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

Options

A. `6.4xx10^(-6)m^(3)//s`
B. `12.6xx10^(-6)m^(3)//s`
C. `8.9xx10^(-6)m^(3)//s`
D. `2.23xx10^(-6)m^(3)//s`

Explanation

To find the rate of flow of water through a small hole at the bottom of a fully filled open tank, we can use Torricelli's theorem, which states that the speed of efflux of a fluid under the force of gravity through an orifice is given by:

\[ v = \sqrt{2gh} \]

where:
- \( v \) is the speed of the fluid,
- \( g \) is the acceleration due to gravity,
- \( h \) is the height of the fluid column above the hole.

### Step 1: Calculate the speed of water exiting the hole

Given:
- \( g = 10 \, \text{m/s}^2 \)
- \( h = 2 \, \text{m} \)

Using the formula:

\[ v = \sqrt{2gh} = \sqrt{2 \times 10 \times 2} = \sqrt{40} \approx 6.32 \, \text{m/s} \]

### Step 2: Convert the area of the hole from mm² to m²

The area of the hole is given as:

\[ \text{Area} = 2 \, \text{mm}^2 \]

To convert mm² to m²:

\[ 1 \, \text{mm} = 10^{-3} \, \text{m} \implies 1 \, \text{mm}^2 = (10^{-3})^2 \, \text{m}^2 = 10^{-6} \, \text{m}^2 \]

Thus,

\[ \text{Area} = 2 \, \text{mm}^2 = 2 \times 10^{-6} \, \text{m}^2 \]

### Step 3: Calculate the rate of flow of water through the hole

The rate of flow (Q) can be calculated using the formula:

\[ Q = \text{Area} \times \text{Velocity} \]

Substituting the values:

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

Calculating this gives:

\[ Q = 12.64 \times 10^{-6} \, \text{m}^3/\text{s} \approx 12.6 \times 10^{-6} \, \text{m}^3/\text{s} \]

### Final Answer

The rate of flow of water through the open hole is approximately:

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

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