A cylindrical block of wood (density = 650 kg m - 3 ), of base area 30 cm 2 and height 54 cm , floats in a liquid of density 900 kg m - 3 . The block is depressed 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) :

Step-by-step solution

Case-I In equilibrium F up = mg V i ρ L g = m g A h ρ L g = A L ρ s g … … … 1 Case-II Let block is pressed by x more in liquid F n e t = F u p 1 - m g = V i 1 ρ L g - mg F net = A h + x ρ L g - AL ρ s g F net = Ah ρ L g + Ax ρ L g - AL ρ s g . . . . . . . . . . (2) from equation (1) & (2) F net = Ax ρ L g restoring force in upward direction α = F n e t m = A ρ L g A ρ s L x α = ω 2 x for SHM and ω 2 = g l so g l = ρ L g ρ s L so l = ρ s L ρ L = 650 × 54 900 = 39 cm