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Question: A tube filled with water and closed at both ends uniformly rotates in a horizontal plane about the O...

A tube filled with water and closed at both ends uniformly rotates in a horizontal plane about the OO' axis. The manometers A and B fixed in the tube at distances r1{r_1} ​ and r2{r_2} from rotational axis indicate pressure p1{p_1}​ and p2{p_2} ​, respectively. Determine the angular velocity ω\omega of rotation of the tube:

(A) ω=2(p2p1)ρ(r12r22)\omega = \sqrt {\dfrac{{2({p_2} - {p_1})}}{{\rho (r_1^2 - r_2^2)}}}
(B) ω=2(p2p1)ρ(r12+r22)\omega = \sqrt {\dfrac{{2({p_2} - {p_1})}}{{\rho (r_1^2 + r_2^2)}}}
(C) ω=2(r22r12)(p2p1)\omega = \sqrt {\dfrac{{2(r_2^2 - r_1^2)}}{{({p_2} - {p_1})}}}
(D) ω=2(p2p1)ρr1r2\omega = \sqrt {\dfrac{{2({p_2} - {p_1})}}{{\rho {r_1}{r_2}}}}

Explanation

Solution

Hint We know Bernoulli equation relates the speed of the fluid at a point, the pressure at that point and the height of that point above the reference level. Bernoulli equation is given by:
P+ρgh+12ρv2=P + \rho gh + \dfrac{1}{2}\rho {v^2} = constant
Since height is the same for both A and B hence ρgh\rho gh term can be neglected.

Complete Step by step solution
On applying Bernoulli equation separately for A and B we get,
For A, p1{p_1} is the reading shown in the barometer.
Hence, p1+ρgh+12ρv12={p_1} + \rho gh + \dfrac{1}{2}\rho {v_1}^2 = constant…… (1)
For B, p2{p_2} is the reading shown in the barometer.
Hence, p2+ρgh+12ρv22={p_2} + \rho gh + \dfrac{1}{2}\rho v_2^2 = constant…… (2)
From equation (1) and (2) we get

p1+ρgh+12ρv12=p2+ρgh+12ρv22 p1+12ρv12=p2+12ρv22  {p_1} + \rho gh + \dfrac{1}{2}\rho v_1^2 = {p_2} + \rho gh + \dfrac{1}{2}\rho v_2^2 \\\ {p_1} + \dfrac{1}{2}\rho v_1^2 = {p_2} + \dfrac{1}{2}\rho v_2^2 \\\

Now we know that v=rωv = r\omega
Therefore, above equation becomes

p1ρ(r1ω)2=p2+12ρ(r2ω)2 (p2p1)=12ρ(r12r22)ω2 ω2=2(p2p1)ρ(r12r22)   {p_1}\rho {({r_1}\omega )^2} = {p_2} + \dfrac{1}{2}\rho {({r_2}\omega )^2} \\\ ({p_2} - {p_1}) = \dfrac{1}{2}\rho ({r_1}^2 - {r_2}^2){\omega ^2} \\\ \therefore {\omega ^2} = \dfrac{{2({p_2} - {p_1})}}{{\rho ({r_1}^2 - {r_2}^2)}} \\\ \\\

ω=2(p2p1)ρ(r12r22)\omega = \sqrt {\dfrac{{2({p_2} - {p_1})}}{{\rho (r_1^2 - r_2^2)}}}
Hence the required angular velocity is, ω=2(p2p1)ρ(r12r22)\omega = \sqrt {\dfrac{{2({p_2} - {p_1})}}{{\rho (r_1^2 - r_2^2)}}}

Option (A) is correct.

Note Bernoulli equation is just the application of work-energy theorem in the case of fluid flow. In the Bernoulli equation we make a few assumptions like the fluid is ideal i.e. incompressible and nonviscous, it has constant density, both points lie on a streamline, flow is steady and there is no friction.