A water hose pipe of cross-sectional area `5cm^(2)` is used to fill a tank of `120L`. It has been observed that it takes 2 min to fill the tank. Now, a nozzel with an opening of cross-sectional area `1cm^(2)` is attached to the hose. The nozzel is held so that water is projected horizontally from a point 1m above the ground. The horizontal distance over which the water can be projected is (Take `g=10m//s^(2))`.

Options

A. `3m`
B. `8m`
C. `4.47m`
D. `8.64 m`

Explanation

To solve the problem step-by-step, we need to find the horizontal distance over which the water can be projected from the nozzle. Here’s how we can approach this:

### Step 1: Calculate the flow rate of water from the hose pipe.

The flow rate (Q) can be calculated using the formula:
\[ Q = A \times v \]
where:
- \( A \) is the cross-sectional area of the hose,
- \( v \) is the velocity of water.

Given:
- The cross-sectional area of the hose pipe, \( A_1 = 5 \, \text{cm}^2 = 5 \times 10^{-4} \, \text{m}^2 \).
- The volume of the tank, \( V = 120 \, \text{L} = 120 \times 10^{-3} \, \text{m}^3 \).
- The time taken to fill the tank, \( t = 2 \, \text{min} = 120 \, \text{s} \).

Now, we can find the flow rate:
\[ Q = \frac{V}{t} = \frac{120 \times 10^{-3} \, \text{m}^3}{120 \, \text{s}} = 1 \times 10^{-3} \, \text{m}^3/\text{s} \]

### Step 2: Calculate the velocity of water from the hose pipe.

Using the flow rate:
\[ Q = A_1 \times v_1 \]
We can rearrange this to find \( v_1 \):
\[ v_1 = \frac{Q}{A_1} = \frac{1 \times 10^{-3}}{5 \times 10^{-4}} = 2 \, \text{m/s} \]

### Step 3: Calculate the velocity of water from the nozzle.

The nozzle has a smaller cross-sectional area:
- \( A_2 = 1 \, \text{cm}^2 = 1 \times 10^{-4} \, \text{m}^2 \).

Using the principle of conservation of mass (continuity equation):
\[ A_1 v_1 = A_2 v_2 \]
We can solve for \( v_2 \):
\[ v_2 = \frac{A_1}{A_2} v_1 = \frac{5 \times 10^{-4}}{1 \times 10^{-4}} \times 2 = 10 \, \text{m/s} \]

### Step 4: Calculate the time of flight of the water.

The water is projected horizontally from a height \( h = 1 \, \text{m} \). The time of flight \( t_0 \) can be calculated using the formula:
\[ t_0 = \sqrt{\frac{2h}{g}} \]
where \( g = 10 \, \text{m/s}^2 \):
\[ t_0 = \sqrt{\frac{2 \times 1}{10}} = \sqrt{0.2} \approx 0.447 \, \text{s} \]

### Step 5: Calculate the horizontal distance.

The horizontal distance \( R \) can be calculated using:
\[ R = v_2 \times t_0 \]
Substituting the values:
\[ R = 10 \, \text{m/s} \times 0.447 \, \text{s} \approx 4.47 \, \text{m} \]

### Final Answer:
The horizontal distance over which the water can be projected is approximately **4.47 meters**.

---