Question

A water hose pipe of cross-section area $5\,c{m^2}$ is used to fill a tank of 120L. It has been observed that it takes 2min to fill the tank. Now, a nozzle with an opening of cross-section area $1\,c{m^2}$is attached to the hose. The nozzle is held so that the water is projected horizontally from a point 1 m above the ground. The horizontal distance over which the water can be projected is( Take g$ = 10\,m/{s^2}$ )
B. $3m$
C. $8m$
D. $4.47m$
E. $8.64m$

Answer 1

Hint:-The basic approach is to use a one-dimension equation of motion with projectile motion, that is, divide motion into x and y coordinates respectively and then solve equations in the respective directions. Second is the direct use of the equation of continuity which tells us the volume flow rate.

Complete step-by-step solution:
Volume flow rate $ = \dfrac{{volume}}{{time}}$
$ = \dfrac{{Area \times length}}{{time}}$
$ = Area \times Velocity = av$
(As velocity is length divided by time)
So According to equation of continuity the volume flow rate remains constant
Hence ${A_1}$${v_1}$=${A_2}$${v_2}$$ = \dfrac{V}{t}$
Volume=V $ = 120$L
Time = t = $2 \times 6 = 120\,s$
Height = h = 1 m
Let ${A_1}$ and ${A_2}$ be the area of cross-section of pipe and nozzle
Let ${v_1}$ and ${v_2}$ be the velocities at cross-section of pipe and nozzle
As discussed above in equation of continuity the volume flow rate remains constant
Hence ${A_1}$${v_1}$=${A_2}$${v_2}$$ = \dfrac{V}{t}$
Therefore, putting the values in the above equation, ${v_1} = \dfrac{V}{{{A_1}t}} = \dfrac{{120 \times {{10}^{ - 3}}}}{{5 \times {{10}^{ - 4}} \times 2 \times {{60}^{}}}} = 2$ which is the required result
Further simplifying,
Using equation of continuity ${A_1}$${v_1}$=${A_2}$${v_2}$
We get ${v_2} = 5{v_1} = 10$ m/s
Now
$h = ut + \dfrac{1}{2}g{t^2}$ but u = initial velocity is zero hence
${t_o}$$ = \sqrt {\dfrac{{2h}}{g}} $=$\sqrt {\dfrac{{2 \times 1}}{{10}}} = 0.447s$
R$ = {v_2}{t_o}$=4.47 m
Hence the correct option is (C )



Note:- Make sure of dimensional equality on both the side of Equation of continuity.
The direct formula for calculating the range R$ = u\sqrt {\dfrac{{2h}}{g}} $ and time t$ = \sqrt {\dfrac{{2h}}{g}} $
These formulas are derived from again simply 1-D equation of motions
So, what we conclude is Newton's law and equation of motion is the dominating concept in most of the problems.