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 solve the problem of finding the rate of flow of water through a small hole in a tank, we can follow these steps:

### Step 1: Understand the given data
- Height of the water column (h) = 2 m
- Area of the hole (A) = 2 mm² = 2 × 10^(-6) m²
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Use Torricelli's Law
According to Torricelli's law, the velocity (v) of fluid flowing out of an orifice under the influence of gravity is given by:
\[ v = \sqrt{2gh} \]

### Step 3: Calculate the velocity
Substituting the values into the equation:
\[ v = \sqrt{2 \times 10 \, \text{m/s}^2 \times 2 \, \text{m}} \]
\[ v = \sqrt{40} \]
\[ v \approx 6.32 \, \text{m/s} \]

### Step 4: Calculate the rate of flow (Q)
The rate of flow (Q) is given by the product of the area of the hole and the velocity of the fluid:
\[ Q = A \times v \]
Substituting the values:
\[ Q = (2 \times 10^{-6} \, \text{m}^2) \times (6.32 \, \text{m/s}) \]

### Step 5: Perform the multiplication
\[ Q = 12.64 \times 10^{-6} \, \text{m}^3/\text{s} \]

### Step 6: Round the answer
The rate of flow of water through the hole is approximately:
\[ Q \approx 12.6 \times 10^{-6} \, \text{m}^3/\text{s} \]

### Final Answer
The rate of flow of water through the open hole would be nearly \( 12.6 \times 10^{-6} \, \text{m}^3/\text{s} \).

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