A tank is filled with water of density `10^(3)kg//m^(3)` and oil of density `0.9xx10^(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 find the velocity of efflux from an opening at the bottom of a tank filled with water and oil, we can use Torricelli’s theorem, which states that the velocity of efflux \( v \) from an opening is given by:

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

where \( g \) is the acceleration due to gravity and \( h \) is the effective height of the fluid column above the opening.

### Step-by-Step Solution:

1. **Identify the given data:**
- Density of water, \( \rho_w = 10^3 \, \text{kg/m}^3 \)
- Density of oil, \( \rho_o = 0.9 \times 10^3 \, \text{kg/m}^3 \)
- Height of water, \( h_w = 1 \, \text{m} \)
- Height of oil, \( h_o = 4 \, \text{m} \)

2. **Calculate the pressure at the opening due to the water and oil:**
The pressure at the opening can be calculated by considering the contributions from both the water and the oil layers:
\[ P = P_{\text{water}} + P_{\text{oil}} = \rho_w g h_w + \rho_o g h_o \]

3. **Substituting the values:**
- Use \( g \approx 10 \, \text{m/s}^2 \) for simplicity.
\[ P = (10^3 \, \text{kg/m}^3)(10 \, \text{m/s}^2)(1 \, \text{m}) + (0.9 \times 10^3 \, \text{kg/m}^3)(10 \, \text{m/s}^2)(4 \, \text{m}) \]
\[ P = 10^4 \, \text{Pa} + 0.9 \times 10^3 \times 10 \times 4 \, \text{Pa} \]
\[ P = 10^4 \, \text{Pa} + 3600 \, \text{Pa} = 13600 \, \text{Pa} \]

4. **Calculate the effective height \( h \):**
The effective height \( h \) can be calculated from the pressure:
\[ P = \rho_{\text{eff}} g h \]
where \( \rho_{\text{eff}} \) is the effective density of the fluid column above the opening. We can find \( h \) using:
\[ h = \frac{P}{\rho_{\text{eff}} g} \]
Here, \( \rho_{\text{eff}} \) can be calculated as:
\[ \rho_{\text{eff}} = \frac{\rho_o h_o + \rho_w h_w}{h_o + h_w} \]
\[ \rho_{\text{eff}} = \frac{(0.9 \times 10^3)(4) + (10^3)(1)}{4 + 1} \]
\[ \rho_{\text{eff}} = \frac{3600 + 1000}{5} = \frac{4600}{5} = 920 \, \text{kg/m}^3 \]

5. **Calculate the effective height \( h \):**
Now substituting back to find \( h \):
\[ h = \frac{13600 \, \text{Pa}}{920 \, \text{kg/m}^3 \times 10 \, \text{m/s}^2} \]
\[ h = \frac{13600}{9200} = 1.478 \, \text{m} \]

6. **Calculate the velocity of efflux:**
Now we can find the velocity using Torricelli’s theorem:
\[ v = \sqrt{2gh} = \sqrt{2 \times 10 \, \text{m/s}^2 \times 1.478 \, \text{m}} \approx \sqrt{29.56} \approx 5.44 \, \text{m/s} \]

### Final Answer:
The velocity of efflux from the opening at the bottom of the tank is approximately \( 5.44 \, \text{m/s} \).