.5}\times 10^{-6+4} = 4 \times 10^{-2} = 0.04.$$

Now multiply by the beaker volume:

$$V_m = 500 \times 0.04 = 20.$$

Thus

$$V_m \approx 20\ \text{cc}.$$

Hence, the correct answer is Option D (20 cc).