Question

Water stands at a height of $100cm$ in a vessel whose side walls are vertical. $A,{\text{ }}B$ and $C$ are holes at height $80cm$ , $50cm$ , and $20cm$ respectively from the bottom of the vessel. The correct system of water flowing out is:
(A)
(B)
(C)
(D)

Answer 1

We can solve this problem using the help of Torricelli’s law and the speed of efflux. Torricelli’s law describes the relationship between liquid exit velocities and their height in the container. Liquid velocity can be described in simple words as, speed of the fluid leaving the hole in a container. The speed of the efflux is the speed of the fluid by which the fluid flows from the hole of the container.

Complete Step By Step Answer:
We have given four diagrams in four options. We need to find the range of all four options. And then check the options to select the correct option.

Time of flight, $t = \sqrt {\dfrac{{2(H - h)}}{g}}$
Here $H$ is the total height of the water.
$h$ is the respective height of the hole from the container.
$g$ is said to be the acceleration due to gravity.
The formula for the speed of efflux is the speed at which it flows from the hole of the tank.
The velocity of the water ejection $v = \sqrt {2gh}$
Total height of water column $H = 100cm$ .
Now substituting the height of the hole from the container.
For hole $A$ , $h = 20cm$
Therefore the time of flight, $t = \sqrt {\dfrac{{2(100 - 80)}}{g}}$
$\Rightarrow t = \sqrt {\dfrac{{2(80)}}{g}}$
For, velocity, $v = \sqrt {2g(20)}$
The range for the flow of the fluid can be found by using the distance formula.
$R = v \times t$
$\Rightarrow R = \sqrt {2gt} \times \sqrt {\dfrac{{2(H - h)}}{g}}$
$\Rightarrow R = \sqrt {2g(20)} \times \sqrt {\dfrac{{2(80)}}{g}}$
$\Rightarrow R = \sqrt {4 \times 20 \times 80}$ …….. (1)
For hole $B$ , $h = 50cm$
Time of flight, $t = \sqrt {\dfrac{{2(100 - 50)}}{g}}$
$\Rightarrow t = \sqrt {\dfrac{{2(50)}}{g}}$
Velocity $v = \sqrt {2g(50)}$
Range of the fluid $R = \sqrt {2g(50)} \times \sqrt {\dfrac{{2(50)}}{g}}$ …… (2)
For hole $C$ , $h = 80cm$
Time of flight, $t = \sqrt {\dfrac{{2(100 - 80)}}{g}}$
$\Rightarrow t = \sqrt {\dfrac{{2(20)}}{g}}$
Velocity, $v = \sqrt {2g(80)}$
Range of the fluid, $R = \sqrt {2g(80)} \times \sqrt {\dfrac{{2(20)}}{g}}$
$\Rightarrow R = \sqrt {4 \times 80 \times 20}$ ……. (3)
If we compare equations (1) and (3) we can see that they both are equal. Therefore the range of the fluid is the same for $A$ and $C$ .
Correct Answer: Therefore the correct option is C.

Note:
The value of g is the acceleration due to gravity at the surface of the Earth is $9.8$ meters per second. This theorem states that the speed of the water that is flowing through a hole in the tank at a given distance will be the same as that would be attained by a flow of water that is freely falling under the force of gravity alone. The speed of efflux is the speed at which the water flows through the container is independent of the flow of direction.