A tank of height 5 m/s completely filled with water. There is a hole of cross sectional area `1cm^2` near its bottom. What is the initial volume of water that will come out of the hole per second. (Use g=`10m//s^2`)

Options

A. `10^(-3)m^3//s`
B. `2xx10^(-3)m^3//s`
C. `3xx10^(-2)m^3//s`
D. `4xx10^(-3)m^3//s`

Explanation

To solve the problem, we will use Torricelli's theorem, which relates the speed of fluid flowing out of an orifice to the height of the fluid above the hole. Here are the steps to find the initial volume of water that will come out of the hole per second:

### Step 1: Identify the given values
- Height of the water column (h) = 5 m
- Cross-sectional area of the hole (A) = 1 cm² = 1 × 10⁻⁴ m² (conversion from cm² to m²)
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Apply Torricelli's theorem
According to Torricelli's theorem, the velocity (v) of the fluid flowing out of the hole can be calculated using the formula:
\[ v = \sqrt{2gh} \]

### Step 3: Calculate the velocity of water
Substituting the values into the formula:
\[ v = \sqrt{2 \times 10 \, \text{m/s}^2 \times 5 \, \text{m}} \]
\[ v = \sqrt{100} \]
\[ v = 10 \, \text{m/s} \]

### Step 4: Calculate the discharge (Q)
The discharge (Q), which is the volume of water flowing out per second, is given by the formula:
\[ Q = v \times A \]

### Step 5: Substitute the values to find discharge
Substituting the values we have:
\[ Q = 10 \, \text{m/s} \times 1 \times 10^{-4} \, \text{m}^2 \]
\[ Q = 10 \times 10^{-4} \, \text{m}^3/\text{s} \]
\[ Q = 1 \times 10^{-3} \, \text{m}^3/\text{s} \]

### Step 6: Final answer
The initial volume of water that will come out of the hole per second is:
\[ Q = 1 \, \text{litre/s} \] (since 1 m³ = 1000 litres)