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A cylindrical block of wood of density 650 kg m⁻³, base area 30 cm² and height 54 cm floats in a liquid of density 900 kg m⁻³. It is depressed slightly and released. The period of the resulting oscillations equals that of a simple pendulum of length, nearly

Asked in JEE Main Online 2015 · Floating bodies and gas columns

Answer: (3) 39 cm

Step-by-step solution

T=2π√m/(ρ Ag) with m=dAH, so T=2π√(dH)/(ρ g).

Comparing with T=2π√L/g, the equivalent pendulum length is L=(dH)/ρ, which is just the submerged depth.

L=(650×54)/(900)=39 cm. The base area does not enter.

Why the other options are wrong

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