Question

A drum of radius $R$ full of liquid of density $d$ is rotated at $\omega$ rad/s. The increase in pressure at die centre of the drum will be :

• A. $\frac{{\omega }^2{R}^2{d}^2}{2}$
• B. $\frac{\omega R{d}^2}{2}$
• C. $\frac{{\omega }^{2}Rd}{2}$
• D. $\frac{{\omega }^2{R}^{2}d}{2}$

Answer 1

Answer: (D)

When an incompressible and non-viscous liquids flows in stream lined motion from one place to another then at every point of its path the total energy per unit volume is constant.
$P+\frac{1}{2}\rho v^2+\rho gh=$ constant
where $P$ is pressure,
$\rho$ is density,
$v$ is velocity and
$h$ is height.
Also $v=R\omega$
where $R$ is radius and $\omega$ is angular velocity.
Since, velocity at centre is zero and density
$\rho =d$, we have
$P_1+\frac{1}{2}\times 0=P_2+\frac{1}{2}d{v}^2$
$P_1=P_2+\frac{1}{2}d(R\omega {)}^2$
Thus, increase in pressure
$P_1-P_2=\frac{1}{2}d{R}^2{\omega }^2$