Question

When a pipe of radius of cross section ‘r’ is arranged at a height h horizontally the jet of from it touches distance ‘x’ from a point just below the pipe on the ground. If the pipe is partly closed such that its radius of cross section becomes $\dfrac{r}{2}$ and it is arranged at a height of 4h then the horizontal distance at which the waterfall increases by
a) $8x$
b) $2x$
c) $7x$
d) $\dfrac{x}{2}$

Answer 1

In the above question it is asked to determine the horizontal distance covered by the water when the nozzle of the pipe is kept at a distance of 4h and the area of the cross section more specifically the radius is reduced to half. First to determine the change in the velocity of the water we can use the equation of continuity. Further using Newton’s kinematic equation we can determine the distance covered in terms of x.
Formula used:
${{A}_{1}}{{v}_{1}}={{A}_{2}}{{v}_{2}}$
$h=ut+\dfrac{1}{2}g{{t}^{2}}$
$x=vt$

Complete step by step answer:
Initially when the pipe is kept at a height of ‘h’ units from the ground having area ${{A}_{1}}$ , let us say the water comes with a velocity of ${{v}_{1}}$ from the nozzle of the pipe and travels a distance x. Further when the radius of the pipe reduces by half the area of cross section of the pipe is ${{A}_{2}}$ and the speed of water is ${{v}_{2}}$ . hence from the equation of continuity,
$\begin{aligned} & {{A}_{1}}{{v}_{1}}={{A}_{2}}{{v}_{2}}\text{ },\because {{r}_{1}}=\dfrac{{{r}_{1}}}{2} \\\ & \pi {{r}_{1}}^{2}{{v}_{1}}=\pi {{\left( \dfrac{{{r}_{1}}}{2} \right)}^{2}}{{v}_{2}} \\\ & \Rightarrow {{v}_{2}}=4{{v}_{1}} \\\ \end{aligned}$
The distance x covered is equal to the product of horizontal velocity i.e. ${{v}_{1}}$ times the time of flight. When the water comes out of the nozzle it does not have any vertical component of velocity. As soon as the water leaves the nozzle it is under the action of gravity and falls down. Hence from Newton’s second kinematic equation the time of light ‘t’ for water to reach ground can be determined.
Let us say an object falls towards the Earth from a height h and has initial velocity ‘u’. The time ‘t’ taken by the body to reach the ground can be determined by Newton’s second kinematic equation i.e.
$h=ut+\dfrac{1}{2}g{{t}^{2}}$
The initial velocity of water i.e. the vertical component is zero. Hence,
$\begin{aligned} & h=(0)t+\dfrac{1}{2}g{{t}^{2}} \\\ & \Rightarrow {{t}^{2}}=\dfrac{2h}{g} \\\ & \Rightarrow t=\sqrt{\dfrac{2h}{g}} \\\ \end{aligned}$
Hence the horizontal distance covered in the above time is,
$\begin{aligned} & x={{v}_{1}}t \\\ & \Rightarrow x={{v}_{1}}\sqrt{\dfrac{2h}{g}}....(1) \\\ \end{aligned}$
When the radius of the pipe is reduced by half and the pipe is taken up to a height of 4h the horizontal distance covered ‘d’ is
$\begin{aligned} & d={{v}_{2}}\sqrt{\dfrac{2(4h)}{g}},\text{ }\because {{v}_{2}}=4{{v}_{1}} \\\ & \Rightarrow d=2\times 4{{v}_{1}}\sqrt{\dfrac{2h}{g}}=8{{v}_{1}}\sqrt{\dfrac{2h}{g}}.....(2) \\\ \end{aligned}$
Taking the ratio of equation 1 and 2 we get,
$\begin{aligned} & \dfrac{x}{d}=\dfrac{{{v}_{1}}\sqrt{\dfrac{2h}{g}}}{8{{v}_{1}}\sqrt{\dfrac{2h}{g}}}=\dfrac{1}{8} \\\ & \Rightarrow d=8x \\\ \end{aligned}$

So, the correct answer is “Option a”.

Note: The above equation of continuity holds valid only if the pressure inside the pipe remains constant. If the pressure is not constant then the equation of continuity does not hold valid and hence the answer obtained will have no meaning. The path followed by the water is a projectile and hence one can also solve this using the dynamics of projectile motion.