Question

Water from a tap emerges vertically downwards with initial velocity $4m{{s}^{-1}}$. The cross sectional area of the tap is $A$. The flow is steady and pressure is constant throughout the stream of water. The distance $h$ vertically below the tap, where the cross sectional area of the stream becomes $\left( \dfrac{2}{3} \right)A$, is $\left( g=10m/{{s}^{2}} \right)$.
A. $0.5m$
B. $1m$
C. $1.5m$
D. $2.2m$

Answer 1

Hint: This problem can be solved by using the concept of the equation of continuity. It states that the rate of volume of the liquid flowing through different sections of the pipe is the same when the liquid is incompressible and there are no sources or sink. The product of cross sectional area and velocity of fluid will be constant. We can find the relation between the velocity of the fluid just below the tap and distance $h$ below the tap by using the equation of motion for constant acceleration.

Formula used:
According to the equation of continuity,
$Av=\text{constant}$
where $A$ is the cross sectional area of the fluid and $v$ is the velocity of the fluid.
For a body that is undergoing uniformly accelerated motion,
${{v}^{2}}={{u}^{2}}+2as$
where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration, $s$ is the displacement covered by the body.

Complete step by step answer:
We can solve this problem by using the concept of the continuity equation. According to the equation of continuity for an incompressible fluid, the volume of fluid flowing per second in a pipe remains constant when there is no external source or sink.
The volume flow per second can also be written as the product of the cross sectional area of the fluid and the velocity of the fluid.
Hence, according to the equation of continuity,
$Av=\text{constant}$ --(1)
where $A$ is the cross sectional area of the fluid and $v$ is the velocity of the fluid.
We can get a relation between the velocities of the fluid when just out of the tap and when fallen a vertical distance $h$ by making use of the equations for uniformly accelerated motion. The uniform acceleration is the acceleration due to gravity $g$.
Now, let us analyze the question.
Let the initial velocity of water after just exiting the tap be $u=4m/s$.
The initial cross sectional area of the fluid is ${{A}_{1}}=A$.
The final velocity of the water after falling a vertical distance $h$ be $v$.
Final cross sectional area of the stream of water is ${{A}_{2}}=\dfrac{2}{3}A$.
Acceleration of the stream of water is $g$.
Displacement of the stream of water in the direction of acceleration is $h$.
For a body that is undergoing uniformly accelerated motion,
${{v}^{2}}={{u}^{2}}+2as$ --(2)
where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration, $s$ is the displacement covered by the body.
Therefore, using (2) and the information according to the question, we get,
${{v}^{2}}={{u}^{2}}+2gh$ --(3)
Now, using (1), we get,
${{A}_{1}}u={{A}_{2}}v=\text{constant}$
$\therefore Au=\dfrac{2}{3}Av$
$\therefore v=\dfrac{3}{2}u$
Squaring both sides, we get,
${{v}^{2}}={{\left( \dfrac{3}{2}u \right)}^{2}}$
$\therefore {{v}^{2}}=\dfrac{9{{u}^{2}}}{4}$ --(4)
Putting (4) in (3), we get,
$\dfrac{9{{u}^{2}}}{4}={{u}^{2}}+2gh$
$\therefore \dfrac{9{{u}^{2}}}{4}-{{u}^{2}}=2gh$
$\therefore \dfrac{9-4}{4}{{u}^{2}}=2gh$
$\therefore \dfrac{5}{4}{{u}^{2}}=2gh$
$\therefore h=\dfrac{5{{u}^{2}}}{4\times 2g}$
Putting the values of the variables as given in the questions, we get,
$h=\dfrac{5\times {{\left( 4 \right)}^{2}}}{4\times 2\times 10}=\dfrac{5\times 16}{4\times 2\times 10}=1m$
Hence, the required value of $h$ is $1m$.
Therefore, the correct option is B) $1m$.

Note: This is a typical problem which highlights the importance of the equation of continuity in fluid dynamics. One important point to notice is that the question states that the pressure of the fluid remains the same which is essential since it makes the equation of continuity applicable in this case. If there was a pressure change in the stream of the water, then Bernoulli’s equation would have to be used, which makes use of the conservation of energy in the fluid and the whole problem would change. Students must be careful and understand by reading the question carefully, where to use the equation of continuity and where to use the Bernoulli’s equation.