Question

A wooden cube (density of wood ${}^{\prime }{d}^{\prime }$) of side ${}^{\prime }{\ell }^{\prime }$ floats in a liquid of density ${}^{\prime }{\rho }^{\prime }$ with its upper and lower surfaces horizontal. If the cube is pushed slightly down and released, it performs simple harmonic motion of period ${}^{\prime }{T}^{\prime }$. Then, ${}^{\prime }{T}^{\prime }$ is equal to :

• A. $2\pi \sqrt{\frac{\ell d}{\rho g}}$
• B. $2\pi \sqrt{\frac{\ell \rho }{dg}}$
• C. $2\pi \sqrt{\frac{\ell d}{\left(\rho -d\right)g}}$
• D. $2\pi \sqrt{\frac{\ell \rho }{\left(\rho -d\right)g}}$

Answer 1

Answer: (A)

At equilibrium
$F_b=mg$
$\rho A{\ell }_{0}g=dA\ell g...........\left(i\right)$
Restoring force,
$F=mg-F_b^{\prime }$
$F=mg-\rho A\left({\ell }_0+x\right)g$
$dA\ell a=dA\ell g-\rho A{\ell }_{0}g-\rho gAx$
$a=\frac{\rho g}{d\ell }x$
$\omega =\sqrt{\frac{\rho g}{d\ell }}$
$T=2\pi \sqrt{\frac{\ell d}{\rho g}}...........\left(i\right)$