1. Condition for steady height

If the water level remains constant, then

$\text{inflow rate} = \text{outflow rate}.$

Given inflow rate,

$Q_{\text{in}} = 10^{-4}\ \text{m}^3\text{s}^{-1}.$

2. Use Torricelli’s theorem for outflow speed

For a hole at the bottom of a large tank, the speed of efflux is

$v = \sqrt{2gh}$

where $h$ is the height of water above the hole.

Hence the outflow rate is

$Q_{\text{out}} = a\sqrt{2gh}$

where hole area

$a = 1\ \text{cm}^2 = 10^{-4}\ \text{m}^2.$

3. Equate inflow and outflow

$10^{-4} = 10^{-4}\sqrt{2gh}$

So,

$\sqrt{2gh} = 1$

Squaring both sides,

$2gh = 1$

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

Taking $g = 9.8\ \text{m/s}^2$,

$h = \frac{1}{19.6} \approx 0.051\ \text{m}.$

Converting to cm,

$0.051\ \text{m} = 5.1\ \text{cm}.$

4. Check options

• A: $2.9\ \text{cm}$
• B: $5.1\ \text{cm}$
• C: $4\ \text{cm}$
• D: $1.7\ \text{cm}$

Thus the correct option is

$\boxed{\text{B } (5.1\ \text{cm})}.$