1. Use Torricelli’s theorem

For a tank open to air with maintained water level, the speed of efflux from a small opening at depth $h$ is

$v = \sqrt{2gh}$

Also, discharge rate is

$Q = Av$

where $A$ is the area of the opening.

2. Convert the given flow rate

Given:

$Q = 0.74\ \text{m}^3/\text{min}$

Convert to SI units:

$Q = \frac{0.74}{60}\ \text{m}^3/\text{s} = 0.01233\ \text{m}^3/\text{s}$

3. Find area of the circular opening

Radius of opening:

$r = 2\ \text{cm} = 0.02\ \text{m}$

So,

$A = \pi r^2 = \pi(0.02)^2 = 0.0004\pi \approx 1.256 \times 10^{-3}\ \text{m}^2$

4. Find speed of efflux

Using $Q=Av$:

$v = \frac{Q}{A} = \frac{0.01233}{1.256\times 10^{-3}} \approx 9.82\ \text{m/s}$

5. Find depth $h$

Using

$v = \sqrt{2gh}$

$h = \frac{v^2}{2g}$

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

$h = \frac{(9.82)^2}{2\times 9.8} \approx \frac{96.4}{19.6} \approx 4.92\ \text{m}$

This is closest to $4.8\ \text{m}$.

6. Check options

• A: $6.0\ \text{m}$ ✗
• B: $4.8\ \text{m}$ ✓
• C: $9.6\ \text{m}$ ✗
• D: $2.9\ \text{m}$ ✗

Therefore, the correct option is B.