question_answer

A wooden cube (density of wood 'd') of side \['\ell '\] floats in a liquid of density\['\rho '\] with its upper and lower surfaces horizontal. If the cube is pushed slightly down and released, it performs simple harmonic motion of period T'. Then, T is equal to :     AIEEE  Solved  Paper (Held On 11 May  2011)

A) \[2\pi \sqrt{\frac{\ell d}{\rho g}}\]

B) \[2\pi \sqrt{\frac{\ell \rho }{dg}}\]

C) \[2\pi \sqrt{\frac{\ell d}{(\rho -d)g}}\]

D) \[2\pi \sqrt{\frac{\ell \rho }{(\rho -d)g}}\]

Correct Answer: A

Solution :

                At equilibrium \[{{F}_{b}}=mg\] \[\rho A{{\ell }_{0}}g=dA\ell g\]                                                                                                                                                ??(i) Restoring force, \[F=mg-{{F}_{b}}'\] \[F=mg-\rho A({{\ell }_{0}}+x)g\] \[dA\ell a=dA\ell g-\rho A{{\ell }_{0}}g-\rho gAx\] \[a=-\frac{\rho g}{d\ell }x\] \[\omega =\sqrt{\frac{\rho g}{d\ell }}\] \[T=2\pi \sqrt{\frac{\ell d}{\rho g}}\]                                                                                                                                      ?.(i)