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# 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 ${r_1}$ ​ and ${r_2}$ from rotational axis indicate pressure ${p_1}$​ and ${p_2}$ ​, respectively. Determine the angular velocity $\omega$ of rotation of the tube:(A) $\omega = \sqrt {\dfrac{{2({p_2} - {p_1})}}{{\rho (r_1^2 - r_2^2)}}}$(B) $\omega = \sqrt {\dfrac{{2({p_2} - {p_1})}}{{\rho (r_1^2 + r_2^2)}}}$(C) $\omega = \sqrt {\dfrac{{2(r_2^2 - r_1^2)}}{{({p_2} - {p_1})}}}$(D) $\omega = \sqrt {\dfrac{{2({p_2} - {p_1})}}{{\rho {r_1}{r_2}}}}$

Last updated date: 13th Jun 2024
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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:
$P + \rho gh + \dfrac{1}{2}\rho {v^2} =$ constant
Since height is the same for both A and B hence $\rho gh$ term can be neglected.

Complete Step by step solution
On applying Bernoulli equation separately for A and B we get,
For A, ${p_1}$ is the reading shown in the barometer.
Hence, ${p_1} + \rho gh + \dfrac{1}{2}\rho {v_1}^2 =$ constant…… (1)
For B, ${p_2}$ is the reading shown in the barometer.
Hence, ${p_2} + \rho gh + \dfrac{1}{2}\rho v_2^2 =$ constant…… (2)
From equation (1) and (2) we get
${p_1} + \rho gh + \dfrac{1}{2}\rho v_1^2 = {p_2} + \rho gh + \dfrac{1}{2}\rho v_2^2 \\ {p_1} + \dfrac{1}{2}\rho v_1^2 = {p_2} + \dfrac{1}{2}\rho v_2^2 \\$
Now we know that $v = r\omega$
Therefore, above equation becomes
${p_1}\rho {({r_1}\omega )^2} = {p_2} + \dfrac{1}{2}\rho {({r_2}\omega )^2} \\ ({p_2} - {p_1}) = \dfrac{1}{2}\rho ({r_1}^2 - {r_2}^2){\omega ^2} \\ \therefore {\omega ^2} = \dfrac{{2({p_2} - {p_1})}}{{\rho ({r_1}^2 - {r_2}^2)}} \\ \\$
$\omega = \sqrt {\dfrac{{2({p_2} - {p_1})}}{{\rho (r_1^2 - r_2^2)}}}$
Hence the required angular velocity is, $\omega = \sqrt {\dfrac{{2({p_2} - {p_1})}}{{\rho (r_1^2 - r_2^2)}}}$

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.