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Question: A water hose pipe of cross-section area \(5\,c{m^2}\) is used to fill a tank of 120L. It has been ob...

A water hose pipe of cross-section area 5 cm25\,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 cm21\,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/s2 = 10\,m/{s^2} )
B. 3m3m
C. 8m8m
D. 4.47m4.47m
E. 8.64m8.64m

Explanation

Solution

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 =volumetime = \dfrac{{volume}}{{time}}
=Area×lengthtime= \dfrac{{Area \times length}}{{time}}
=Area×Velocity=av= Area \times Velocity = av
(As velocity is length divided by time)
So According to equation of continuity the volume flow rate remains constant
Hence A1{A_1} v1{v_1}=A2{A_2} v2{v_2} =Vt = \dfrac{V}{t}
Volume=V =120 = 120L
Time = t = 2×6=120 s2 \times 6 = 120\,s
Height = h = 1 m
Let A1{A_1} and A2{A_2} be the area of cross-section of pipe and nozzle
Let v1{v_1} and v2{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 A1{A_1} v1{v_1}=A2{A_2} v2{v_2} =Vt = \dfrac{V}{t}
Therefore, putting the values in the above equation, v1=VA1t=120×10−35×10−4×2×60=2{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 A1{A_1} v1{v_1}=A2{A_2} v2{v_2}
We get v2=5v1=10{v_2} = 5{v_1} = 10 m/s
Now
h=ut+12gt2h = ut + \dfrac{1}{2}g{t^2} but u = initial velocity is zero hence
to{t_o} =2hg= \sqrt {\dfrac{{2h}}{g}}=2×110=0.447s\sqrt {\dfrac{{2 \times 1}}{{10}}} = 0.447s
R=v2to = {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=u2hg= u\sqrt {\dfrac{{2h}}{g}} and time t=2hg= \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.