Question

At what velocity does water emerge from an orifice in a tank in which gauge pressure is
\[3 \times {10^5}N{m^{ - 2}}\;\] before the flow starts? (Take the density of water \[ = {\text{ }}1000{\text{ }}kg{\text{ }}{m^{ - 3}}\])
A. $24.5m{s^{ - 1}}$
B. $14.5m{s^{ - 1}}$
C. $34.5m{s^{ - 1}}$
D. $44.5m{s^{ - 1}}$

Answer 1

Hint:use the formula of pressure exerted by fluid, that is $p = h\rho g$, where $p$ represents the gauge pressure, $\rho $ represents density of water, and $h$ represents the height to which water is filled in a container and $g$ represents the acceleration due to gravity. Using the above formula, you will get the value of $h$. Now use the efflux velocity formula $v = \sqrt {2gh} $ and you will get the answer.

Complete step-by-step solution:
Formula for pressure exerted by a fluid in any container is given by : $p = h\rho g - - - (1)$
Where $p = $ pressure exerted by fluid in any container
$h = $ Height to which water is filled in a container
$\rho = $ Density of water
$g = $Acceleration due to gravity = $9.8m{s^{ - 2}}$
According to question, $p = 3 \times {10^5}N{m^{ - 2}}$
$\rho = 1000kg{m^{ - 3}}$
$g = 9.8m{s^{ - 2}}$
Putting these values in $(1)$
We get the value of $h$
\[\therefore h = \dfrac{p}{{\rho g}} = \dfrac{{3 \times {{10}^5}}}{{1000 \times 9.8}}m - - - (2)\]
So the height to which water is filled is given by $h$
Now, as per the question we need to find the speed of the water at which it will emerge from the orifice.
For this we have to first understand what velocity of efflux is:
Velocity of efflux: the average flow rate of liquid emitted out of the source.
Velocity of efflux is given by: $v = \sqrt {2gh} - - - (3)$
Putting the value of $h$ from $(2)$ in $(3)$
We get:
$
  v = \sqrt {\dfrac{{2 \times 9.8 \times 3 \times {{10}^5}}}{{1000 \times 9.8}}} \\
  v = \sqrt {600} \\
 $
Solving this we get:
$v = 24.495m{s^{ - 1}}$
$ \approx 24.5m{s^{ - 1}}$
So the correct answer is option A.

Additional information:
Velocity of efflux is same as object falling from rest, from a height $h$, it would be derived as:
Using ${1^{st}}$ equation of motion
$v = u + gt$
Where $u = $initial speed
$v = $Final speed
Since object was in rest initially $u = 0$
We get: $v = gt - - - (4)$
Using ${2^{nd}}$equation of motion $s = ut + \dfrac{1}{2}g{t^2}$
$
  h = ut + \dfrac{1}{2}g{t^2} \\
  u = 0 \\
 $
We get $h = \dfrac{1}{2}g{t^2}$
Solving for $t$:
We get $t = \sqrt {\dfrac{{2h}}{g}} - - - (5)$
Putting this in equation $(4)$ we get:
$v = \sqrt {2gh} $

Note:- Remember always take care of units. Remember that density of liquid varies from one liquid to another and pressure is a measurement force exerted per unit area, this means pressure is force applied.