Question

A solid cylinder of density ${\rho }_0$, cross-section area A and length l floats in a liquid of density $\rho (>{\rho }_0)$ with its axis vertical, as shown. If it is slightly displaced downward and released, the time period will be.

• A. $2\pi \sqrt{\frac{l}{g}}$
• B. $2\pi \sqrt{\frac{{\rho }_{0}l}{\rho g}}$
• C. $2\pi \sqrt{\frac{\rho l}{{\rho }_{0}g}}$
• D. $2\pi \sqrt{\frac{l}{2g}}$

Answer 1

The correct option is B $2\pi \sqrt{\frac{{\rho }_{0}l}{\rho g}}$

At equilibrium the weight of the body and buoyant force balance each other.

When body is displace down by $x$, then the upward force is due to increase in buoyant force due to more displacement of liquid.

$F=-V\rho g=-\rho Axg$ $-ve$ sign is because displacement and force are in opposite direction

Acceleration=$a=\frac{-rhoAxg}{M}=\frac{-\rho Ag}{{\rho }_{o}AL}x=-{\omega }^{2}x$

$\omega =\sqrt{\frac{\rho g}{{\rho }_{o}L}}$

Time period $=T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{{\rho }_{o}L}{\rho g}}$

Answer-(B)