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 $$y\pi\times10^{-2}s$$, where the value of y is (Acceleration due to gravity, $$g=10 m/s^{2}$$, density of water $$=10^{3}kg/m^{3}$$

A light hollow cube of side $$a = 10$$ cm = 0.1 m and mass $$m = 10$$ g = 0.01 kg floats in water. We need to find the time period of vertical SHM oscillations.

At equilibrium the weight of the cube equals the buoyant force; if $$h_0$$ is the depth of immersion, then $$mg = \rho_{water} \cdot a^2 \cdot h_0 \cdot g$$ from which $$h_0 = \frac{m}{\rho_{water} \cdot a^2} = \frac{0.01}{1000 \times 0.01} = \frac{0.01}{10} = 0.001 \text{ m}.$$

When the cube is pushed down by a small distance $$x$$ from equilibrium, the additional buoyant force provides a restoring force given by $$F_{restoring} = -\rho_{water} \cdot a^2 \cdot x \cdot g,$$ where the negative sign indicates the force opposes the displacement.

Comparing this with $$F = -kx$$ shows that the effective spring constant is $$k = \rho_{water} \cdot a^2 \cdot g = 1000 \times (0.1)^2 \times 10 = 1000 \times 0.01 \times 10 = 100 \text{ N/m}.$$

The time period of oscillation is therefore $$T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.01}{100}} = 2\pi\sqrt{10^{-4}} = 2\pi \times 10^{-2} \text{ s}.$$ This corresponds to $$y = 2$$, so the correct answer is Option 2: $$y = 2$$.