The level of water in a tank is 5 m high. A hole of area of cross section 1 `cm^(2)` is made at the bottom of the tank. The rate of leakage of water for the hole in `m^(3)s^(-1)` is `(g=10ms^(-2))`

Options

A. `10^(-3) m^(3)s^(-1)`
B. `10^(-4) m^(3)s^(-1)`
C. `10 m^(3)s^(-1)`
D. `10^(-2) m^(3)s^(-1)`

Explanation

To solve the problem of determining the rate of leakage of water from a hole in a tank, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Height of water in the tank (h) = 5 m
- Area of the hole (A) = 1 cm² = \(1 \times 10^{-4}\) m² (conversion from cm² to m²)
- Acceleration due to gravity (g) = 10 m/s²

2. **Use the Formula for Velocity of Efflux**:
The velocity of efflux (v) through the hole can be calculated using Torricelli's law:
\[ v = \sqrt{2gh} \]
Substituting the values:
\[ v = \sqrt{2 \times 10 \, \text{m/s}^2 \times 5 \, \text{m}} = \sqrt{100} = 10 \, \text{m/s} \]

3. **Calculate the Rate of Leakage (Q)**:
The rate of leakage (Q) can be calculated using the formula:
\[ Q = A \times v \]
Substituting the area and the velocity:
\[ Q = (1 \times 10^{-4} \, \text{m}^2) \times (10 \, \text{m/s}) = 1 \times 10^{-3} \, \text{m}^3/\text{s} \]

4. **Final Answer**:
The rate of leakage of water from the hole is:
\[ Q = 1 \times 10^{-3} \, \text{m}^3/\text{s} \]