Question

A hole of area 1 mm\(^2\) opens in the pipe near the lower end of a large water storage tank, and a stream of water shoots from it. If the top of the water in the tank is 20 m above the point of the leak, the amount of water escapes in 1 s is:

• A. 87.5 cm\(^3\)/s
• B. 43.1 cm\(^3\)/s
• C. 27.5 cm\(^3\)/s
• D. 19.8 cm\(^3\)/s

Hint:

Use Torricelli’s Law to calculate the speed of a fluid exiting a hole under the influence of gravity.

Answer 1

The Correct Option is A

Water Escape Calculation 

We need to calculate the amount of water that escapes from a hole in a pipe near the lower end of a large water storage tank.

The hole has an area of \( 1 \, \text{mm}^2 \), and the height of the water in the tank is \( 20 \, \text{m} \) above the point of the leak.

Step 1: Calculate the Speed of Water

Using Torricelli's Law, the speed \( v \) of the water is given by the equation:

\( v = \sqrt{2gh} \)

where:

• \( g \approx 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity,
• \( h = 20 \, \text{m} \) is the height of the water above the hole.

Substituting the values:

\( v = \sqrt{2 \times 9.81 \times 20} = \sqrt{392.4} \approx 19.8 \, \text{m/s} \)

Step 2: Calculate the Volume of Water Escaping Per Second

The flow rate \( Q \) is given by:

\( Q = A \times v \)

where:

• \( A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2 \) is the area of the hole,
• \( v = 19.8 \, \text{m/s} \) is the velocity of the water.

Substituting the values:

\( Q = (1 \times 10^{-6} \, \text{m}^2) \times 19.8 \, \text{m/s} = 1.98 \times 10^{-5} \, \text{m}^3/\text{s} \)

Thus, the amount of water escaping in 1 second is:

\( \boxed{1.98 \times 10^{-5} \, \text{m}^3} = 19.8 \, \text{cm}^3\)

Answer 2

Water escapes through a small hole of area \( 1 \, \text{mm}^2 \) near the base of a tank, with the water surface \( 20 \, \text{m} \) above the hole. We want the volume flow rate out of the hole, and we'll check it against each option using energy conservation directly (rather than quoting Torricelli's formula outright).

Consider a small parcel of water at the surface and the same parcel as it exits the hole. Since the tank is large, the surface drops negligibly slowly, so we can treat the surface as essentially at rest while the exiting water has spdeed \( v \). Applying Bernoulli's equation between the surface (atmospheric pressure, height \( h = 20 \, \text{m} \), speed \( \approx 0 \)) and the hole (atmospheric pressure, height \( 0 \), speed \( v \)):

\[ P_{atm} + \rho g h + 0 = P_{atm} + 0 + \frac{1}{2}\rho v^2 \]

\[ v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 20} = \sqrt{392.4} \approx 19.81 \, \text{m/s} \]

The volume flow rate is the exit speed times the hole's cross-sectional area:

\[ Q = Av = (1 \times 10^{-6} \, \text{m}^2)(19.81 \, \text{m/s}) \approx 1.981 \times 10^{-5} \, \text{m}^3/\text{s} = 19.8 \, \text{cm}^3/\text{s} \]

Checking each option:

1. Option A, \( 87.5 \, \text{cm}^3/\text{s} \): would require an exit speed of roughly \( 87.5 \, \text{m/s} \) (since the hole area is \( 10^{-2} \, \text{cm}^2 \), giving \( v = Q/A \)), which would need a head of about \( \frac{87.5^2}{2\times9.81} \approx 390 \, \text{m} \), far more than the \( 20 \, \text{m} \) given.
2. Option B, \( 43.1 \, \text{cm}^3/\text{s} \): corresponds to an exit speed of about \( 43.1 \, \text{m/s} \), needing a head of roughly \( 94.7 \, \text{m} \), again far more than the given \( 20 \, \text{m} \).
3. Option C, \( 27.5 \, \text{cm}^3/\text{s} \): corresponds to a speed of about \( 27.5 \, \text{m/s} \), needing a head of about \( 38.5 \, \text{m} \), still nearly double the actual height.
4. Option D, \( 19.8 \, \text{cm}^3/\text{s} \): corresponds to an exit speed of about \( 19.8 \, \text{m/s} \), which is exactly what \( \sqrt{2gh} \) gives for \( h = 20 \, \text{m} \).

Only option D is consistent with the actual \( 20 \, \text{m} \) head of water.

Therefore, the correct answer is 19.8 cm³/s.