A light hollow cube of side length 10 cm and mass 10 g , is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is y \pi \times 10^-2 ~s , where the value of y is (Acceleration due to gravity, g=10 ~m / s^2 , density of water =10^3 ~kg / m^3 )

Simple Harmonic Motion PYQ — JEE Main 2025 · 23 Jan · Shift 1 · Q60 with Solution

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Correct answer: A

We need the time period of small vertical oscillations of the floating cube.

If the cube is pushed down by a small distance $x$ , the additional volume submerged is

\Delta V = a^2 x

So the additional buoyant force is

\Delta F = \rho g \Delta V = \rho g a^2 x

This force acts upward, opposite to displacement, so restoring force is

F = -\rho g a^2 x

Comparing with SHM form $F=-kx$ , the effective spring constant is

k = \rho g a^2

For SHM,

T = 2\pi \sqrt{\frac{m}{k}}

Substitute $k = \rho g a^2$ :

T = 2\pi \sqrt{\frac{m}{\rho g a^2}}

Now put the values:

a^2 = (0.1)^2 = 0.01

\rho g a^2 = 10^3 \times 10 \times 0.01 = 100

Thus,

T = 2\pi \sqrt{\frac{0.01}{100}}

T = 2\pi \sqrt{10^{-4}} = 2\pi \times 10^{-2}\text{ s}

So,

T = 2\pi \times 10^{-2}\text{ s}

Comparing with $y\pi \times 10^{-2}\text{ s}$ ,

y = 2

Therefore, the correct option is A.