A cylindrical tank is filled with water to a level of 3m. A hole is opened at a height of 52.5 cm from bottom. The ratio of the area of the hole to that of cross-sectional area of the cylinder is 0.1. The square of the speed with which water is coming out from the orifice is (Take `g= 10 m//s^(2)`)

Options

A. `50 m^(2)//s^(2)`
B. `40 m^(2)//s^(2)`
C. `51.5 m^(2)//s^(2)`
D. `50.5m^(2)//s^(2)`

Explanation

To find the square of the speed with which water is coming out from the orifice of the cylindrical tank, we can use Torricelli's theorem, which states that the speed \( v \) of efflux of a fluid under the force of gravity through an orifice is given by:

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

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

### Step 1: Determine the height \( h \) of the water column above the hole.
The total height of the water in the tank is 3 m, and the hole is located at a height of 52.5 cm (or 0.525 m) from the bottom. Therefore, the height \( h \) of the water column above the hole is:

\[ h = 3 \, \text{m} - 0.525 \, \text{m} = 2.475 \, \text{m} \]

### Step 2: Substitute the values into the equation for speed \( v \).
Now we can substitute \( h \) into the equation for \( v \):

\[ v = \sqrt{2gh} = \sqrt{2 \times 10 \, \text{m/s}^2 \times 2.475 \, \text{m}} \]

### Step 3: Calculate \( v^2 \).
We need to find \( v^2 \):

\[ v^2 = 2gh = 2 \times 10 \times 2.475 \]

Calculating this gives:

\[ v^2 = 20 \times 2.475 = 49.5 \, \text{m}^2/\text{s}^2 \]

### Step 4: Adjust for the area ratio.
The problem states that the ratio of the area of the hole \( A_1 \) to the cross-sectional area of the cylinder \( A_2 \) is given as 0.1. Therefore, we need to adjust the speed using the formula:

\[ v^2 = \frac{2gh}{1 - \frac{A_1^2}{A_2^2}} \]

Given that \( \frac{A_1}{A_2} = 0.1 \), we have:

\[ \frac{A_1^2}{A_2^2} = (0.1)^2 = 0.01 \]

Thus, we can substitute this into our equation:

\[ v^2 = \frac{49.5}{1 - 0.01} = \frac{49.5}{0.99} \]

Calculating this gives:

\[ v^2 \approx 50.0 \, \text{m}^2/\text{s}^2 \]

### Final Answer:
The square of the speed with which water is coming out from the orifice is approximately:

\[ \boxed{50.0 \, \text{m}^2/\text{s}^2} \]