The surface of water in a water tank of cross section area $$750 \text{ cm}^2$$ on the top of a house is $$h$$ m. above the tap level. The speed of water coming out through the tap of cross section area $$500 \text{ mm}^2$$ is $$30 \text{ cm s}^{-1}$$. At that instant, $$\frac{dh}{dt}$$ is $$x \times 10^{-3} \text{ m s}^{-1}$$. The value of $$x$$ will be ______.

We are given a water tank with cross-sectional area $$750 \text{ cm}^2$$ and a tap with cross-sectional area $$500 \text{ mm}^2$$. The speed of water at the tap is $$30 \text{ cm/s}$$, and we wish to determine $$\frac{dh}{dt}$$.

First, it is convenient to convert all quantities to SI units. The tank cross-sectional area becomes $$A_{\text{tank}} = 750 \text{ cm}^2 = 750 \times 10^{-4} \text{ m}^2$$, the tap cross-sectional area becomes $$A_{\text{tap}} = 500 \text{ mm}^2 = 500 \times 10^{-6} \text{ m}^2$$, and the speed of water at the tap becomes $$v_{\text{tap}} = 30 \text{ cm/s} = 0.3 \text{ m/s}$$.

Since the volume flow rate out of the tap must match the rate at which the volume in the tank decreases, we apply the equation of continuity: $$A_{\text{tap}} \times v_{\text{tap}} = A_{\text{tank}} \times \left|\frac{dh}{dt}\right|$$.

Next, we compute the outflow rate: $$Q = A_{\text{tap}} \times v_{\text{tap}} = 500 \times 10^{-6} \times 0.3 = 1.5 \times 10^{-4} \text{ m}^3/\text{s}$$.

Substituting this into the continuity equation gives $$\left|\frac{dh}{dt}\right| = \frac{Q}{A_{\text{tank}}} = \frac{1.5 \times 10^{-4}}{750 \times 10^{-4}} = \frac{1.5 \times 10^{-4}}{7.5 \times 10^{-2}} = 2 \times 10^{-3} \text{ m/s}$$.

Since the water level is falling, $$\frac{dh}{dt} = -2 \times 10^{-3} \text{ m/s}$$. Writing this in the form $$\frac{dh}{dt} = x \times 10^{-3} \text{ m/s}$$ shows that $$x = 2$$. Therefore, the final answer is $$2$$.