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)You need to login to perform this action.
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