Question

A cylindrical tank of radius $20\text{cm}$ and height $50\text{cm}$ has water upto $30\text{cm}$ of height. Frequency of rotation when water just starts spilling over the sides of the vessel is given by $\frac{N\sqrt{2}}{\pi }{s}^{-1}$. Then, the value of $N$ is

• A. $5$
• B. $3$
• C. $7$
• D. $2$

Answer 1

The correct option is A $5$

Let us suppose, angular velocity of rotation is $\omega$ when water just starts spilling.
According to question,


We know that for rotating fluid
$\Delta P=P_B-P_0=\frac{1}{4}\rho {\omega }^2{x}^2----\left(1\right)$
Also pressure difference in vertical direction
$\Delta P-P_B-P_A=\rho gh$
$\Rightarrow P_B-P_0=\rho gh--------\left(2\right)$
$[\because P_A=P_0=\text{Atnespheric Pressure}]$
From (1) and (2),
$\rho gh=\frac{1}{4}\rho {\omega }^2{x}^2$
$\Rightarrow \omega =\sqrt{\frac{4gh}{x^2}}$
$=\sqrt{\frac{4\times 10\times 0.2}{{0.2}^2}}\left[\begin{array}{c}h=20\text{cm}=0.2m\\ x=20\text{cm}=0.2m\end{array}\right]$
[because when water just starts spilling, height difference between the centre and edge $=50-30=20\text{cm}$]
$\Rightarrow \omega =10\sqrt{2}\text{rad/s}----\left(3\right)$
We know,
$\omega =2\pi f$
$\Rightarrow f=\frac{\omega }{2\pi }$
$\Rightarrow f=\frac{10\sqrt{2}}{2\pi }$ [From (3)]
$\Rightarrow f=\frac{5\sqrt{2}}{\pi }{\text{s}}^{-1}$
From $f=\frac{N\sqrt{2}}{\pi }{\text{s}}^{-1}\left[\text{given}\right]$,
value of $N$ is $5$.