Water flows into a large tank with flat bottom at the rate of $$10^{-4}$$ m$$^3$$s$$^{-1}$$. Water is also leaking out of a hole of area 1 cm$$^2$$ at its bottom. If the height of the water in the tank remains steady then this height is:

We are told that water enters the tank with a volume flow rate $$Q_{\text{in}} = 10^{-4}\, \text{m}^3\text{s}^{-1}$$. At the same time, water leaves through a small hole in the bottom.

The cross-sectional area of this hole is given as $$1\; \text{cm}^2$$. Converting to SI units, $$1\; \text{cm}^2 = 1 \times 10^{-4}\; \text{m}^2$$.

For the speed of efflux of a liquid through a small orifice at depth $$h$$ below the free surface, we use Torricelli’s theorem, which states:

$$v = \sqrt{2gh},$$

where $$g$$ is the acceleration due to gravity (take $$g = 9.8\; \text{m s}^{-2}$$) and $$h$$ is the height of the liquid column above the hole.

The volume flow rate of water leaking out is therefore

$$Q_{\text{out}} = A\,v = (1 \times 10^{-4}) \sqrt{2gh}\; \text{m}^3\text{s}^{-1}.$$

Because the water level is steady, the inflow and outflow rates are equal:

$$Q_{\text{in}} = Q_{\text{out}}.$$

Substituting the expressions for the two flow rates, we have

$$10^{-4} = (1 \times 10^{-4}) \sqrt{2gh}.$$

Dividing both sides by $$1 \times 10^{-4}$$ gives

$$1 = \sqrt{2gh}.$$

Now we square both sides to remove the square root:

$$1^2 = ( \sqrt{2gh} )^2 \;\;\Longrightarrow\;\; 1 = 2gh.$$

Solving for $$h$$:

$$h = \frac{1}{2g}.$$

Substituting $$g = 9.8\; \text{m s}^{-2}$$, we obtain

$$h = \frac{1}{2 \times 9.8} = \frac{1}{19.6} \approx 0.05102\; \text{m}.$$

Converting meters to centimeters:

$$0.05102\; \text{m} = 0.05102 \times 100\; \text{cm} \approx 5.1\; \text{cm}.$$

Hence, the correct answer is Option A.