A drum of radius ' $R$ ' full of liquid of density ' $d$ ' is rotated at angular velocity ' $\omega$ ' rad/s…
A drum of radius ' $R$ ' full of liquid of density ' $d$ ' is rotated at angular velocity ' $\omega$ ' rad/s. The increase in pressure at the centre of the drum will be
$\frac{\omega^2 R^2 d}{2}$
$\frac{\omega^2 \mathrm{Rd}}{2}$
$\frac{\omega R d^2}{2}$
$\frac{\omega^2 \mathrm{R}^2 \mathrm{~d}^2}{2}$
Solution
$\begin{aligned}
& P_1+\frac{1}{2} \rho v^2+\rho g h=\text { constant } \quad...(i)
& \text { and } v=R \omega \\
& \text {At centre } R=0 \Rightarrow v_1=0 \quad...(ii)
\therefore \quad & \text { From (i), } \\
& P_1+\frac{1}{2} \rho v_1^2+\rho g h=P_2+\frac{1}{2} \rho v_1^2+\rho g h \\
& P_1+0=P_2+\frac{1}{2} \rho v_1^2 \\
\therefore \quad & P_1=P_2+\frac{1}{2} \rho(R \omega)^2 \\
\therefore \quad & P_1-P_2=\frac{\omega^2 R^2 d}{2} \quad \ldots(\because r o m(i i)
\end{aligned}$