
A liquid is kept in a cylindrical jar, which is rotated about the cylindrical axis. The liquid rises as it sends. The radius of the jar is $r$ and speed of rotation is \[~\mathbf{\omega }\]. The difference in height at the centre and the sides of jar is:
A. \[\dfrac{{{r}^{2}}{{w}^{2}}}{g}\]
B. \[\dfrac{{{r}^{2}}{{w}^{2}}}{2g}\]
C. \[\dfrac{g}{{{r}^{2}}{{w}^{2}}}\]
D. \[\dfrac{2g}{{{r}^{2}}{{w}^{2}}}\]
Answer
566.1k+ views
Hint: If we take liquid in any container and give an acceleration to the container, the shape of the liquid changes. It becomes inclined. We use Bernoulli’s principle here, i.e. during rotation, the pressure at the centre and the sides will be the same as atmospheric pressure. The centre experiences kinetic energy and the top experiences potential energy.
Formula used:
Bernoulli’s principle:
${{P}_{1}}+\dfrac{1}{2}\rho v{{1}^{2}}+\rho g{{h}_{1}}={{P}_{2}}+\dfrac{1}{2}\rho v{{2}^{2}}+\rho g{{h}_{2}}$
Where, $\rho $ is the fluid density, g is the acceleration due to gravity,
$P_1$, $V_1$ and $h_1$ are the pressure, velocity and height at elevation 1
And $P_2$, $V_2$ and $h_2$ are the pressure, velocity and height at elevation 2
Complete step by step answer:
Bernoulli’s principle in fluid dynamics states that the increase in the speed of the fluid leads to the decrease in static pressure or the fluid potential energy.
In our given condition the potential energy at the top $P_1$ is $\rho gh$
Also the centre $P_2$ experiences a kinetic energy of $\dfrac{1}{2}\rho {{v}^{2}}$
When we apply Bernoulli’s theorem, we can infer that the energies at all the points will be equal.
Only potential energy acts at $P_1$ and kinetic energy at $P_2$.
Therefore, $
\rho gh=\dfrac{1}{2}\rho {{v}^{2}}
\implies h=\dfrac{{{v}^{2}}}{2g} $
We know that velocity $v=r\omega $
Hence, $h=\dfrac{{{r}^{2}}{{\omega }^{2}}}{2g}$
Therefore the difference in height at the centre and the sides of jar is $\dfrac{{{r}^{2}}{{\omega }^{2}}}{2g}$
So, the correct answer is “Option B”.
Note:
Applications of Bernoulli’s theorem include the Bunsen burner, aerofoil lift and venturimeter.
It is also used in other applications like automobiles, filter pumps, atomizers and sprays.
Formula used:
Bernoulli’s principle:
${{P}_{1}}+\dfrac{1}{2}\rho v{{1}^{2}}+\rho g{{h}_{1}}={{P}_{2}}+\dfrac{1}{2}\rho v{{2}^{2}}+\rho g{{h}_{2}}$
Where, $\rho $ is the fluid density, g is the acceleration due to gravity,
$P_1$, $V_1$ and $h_1$ are the pressure, velocity and height at elevation 1
And $P_2$, $V_2$ and $h_2$ are the pressure, velocity and height at elevation 2
Complete step by step answer:
Bernoulli’s principle in fluid dynamics states that the increase in the speed of the fluid leads to the decrease in static pressure or the fluid potential energy.
In our given condition the potential energy at the top $P_1$ is $\rho gh$
Also the centre $P_2$ experiences a kinetic energy of $\dfrac{1}{2}\rho {{v}^{2}}$
When we apply Bernoulli’s theorem, we can infer that the energies at all the points will be equal.
Only potential energy acts at $P_1$ and kinetic energy at $P_2$.
Therefore, $
\rho gh=\dfrac{1}{2}\rho {{v}^{2}}
\implies h=\dfrac{{{v}^{2}}}{2g} $
We know that velocity $v=r\omega $
Hence, $h=\dfrac{{{r}^{2}}{{\omega }^{2}}}{2g}$
Therefore the difference in height at the centre and the sides of jar is $\dfrac{{{r}^{2}}{{\omega }^{2}}}{2g}$
So, the correct answer is “Option B”.
Note:
Applications of Bernoulli’s theorem include the Bunsen burner, aerofoil lift and venturimeter.
It is also used in other applications like automobiles, filter pumps, atomizers and sprays.
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