A tank is filled with water of density `1 g cm^(-3)` and oil of density `0.9 g cm^(-3)` . The height of the water layer is 100 cm and of the oil layer is 400 cm. If g = 980 `cm s^(-2)` , then the velocity of efflux from an opening in the bottom of the tank is

Options

A. `sqrt(920 xx980) " cm s"^(-1)`
B. `sqrt(900 xx980) " cm s"^(-1)`
C. `sqrt(1000 xx980) " cm s"^(-1)`
D. `sqrt(92 xx980) " cm s"^(-1)`

Answer

To solve the problem, we need to find the velocity of efflux from an opening at the bottom of a tank filled with water and oil. We will use Torricelli's theorem, which states that the velocity of efflux (v) from an opening under the influence of gravity 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 densities and heights:**
- Density of water (\( \rho_w \)) = 1 g/cm³
- Density of oil (\( \rho_o \)) = 0.9 g/cm³
- Height of water (\( h_w \)) = 100 cm
- Height of oil (\( h_o \)) = 400 cm
- Acceleration due to gravity (\( g \)) = 980 cm/s²

2. **Calculate the pressure at the bottom of the tank:**
The pressure at the bottom of the tank is due to both the water and the oil layers. The pressure due to a fluid column is given by:

\[ P = \rho g h \]

Therefore, the total pressure at the bottom (\( P_{total} \)) is the sum of the pressures due to the water and the oil:

\[ P_{total} = P_{water} + P_{oil} = \rho_w g h_w + \rho_o g h_o \]

Substituting the values:

\[ P_{total} = (1 \, \text{g/cm}^3)(980 \, \text{cm/s}^2)(100 \, \text{cm}) + (0.9 \, \text{g/cm}^3)(980 \, \text{cm/s}^2)(400 \, \text{cm}) \]

3. **Calculate each pressure:**
- Pressure due to water:
\[ P_{water} = 1 \times 980 \times 100 = 98000 \, \text{dyne/cm}^2 \]
- Pressure due to oil:
\[ P_{oil} = 0.9 \times 980 \times 400 = 352800 \, \text{dyne/cm}^2 \]

4. **Total pressure at the bottom:**
\[ P_{total} = 98000 + 352800 = 450800 \, \text{dyne/cm}^2 \]

5. **Calculate the effective height (h) for the velocity of efflux:**
To find the effective height \( h \) that corresponds to the total pressure, we can use the formula for pressure again:

\[ P_{total} = \rho_w g h \]

Rearranging gives:

\[ h = \frac{P_{total}}{\rho_w g} = \frac{450800}{1 \times 980} = 460 \, \text{cm} \]

6. **Calculate the velocity of efflux using Torricelli's theorem:**
Now we can find the velocity of efflux:

\[ v = \sqrt{2gh} = \sqrt{2 \times 980 \times 460} \]

Calculate:

\[ v = \sqrt{902800} \approx 950.84 \, \text{cm/s} \]

### Final Answer:
The velocity of efflux from the opening in the bottom of the tank is approximately \( 950.84 \, \text{cm/s} \).