1. Given data

• Capacity of beaker at $30^\circ\text{C}$: $V_b = 500\ \text{cc}$
• Volume of mercury filled at $30^\circ\text{C}$: $V_m = ?$
• Coefficient of volume expansion of beaker: $\gamma_b = 6\times 10^{-6}\ ^\circ\text{C}^{-1}$
• Coefficient of volume expansion of mercury: $\gamma_m = 1.5\times 10^{-4}\ ^\circ\text{C}^{-1}$

The unfilled volume remains constant when temperature changes.

2. Expression for unfilled volume

At temperature change $\Delta T$, the beaker capacity becomes $V_b' = 500(1+\gamma_b \Delta T)$

The mercury volume becomes $V_m' = V_m(1+\gamma_m \Delta T)$

So the unfilled volume is $V_{\text{empty}} = V_b' - V_m'$ $V_{\text{empty}} = 500(1+\gamma_b\Delta T) - V_m(1+\gamma_m\Delta T)$

We are told this remains constant as temperature varies. Therefore, the coefficient of $\Delta T$ must be zero.

3. Set temperature-dependent part to zero

From $V_{\text{empty}} = (500 - V_m) + (500\gamma_b - V_m\gamma_m)\Delta T$

For this to be independent of temperature, $500\gamma_b - V_m\gamma_m = 0$

So, $V_m = \frac{500\gamma_b}{\gamma_m}$

Substitute values: $V_m = \frac{500\times 6\times 10^{-6}}{1.5\times 10^{-4}}$

4. Calculate

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

Thus, $V_m = 500\times 0.04 = 20\ \text{cc}$

5. Final answer

$\boxed{20}$

The required volume of mercury is 20 cc.