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

Hint: The formula discharge can be used to solve this problem. We can also observe that the total volume of the container is fixed therefore the volume inlet per second is equal to the volume outlet per second. We must also consider that the acceleration in the horizontal direction is zero, so a simple formula of speed can be applied to find the horizontal distance at which the water will be projected.

Formula used:
The formula of the discharge is given by $Q = A \cdot v$ where $Q$ is discharge $A$ is the area of cross-section and $v$ is the velocity of the water.
The formula of time taken by a body to fall from height h with some initial velocity in the horizontal direction is given by $t = \sqrt {\dfrac{{2 \cdot h}}{g}} $where g is the acceleration due to gravity.

Complete step by step answer:
The discharge is given by $Q = A \cdot v$ where Q is the discharge, A is the area of cross section and v is the velocity of flow.
The velocity is given by $v = \dfrac{Q}{A}$.
$ \Rightarrow {v_1} = \dfrac{{120 \times {{10}^{ - 3}}}}{{60 \times 2 \times 5 \times {{10}^{ - 4}}}}$
$ \Rightarrow {v_1} = 2\dfrac{m}{s}$
As the volume is fixed therefore the volume inlet per second is equal to the volume output per second.
Therefore,
$ \Rightarrow {Q_1} = {Q_2}$
$ \Rightarrow {A_1}{v_1} = {A_2}{v_2}$
$ \Rightarrow 2 \cdot {A_1} = {A_2}{v_2}$
Since, ${A_2} = \dfrac{{{A_1}}}{5}$
$ \Rightarrow 2 \cdot {A_1} = \left( {\dfrac{{{A_1}}}{5}} \right){v_2}$
$ \Rightarrow {v_2} = 10\dfrac{m}{s}$
Since,
$ \Rightarrow t = \sqrt {\dfrac{{2 \cdot h}}{g}} $ Where $t$ is the time taken $g$ is the acceleration due to gravity and $h$ is height.
$ \Rightarrow t = \sqrt {\dfrac{{2 \times 1}}{{10}}} $
$ \Rightarrow t = \sqrt {0.2} $
$ \Rightarrow t = 0.447s$
Since,
${\text{Distance}} = {\text{velocity}} \times {\text{time}}$.
$ \Rightarrow d = 10 \times 0. 447$
$ \Rightarrow d = 4.47$.

The horizontal distance at which the water will be projected is equal to $d = 4.47m$.

Note:
It is advisable to students to remember the formula of time taken by a body down from height h having some initial velocity as it can be very useful in solving these types of problems. The sign of acceleration due to gravity is positive because as the water is flowing down so flow is in the direction of acceleration.