A light hollow cube of side length 10 cm and mass 10g, is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is y$\mathrm{\pi }$ × 10^–2 s, where the value of y is
(Acceleration due to gravity, g = 10 m/s², density of water = 10³ kg/m³)

We will determine the time period $T$ of simple harmonic motion (SHM) for the floating hollow cube when it is displaced and released.

Step 1: Understanding the Restoring Force

When the cube is floating in water, the buoyant force balances the weight. If it is displaced downward, an additional restoring force arises due to the increased buoyant force, leading to SHM.

For a floating body undergoing vertical oscillations:

$$T = 2\pi \sqrt{\frac{m}{\text{effective restoring force per unit displacement}}}$$

Step 2: Calculating the Restoring Force

The restoring force is due to the buoyancy change when the cube is displaced by $x$. The additional buoyant force is given by:

$$F = (\text{density of water}) \times (\text{displaced volume}) \times g$$

For a small displacement $x$:

$$F = \rho_{\text{water}} \cdot A \cdot x \cdot g$$

where:

$\rho_{\text{water}} = 10^3$ kg/m³

$A = (0.1 \times 0.1)$ m² = $10^{-2}$ m² (cross-sectional area of the cube)

$g = 10$ m/s²

Thus,

$$F = 10^3 \times 10^{-2} \times x \times 10 = 100x$$

This behaves like a restoring force in SHM:

$$F = - kx$$

Thus, the effective force constant:

$$k = 100 \text{ N/m}$$

Step 3: Calculating the Time Period

The formula for the time period of SHM is:

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

Given:

$m = 10$ g = $10^{-2}$ kg

$k = 100$ N/m

$$T = 2\pi \sqrt{\frac{10^{-2}}{100}}$$

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

Comparing with $y\pi \times 10^{-2}$, we get $y = 2$.

Final Answer:

2

In Short: 

$\begin{array}{l}{\mathrm{a}}^2\mathrm{x\rho g}={\mathrm{ma}}_{\text{net }}\\ \frac{{\mathrm{L}}^2\mathrm{\rho g}}{\mathrm{m}}\mathrm{x}={\mathrm{a}}_{\text{net }}\\ \mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{{\mathrm{L}}^2\mathrm{\rho g}}}\end{array}$
where m = 10g, L = 10 cm, $\mathrm{\rho }$ = 1000 kg/m³