A uniform cube of side 10 cm and…

A uniform cube of side 10 cm and density 900 kg m^-3 floats in a tank with its faces horizontal and vertical, partly in water (density 1000 kg m^-3)…

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A uniform cube of side

10\ \mathrm{cm}

and density

900\ \mathrm{kg\,m^{-3}}

floats in a tank with its faces horizontal and vertical, partly in water (density

1000\ \mathrm{kg\,m^{-3}}

) and partly in a layer of oil (density

800\ \mathrm{kg\,m^{-3}}

) lying on top of the water, the oil layer being thick enough that the cube always remains completely immersed in the two liquids. The cube is pushed slightly down along the vertical and released. Take

g=10\ \mathrm{m\,s^{-2}}

and neglect viscosity and surface effects. The time period of the small vertical oscillations is closest to

A. $0.60\ \mathrm{s}$
B. $0.67\ \mathrm{s}$
C. $1.33\ \mathrm{s}$
D. $0.63\ \mathrm{s}$

Show answer and solution

Answer: (C) $1.33\ \mathrm{s}$

First find the equilibrium: if a depth

x

of the cube lies in water,

900a=1000x+800(a-x)

, so

x=a/2

. Displace the cube down by

y

: a length

y

of oil is replaced by water, so the buoyant force increases by

(\rho_w-\rho_o)gAy

with

A=a^2

. The restoring force is therefore

F=-(\rho_w-\rho_o)gA\,y

and the mass is

m=\rho Aa

. Hence

\omega^2=\dfrac{(\rho_w-\rho_o)g}{\rho a}=\dfrac{200\times10}{900\times0.1}=\dfrac{200}{9}

and

T=2\pi/\omega=2\pi\sqrt{0.045}=1.33\ \mathrm{s}

. Taking the cube as floating only in water gives

0.60\ \mathrm{s}

, only in oil

0.67\ \mathrm{s}

, and using the average liquid density gives

0.63\ \mathrm{s}

. All these ignore that the restoring force depends on the density difference.

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Original question written in the JEE Advanced 2025 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake