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Step-by-Step Solution
Step 1: Identify the Known Information
• The rate of flow of water into the tank is
$10^{-4}\,\text{m}^3\text{/s}$.
• The area of the leakage hole at the bottom is
$1\,\text{cm}^2 = 1\times10^{-4}\,\text{m}^2$.
• At steady height of water $H$, the inflow rate equals the outflow rate.
Step 2: Write the Expression for Outflow Rate
According to Torricelli’s theorem, the speed $v$ of efflux of a fluid from a hole under a height of water $H$ is
$v = \sqrt{2\,g\,H}$, where $g$ is the acceleration due to gravity ($\approx 9.8\,\text{m/s}^2$).
Hence, the volumetric outflow rate $Q_\text{out}$ from the hole is given by
$Q_\text{out} = (\text{area of hole}) \times (\text{speed of efflux})$.
Therefore,
$$
Q_\text{out}
= 1\times10^{-4}\,\text{m}^2 \times \sqrt{2\,g\,H}.
$$
Step 3: Set Inflow = Outflow at Steady State
Since the inflow rate is $10^{-4}\,\text{m}^3\text{/s}$ and at steady height these rates must be equal:
$$
10^{-4}\,\text{m}^3\text{/s}
= 1\times10^{-4}\,\text{m}^2 \times \sqrt{2\,g\,H}.
$$
Simplifying, we get:
$$
1
= \sqrt{2\,g\,H}.
$$
Step 4: Solve for Height H
Squaring both sides:
$$
1^2 = 2\,g\,H.
$$
Thus,
$$
H = \frac{1}{2\,g}.
$$
Substituting $g \approx 9.8\,\text{m/s}^2$:
$$
H = \frac{1}{2 \times 9.8}
= \frac{1}{19.6}
= 0.051\,\text{m}
= 5.1\,\text{cm}.
$$
Step 5: Conclusion
Therefore, the steady height of water in the tank is 5.1 cm.
