Question

A cylindrical tank has a hole of \[1\,{\text{c}}{{\text{m}}^2}\] m bottom. If the water is allowed to flow into the tank from a tube above it at the rate of \[70\,{\text{c}}{{\text{m}}^3}/{\text{s}}\], then the maximum height up to which water can rise in the tank is?
A. \[2.5\,{\text{cm}}\]
B. \[5\,{\text{cm}}\]
C. \[10\,{\text{cm}}\]
D. \[0.25\,{\text{cm}}\]

Answer 1

Hint: Use the equation for the volume flow rate of the liquid. Apply Bernoulli’s theorem to the water flowing in from the tank and the water flowing out from the hole to determine the velocity of the liquid flowing from the tube. Use the condition to attain the maximum height for the liquid in the tank.

Formulae used:
The volume flow rate \[Q\] of the liquid is given by
\[Q = Av\] …… (1)
Here, \[A\] is the area and \[v\] is the flow velocity of the liquid.
The expression for Bernoulli’s theorem is
\[P + \dfrac{1}{2}\rho {v^2} + \rho gh = \operatorname{constant} \] …… (2)
Here, is the pressure energy of the liquid, is the density of the liquid, is the velocity of the liquid, is the acceleration due to gravity and is the height of the liquid.

Complete step by step answer:
We have given that a tank has a hole of area \[1\,{\text{c}}{{\text{m}}^2}\] at its bottom from which the water from the tank flows out of the tank. The water from a tube flows into the tank with a volume flow rate of \[70\,{\text{c}}{{\text{m}}^3}/{\text{s}}\].
\[{A_{out}} = 1\,{\text{c}}{{\text{m}}^2}\]
\[{Q_{in}} = 70\,{\text{c}}{{\text{m}}^3}/{\text{s}}\]
For the water height to rise to the maximum height in the tank, the volume flow rate \[{Q_{in}}\] of water in the tank and \[{Q_{out}}\] out of the tank must be equal.
\[{Q_{in}} = {Q_{out}}\]
Substitute \[70\,{\text{c}}{{\text{m}}^3}/{\text{s}}\]\[{A_{in}}{v_{in}}\] for \[{Q_{in}}\] and \[{A_{out}}{v_{out}}\] for \[{Q_{out}}\] in the above equation.
\[70\,{\text{c}}{{\text{m}}^3}/{\text{s}} = {A_{out}}{v_{out}}\] …… (3)
Let us derive the expression for the velocity of the water flowing out of the tank from the hole.The pressure on the water flowing in the tank from the tube and out of the tank from hole is equal to atmospheric pressure \[{P_0}\].The kinetic energy of the water flowing into the tank is zero as its velocity can be considered zero when compared to large amounts of water in the large size of the tank.The potential energy of the water coming out of the tank is zero as its height from the ground is zero.Apply Bernoulli’s theorem to the water flowing from the tube and coming out of the hole from the tank.
\[ {P_0} + 0 + \rho gh = {P_0} + \dfrac{1}{2}\rho v_{out}^2 + 0\]
\[ \Rightarrow {v_{out}} = \sqrt {2gh} \]
Here, \[h\] is the maximum height of water in the tank.
Substitute \[\sqrt {2gh} \] for \[{v_{out}}\] in equation (3).
\[70 = {A_{out}}\sqrt {2gh} \]
Take the square on both sides of the above equation and rearrange it for \[h\].
\[{\left( {70} \right)^2} = 2ghA_{out}^2\]
\[h = \dfrac{{4900}}{{2gA_{out}^2}}\]
Substitute \[980\,{\text{cm/}}{{\text{s}}^2}\] for \[g\] and \[1\,{\text{c}}{{\text{m}}^2}\] for \[{A_{out}}\] in the above equation.
\[h = \dfrac{{4900}}{{2\left( {980\,{\text{cm/}}{{\text{s}}^2}} \right){{\left( {1\,{\text{c}}{{\text{m}}^2}} \right)}^2}}}\]
\[ \Rightarrow h = \dfrac{{4900}}{{2\left( {980\,{\text{cm/}}{{\text{s}}^2}} \right)\left( 1 \right)}}\]
\[ \therefore h = 2.5\,{\text{cm}}\]

Therefore, the maximum height of water in the tank is \[2.5\,{\text{cm}}\].Hence, the correct option is A.

Note: The students should keep in mind that the value of acceleration due to gravity should be used in the CGS system of units and not in the SI system of units because all the terms used in the formula are in the CGS system of units. Otherwise, the final result for the maximum height of the water in the tank will be incorrect.