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:

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