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Question: A liquid is kept in a cylindrical jar, which is rotated about the cylindrical axis. The liquid rises...

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 rr and speed of rotation is  ω~\mathbf{\omega }. The difference in height at the centre and the sides of jar is:
A. r2w2g\dfrac{{{r}^{2}}{{w}^{2}}}{g}
B. r2w22g\dfrac{{{r}^{2}}{{w}^{2}}}{2g}
C. gr2w2\dfrac{g}{{{r}^{2}}{{w}^{2}}}
D. 2gr2w2\dfrac{2g}{{{r}^{2}}{{w}^{2}}}

Explanation

Solution

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:
P1+12ρv12+ρgh1=P2+12ρv22+ρgh2{{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, ρ\rho is the fluid density, g is the acceleration due to gravity,
P1P_1, V1V_1 and h1h_1 are the pressure, velocity and height at elevation 1
And P2P_2, V2V_2 and h2h_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 P1P_1 is ρgh\rho gh
Also the centre P2P_2 experiences a kinetic energy of 12ρv2\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 P1P_1 and kinetic energy at P2P_2.
Therefore, ρgh=12ρv2    h=v22g \rho gh=\dfrac{1}{2}\rho {{v}^{2}} \implies h=\dfrac{{{v}^{2}}}{2g}
We know that velocity v=rωv=r\omega
Hence, h=r2ω22gh=\dfrac{{{r}^{2}}{{\omega }^{2}}}{2g}
Therefore the difference in height at the centre and the sides of jar is r2ω22g\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.