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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.

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