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. `50`
B. `51`
C. `48`
D. `51.5`

Explanation

To solve the problem, we will use Bernoulli's principle and the equation of continuity. Here are the steps to find \( v^2 \), the square of the velocity of water coming out of the hole.

### Step 1: Understand the given data
- Height of water in the container, \( H = 3 \, \text{m} = 300 \, \text{cm} \)
- Height of the hole from the bottom, \( h = 52.5 \, \text{cm} \)
- Area of the hole, \( a \)
- Cross-sectional area of the container, \( A \)
- Given ratio \( \frac{a}{A} = 0.1 \)

### Step 2: Calculate the effective height of water above the hole
The effective height \( h' \) of water above the hole is calculated as:
\[ h' = H - h = 300 \, \text{cm} - 52.5 \, \text{cm} = 247.5 \, \text{cm} = 2.475 \, \text{m} \]

### Step 3: Apply Bernoulli's equation
According to Torricelli's law, the velocity \( v \) of efflux from the hole can be given by:
\[ v = \sqrt{2gh'} \]
where \( g \) is the acceleration due to gravity, approximately \( 10 \, \text{m/s}^2 \).

### Step 4: Substitute the values into the equation
Substituting \( g = 10 \, \text{m/s}^2 \) and \( h' = 2.475 \, \text{m} \):
\[ v = \sqrt{2 \cdot 10 \cdot 2.475} \]

### Step 5: Calculate \( v^2 \)
Calculating \( v^2 \):
\[ v^2 = 2 \cdot 10 \cdot 2.475 = 49.5 \, \text{m}^2/\text{s}^2 \]

### Step 6: Adjust for the area ratio
Since the area of the hole is much smaller than the cross-sectional area of the container, we need to consider the effect of the area ratio. The formula for the velocity considering the area ratio is:
\[ v^2 = \frac{2gh'}{1 - \frac{a}{A}}^2 \]
Substituting \( \frac{a}{A} = 0.1 \):
\[ v^2 = \frac{49.5}{(1 - 0.1)^2} = \frac{49.5}{(0.9)^2} = \frac{49.5}{0.81} \approx 61.11 \, \text{m}^2/\text{s}^2 \]

### Final Result
Thus, the square of the velocity of water coming out of the hole is approximately:
\[ v^2 \approx 61.11 \, \text{m}^2/\text{s}^2 \]