Answer
374.3k+ views
Hint:-This problem can be solved by direct application of Bernoulli's principle. Bernoulli's principle establishes a relationship between the various pressures acting on a fluid during the incompressible flow of a fluid. It states that,
$P + \dfrac{1}{2}\rho {v^2} + \rho gh = C$
where P is the static pressure, $\rho $ is the density of the fluid, $v$ is the velocity of fluid flow, g is the acceleration due to gravity and h is the height at which the flow takes place.
Complete step-by-step solution:-
Whenever a fluid is under motion, there are three kinds of pressure acting on it. They are:
i) Static pressure, which is related to the mechanical force.
ii) Pressure due to flow of the fluid
iii) Pressure due to the height at which the fluid is flowing from the ground.
As per Bernoulli's principle, the sum of the above three mentioned pressures is constant at every point in the fluid.
$P + \dfrac{1}{2}\rho {v^2} + \rho gh = C$
where P is the static pressure, $\rho $ is the density of the fluid, $v$ is the velocity of fluid flow, g is the acceleration due to gravity and h is the height at which the flow takes place.
Consider the tank filled with water and a hole is made in the bottom. Let us apply Bernoulli’s principle for the surface of the tank and the hole.
${P_1} + \dfrac{1}{2}\rho {v_1}^2 + \rho g{h_1} = {P_2} + \dfrac{1}{2}\rho {v_2}^2 + \rho g{h_2}$
Since the fluid is incompressible, there will be no change in the static pressures of the fluid. This means that:
${P_1} = {P_2}$
Since, the water at the surface level of the tank is constant, ${v_1} = 0$
To calculate, Velocity at the hole of the tank – ${v_2}$
Applying the Bernoulli’s equation to this case by substituting, we get –
$\dfrac{1}{2}\rho {v_2}^2 - \dfrac{1}{2}\rho {v_1}^2 = {P_2} - {P_1} + \rho g{h_2} - \rho g{h_1}$
$\dfrac{1}{2}\rho {v_2}^2 = \rho gh$
${v_2}^2 = 2gh$
To find h –
The height difference can be calculated with the help of pressure difference between the surface and at the hole by using the formula below –
$p = \rho gh$
Pressure at surface will be equal to the atmospheric pressure
${p_1} = \rho g{h_1} = 1atm = {10^5}N{m^{ - 2}}$
Pressure at hole as given
${p_2} = \rho g{h_2} = 3atm = 3 \times {10^5}N{m^{ - 2}}$
Density of water, $\rho = 1000kg{m^{ - 3}}$
Assuming the acceleration due to gravity, $g = 10m{s^{ - 2}}$
${p_2} - {p_1} = \rho g\left( {{h_2} - {h_1}} \right) = \rho gh$
Hence,
$h = \dfrac{{{p_2} - {p_1}}}{{\rho g}} = \dfrac{{3 \times {{10}^5} - {{10}^5}}}{{1000 \times 10}} = 20m$
Substituting in the equation for velocity,
${v_2}^2 = 2gh$
${v_2}^2 = 2 \times 10 \times 20 = 400$
$ \Rightarrow {v_2} = \sqrt {400} m{s^{ - 1}}$
Hence, the correct option is Option A.
Note:- This is the most fundamental method to solve the problem. However, there is a shortcut to solve this problem. For that, the concept of Torricelli’s law should be applied –
The Torricelli’s law gives the velocity of efflux of water jet from a hole which is at a height h from the surface.
Velocity, $v = \sqrt {2gh} $
So, you can see that instead of deriving the above equation from Bernoulli’s principle, you can directly use Torricelli's equation to find the velocity of efflux.
$P + \dfrac{1}{2}\rho {v^2} + \rho gh = C$
where P is the static pressure, $\rho $ is the density of the fluid, $v$ is the velocity of fluid flow, g is the acceleration due to gravity and h is the height at which the flow takes place.
Complete step-by-step solution:-
Whenever a fluid is under motion, there are three kinds of pressure acting on it. They are:
i) Static pressure, which is related to the mechanical force.
ii) Pressure due to flow of the fluid
iii) Pressure due to the height at which the fluid is flowing from the ground.
As per Bernoulli's principle, the sum of the above three mentioned pressures is constant at every point in the fluid.
$P + \dfrac{1}{2}\rho {v^2} + \rho gh = C$
where P is the static pressure, $\rho $ is the density of the fluid, $v$ is the velocity of fluid flow, g is the acceleration due to gravity and h is the height at which the flow takes place.
Consider the tank filled with water and a hole is made in the bottom. Let us apply Bernoulli’s principle for the surface of the tank and the hole.
${P_1} + \dfrac{1}{2}\rho {v_1}^2 + \rho g{h_1} = {P_2} + \dfrac{1}{2}\rho {v_2}^2 + \rho g{h_2}$
Since the fluid is incompressible, there will be no change in the static pressures of the fluid. This means that:
${P_1} = {P_2}$
Since, the water at the surface level of the tank is constant, ${v_1} = 0$
To calculate, Velocity at the hole of the tank – ${v_2}$
Applying the Bernoulli’s equation to this case by substituting, we get –
$\dfrac{1}{2}\rho {v_2}^2 - \dfrac{1}{2}\rho {v_1}^2 = {P_2} - {P_1} + \rho g{h_2} - \rho g{h_1}$
$\dfrac{1}{2}\rho {v_2}^2 = \rho gh$
${v_2}^2 = 2gh$
To find h –
The height difference can be calculated with the help of pressure difference between the surface and at the hole by using the formula below –
$p = \rho gh$
Pressure at surface will be equal to the atmospheric pressure
${p_1} = \rho g{h_1} = 1atm = {10^5}N{m^{ - 2}}$
Pressure at hole as given
${p_2} = \rho g{h_2} = 3atm = 3 \times {10^5}N{m^{ - 2}}$
Density of water, $\rho = 1000kg{m^{ - 3}}$
Assuming the acceleration due to gravity, $g = 10m{s^{ - 2}}$
${p_2} - {p_1} = \rho g\left( {{h_2} - {h_1}} \right) = \rho gh$
Hence,
$h = \dfrac{{{p_2} - {p_1}}}{{\rho g}} = \dfrac{{3 \times {{10}^5} - {{10}^5}}}{{1000 \times 10}} = 20m$
Substituting in the equation for velocity,
${v_2}^2 = 2gh$
${v_2}^2 = 2 \times 10 \times 20 = 400$
$ \Rightarrow {v_2} = \sqrt {400} m{s^{ - 1}}$
Hence, the correct option is Option A.
Note:- This is the most fundamental method to solve the problem. However, there is a shortcut to solve this problem. For that, the concept of Torricelli’s law should be applied –
The Torricelli’s law gives the velocity of efflux of water jet from a hole which is at a height h from the surface.
Velocity, $v = \sqrt {2gh} $
So, you can see that instead of deriving the above equation from Bernoulli’s principle, you can directly use Torricelli's equation to find the velocity of efflux.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)