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 Rd^{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.
$P+\frac{1}{2} \rho v^{2}+\rho g h=$ 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
$P_{1}-P_{2}=\frac{1}{2} d R^{2} \omega^{2}$