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Step-by-Step Detailed Solution
Step 1: Understand the Problem
The question provides:
A water tank with its top open to the atmosphere.
A circular opening (radius 2 cm) in its wall from which water flows out.
The flow rate (volume of water flowing out per minute) is 0.74 m³/min.
We need to find the depth of the center of the opening below the water surface.
Since the tank is open at the top and water level is maintained (so pressure at the top is atmospheric and the height remains constant), we can use Torricelli’s theorem to connect the velocity of efflux of water with the depth under the surface.
Step 2: Convert Given Flow Rate to SI Units
Flow rate is given as 0.74 m³ per minute. We convert this to m³/s:
\[
Q = \frac{0.74 \text{ m}^3}{1 \text{ min}} = \frac{0.74 \text{ m}^3}{60 \text{ s}} \approx 0.01233 \text{ m}^3/\text{s}.
\]
Step 3: Calculate the Cross-sectional Area of the Orifice
The opening is circular with radius 2 cm = 0.02 m. Hence, the area of cross-section:
\[
A = \pi r^2 = \pi \times (0.02 \text{ m})^2 = \pi \times 0.0004 \text{ m}^2 \approx 0.0012566 \text{ m}^2.
\]
Step 4: Find the Speed of Efflux
The speed of water flowing out, given the volumetric flow rate and cross-sectional area, is:
\[
v = \frac{Q}{A} = \frac{0.01233 \text{ m}^3/\text{s}}{0.0012566 \text{ m}^2} \approx 9.8 \text{ m/s}.
\]
Step 5: Apply Torricelli's Theorem to Find the Depth
According to Torricelli's theorem, under ideal conditions (neglecting viscosity and assuming steady flow),
\[
v = \sqrt{2gh},
\]
where
\[
g \approx 9.8 \text{ m/s}^2, \quad h \text{ is the depth below the water surface}.
\]
Thus,
\[
h = \frac{v^2}{2g} = \frac{(9.8)^2}{2 \times 9.8} \approx \frac{96.04}{19.6} \approx 4.9 \text{ m}.
\]
This value is close to 4.8 m, which matches the given correct answer.
Step 6: Final Answer
The depth of the center of the opening below the water surface is approximately 4.8 m.
