The surface of water in a water tank of cross section area $$750$$ 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 $$500$$ mm$$^2$$ is $$30$$ cm s$$^{-1}$$. At that instant, $$\frac{dh}{dt}$$ is $$x \times 10^{-3}$$ m s$$^{-1}$$. The value of $$x$$ will be ______.

We need to find the value of $$x$$ where $$\frac{dh}{dt} = x \times 10^{-3}$$ m/s.

The cross-sectional area of the tank is $$A = 750$$ cm$$^2 = 750 \times 10^{-4}$$ m$$^2$$, the cross-sectional area of the tap is $$a = 500$$ mm$$^2 = 500 \times 10^{-6}$$ m$$^2$$, and the speed of water at the tap is $$v = 30$$ cm/s $$= 0.30$$ m/s.

By the equation of continuity, the volume flow rate out of the tap equals the rate of decrease of water volume in the tank: $$A \cdot \left|\frac{dh}{dt}\right| = a \cdot v$$.

Solving for $$\left|\frac{dh}{dt}\right|$$ gives $$\left|\frac{dh}{dt}\right| = \frac{a \cdot v}{A} = \frac{500 \times 10^{-6} \times 0.30}{750 \times 10^{-4}} = \frac{150 \times 10^{-6}}{750 \times 10^{-4}} = \frac{150}{750} \times 10^{-2} = 0.2 \times 10^{-2} = 2 \times 10^{-3} \text{ m/s}$$.

Since $$\frac{dh}{dt} = x \times 10^{-3}$$ m/s, it follows that $$x = \boxed{2}$$.