Question

A cylindrical of wood (density `= 600 kg m^(-3)`) of base area `30 cm^(2)` and height `54 cm`, floats in a liquid of density `900 kg^(-3)` The block is deapressed slightly and then released. The time period of the resulting oscillations of the block would be equal to that of a simple pendulum of length (nearly) :
A. `26 cm`
B. `52 cm`
C. `39cm`
D. `65 cm`

Answer 1

Correct Answer - C
If the block is displaced further by small distance form equilibrium
extra buyont force = mass xx acceleration
`rArr - rho_(L) Axg = rho_(b)Ah(d^(2)x)/(dt^(2))`
`(d^(2)x)/(dt^(2)) = - ((g)/(39))x`
`rArr omega^(2) = (g)/(39)`
for simple pendulum,
`omega^(2) = (g)/(L)`
comparing we get., `L = 39 cm`