Water is filled in a container upto height 3m. A small hole of area 'a' is punched in the wall of the container at a height 52.5 cm from the bottom. The cross sectional area of the container is A. If `a//A=0.1` then `v^2` is (where v is the velocity of water coming out of the hole)

Options

A. (a) 50
B. (b) 51
C. (c) 48
D. (d) 51.5

Explanation

To solve the problem, we will use the principles of fluid dynamics, specifically the Bernoulli equation and the equation of continuity.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Height of water in the container, \( H_1 = 3 \, \text{m} \)
- Height of the hole from the bottom, \( H_2 = 52.5 \, \text{cm} = 0.525 \, \text{m} \)
- The ratio of the areas, \( \frac{a}{A} = 0.1 \)

2. **Calculate the Height Difference:**
- The height difference \( h \) between the water surface and the hole is:
\[ h = H_1 - H_2 = 3 \, \text{m} - 0.525 \, \text{m} = 2.475 \, \text{m} \]

3. **Apply the Bernoulli Equation:**
- According to Bernoulli's principle, for a fluid flowing in a streamline, the following equation holds:
\[ P_1 + \frac{1}{2} \rho v_1^2 + \rho g H_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g H_2 \]
- Since both points are open to the atmosphere, the pressures \( P_1 \) and \( P_2 \) cancel out. Thus, we have:
\[ \frac{1}{2} \rho v_1^2 + \rho g H_1 = \frac{1}{2} \rho v_2^2 + \rho g H_2 \]

4. **Cancel the Density:**
- Dividing the entire equation by \( \rho \) gives:
\[ \frac{1}{2} v_1^2 + g H_1 = \frac{1}{2} v_2^2 + g H_2 \]

5. **Rearranging the Equation:**
- Rearranging gives:
\[ \frac{1}{2} v_2^2 = \frac{1}{2} v_1^2 + g (H_1 - H_2) \]
- Simplifying further:
\[ v_2^2 = v_1^2 + 2g (H_1 - H_2) \]

6. **Using the Continuity Equation:**
- From the continuity equation, we have:
\[ A v_1 = a v_2 \implies v_1 = \frac{a}{A} v_2 \]
- Given \( \frac{a}{A} = 0.1 \), we can write:
\[ v_1 = 0.1 v_2 \]

7. **Substituting \( v_1 \) in the Bernoulli Equation:**
- Substitute \( v_1 \) into the equation:
\[ v_2^2 = (0.1 v_2)^2 + 2g (H_1 - H_2) \]
- Expanding gives:
\[ v_2^2 = 0.01 v_2^2 + 2g (2.475) \]
- Rearranging yields:
\[ v_2^2 - 0.01 v_2^2 = 2g (2.475) \]
- This simplifies to:
\[ 0.99 v_2^2 = 2g (2.475) \]

8. **Calculating \( v_2^2 \):**
- Using \( g = 10 \, \text{m/s}^2 \):
\[ 0.99 v_2^2 = 2 \times 10 \times 2.475 \]
\[ 0.99 v_2^2 = 49.5 \]
\[ v_2^2 = \frac{49.5}{0.99} \approx 50 \, \text{m}^2/\text{s}^2 \]

### Final Answer:
Thus, the velocity squared \( v^2 \) of the water coming out of the hole is:
\[ v^2 = 50 \, \text{m}^2/\text{s}^2 \]