Question

A wooden cube (density of wood is d) of side $‘el{l}^{\prime }$ floats in liquid of density ‘s’ with its upper and lower surface horizontal. If the cube is pushed slightly down and released, it performs simple harmonic motion of period ‘T’. Then T is equal to

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

Answer 1

The correct option is C $2\pi \sqrt{\frac{\ell .d}{s.g}}$

Net force on the cube = upthrust on the submerged part
$\therefore F=-\rho l^{2}x.ga=-\frac{\rho l^{2}g.x}{m}\therefore {\omega }^2=\frac{\rho l^{2}g}{m}or\omega =\sqrt{\frac{\rho l^{2}g}{m}}=\frac{2\pi }{T}\therefore T=2\pi \sqrt{\frac{ld}{\rho g}}$