Q: A wooden cube (density of wood d) of side l floats in a liquid of density ρ 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
(a) $\displaystyle 2\pi \sqrt{\frac{l\rho}{(\rho-d)g}} $
(b) $\displaystyle 2\pi \sqrt{\frac{l d}{\rho g}} $
(c) $ \displaystyle 2\pi \sqrt{\frac{l\rho}{d g}} $
(d) $ \displaystyle 2\pi \sqrt{\frac{l d}{(\rho-d)g}} $
Ans: (b)
Sol: F= -ρ l2 x g
Since , F = -kx
$\displaystyle T = 2\pi \sqrt{\frac{m}{k}} $
$ \displaystyle T = 2\pi \sqrt{\frac{l^3 d}{\rho l^2 g}} $
$ \displaystyle T = 2\pi \sqrt{\frac{l d}{\rho g}} $