JEE Main 2015 · 11 Apr · Shift 1 · Physics Q8

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PhysicsSingle correct+4 / -1Question 8 of 90

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) :

65 cm

52 cm

39 cm

26 cm

Answer and worked solution

Official answer: C

The submerged depth is $54\times\dfrac{650}{900}=39$ cm. A floating block oscillates with $T=2\pi\sqrt{\dfrac{d}g}$ , where $d$ is the submerged depth, so the equivalent pendulum is 39 cm long.

Topics:Simple harmonic motion Buoyancy and Archimedes' principle

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JEE Main 2015 · 11 Apr · Shift 1, Physics question 8