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 $g = 10 \,m/s^{-2}$. the rate of flow of water through the open hole would be nearly:

• A. $2.23 \times 10^{-6}m^3/s^{-1}$
• B. $6.4 \times 10^{-6}m^3/s^{-1}$
• C. $12.6 \times 10^{-6}m^3/s^{-1}$
• D. $8.9 \times 10^{-6}m^3/s^{-1}$

Answer 1

The Correct Option is C

The problem involves calculating the rate of flow of water through a small hole near the bottom of an open tank. This is a classic fluid mechanics problem that can be solved by applying Torricelli's theorem.

Step 1: Understanding Torricelli's Theorem

Torricelli's theorem states that the speed \( v \) of efflux of a fluid under gravity through a hole at the bottom of a tank is given by:

v = \sqrt{2gh}

where:

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

Step 2: Calculate the Speed of Efflux

Substituting the given values into Torricelli's theorem, we find:

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

Approximating further:

v \approx 6.32 \, \text{m/s}

Step 3: Calculate the Rate of Flow

The rate of flow through the hole can be calculated using the formula for volumetric flow rate:

Q = A \times v

where:

• A = 2 \, \text{mm}^2 = 2 \times 10^{-6} \, \text{m}^2 (area of cross-section)
• v = 6.32 \, \text{m/s}

Substituting the values:

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

Conclusion: The approximate rate of flow is:

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

Therefore, the correct option is 12.6 \times 10^{-6} \, \text{m}^3/\text{s}.