Question

A small hole of area of cross-section \(2 mm^2\) is present near the bottom of a fully filled open tank of height \(2 m\). Taking \(g=10 m/s2\), the rate of flow of water through the open hole would be nearly:

• A. \(12.6 \times 10^{-6} m^3/s \)
• B. \(8.9 \times 10^{-6} m^3/s \)
• C. \(2.23\times 10^{-6} m^3/s \)
• D. \(6.4 \times 10^{-6} m^3/s \)

Answer 1

The Correct Option is A

 To solve this problem, we'll use Torricelli's law, which provides a way to determine the speed at which fluid exits a hole in a container under the influence of gravity. The formula derived from Torricelli's theorem is:

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

where:

• \(v\) = velocity of the fluid exiting the hole
• \(g\) = acceleration due to gravity (\(10 \, \text{m/s}^2\))
• \(h\) = height of the fluid above the hole (\(2 \, \text{m}\))

 

First, let's calculate the velocity of the water exiting the hole:

\(v = \sqrt{2 \times 10 \times 2} = \sqrt{40} = 6.32 \, \text{m/s}\)

Next, we calculate the rate of flow of water through the hole. The rate of flow (or discharge) can be determined using the formula:

\(Q = A \times v\)

where:

• \(Q\) = rate of flow
• \(A\) = cross-sectional area of the hole (\(2 \, \text{mm}^2 = 2 \times 10^{-6} \, \text{m}^2\))
• \(v\) = velocity of the fluid (\(6.32 \, \text{m/s}\))

 

Now, substitute the known values:

\(Q = 2 \times 10^{-6} \times 6.32 = 12.64 \times 10^{-6} \, \text{m}^3/\text{s}\)

Upon rounding this to two decimal places, we get:

\(Q \approx 12.6 \times 10^{-6} \, \text{m}^3/\text{s}\)

Thus, the correct answer is approximately \(12.6 \times 10^{-6} \, \text{m}^3/\text{s}\), which matches with the given option \(12.6 \times 10^{-6} \, \text{m}^3/\text{s}\).