A tank is filled with water of density `10^(3)kg//m^(3)` and oil of density `9xx10^(3)kg//m^(3)`.The height of water layer is `1 m` and that of the oil layer is `4 M`. The velocity of efflux front an opening in the bottom of the tank is

Options

A. `sqrt(85)m//s`
B. `sqrt(88)m//s`
C. `sqrt(92)m//s`
D. `sqrt(98)m//s`

Explanation

To determine the velocity of efflux from an opening at the bottom of the tank, we can use Torricelli's theorem, which states that the speed of efflux \( v \) of a fluid under the force of gravity through an opening is given by:

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

where:
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( h \) is the effective height of the fluid column above the opening.

### Step-by-Step Solution:

1. **Identify the heights of the fluid layers**:
- The height of the water layer \( h_w = 1 \, \text{m} \).
- The height of the oil layer \( h_o = 4 \, \text{m} \).

2. **Calculate the effective height \( h \)**:
- The effective height \( h \) for the fluid column above the opening is the total height of the oil plus the height of the water, since the water is below the oil.
- Therefore, the total height \( h \) is:
\[ h = h_o + h_w = 4 \, \text{m} + 1 \, \text{m} = 5 \, \text{m} \]

3. **Substitute the values into Torricelli's theorem**:
- Using \( g = 9.81 \, \text{m/s}^2 \) and \( h = 5 \, \text{m} \):
\[ v = \sqrt{2gh} = \sqrt{2 \times 9.81 \, \text{m/s}^2 \times 5 \, \text{m}} \]

4. **Calculate the value**:
- First, calculate \( 2gh \):
\[ 2gh = 2 \times 9.81 \times 5 = 98.1 \, \text{m}^2/\text{s}^2 \]
- Now, take the square root:
\[ v = \sqrt{98.1} \approx 9.9 \, \text{m/s} \]

5. **Final Answer**:
- The velocity of efflux from the opening at the bottom of the tank is approximately \( 9.9 \, \text{m/s} \).