QJEEM25Ph9b3b: JEE Main 2025 Physics - Class 11 JEEM25Ph9b3b Solution: A light hollow cube of side length 10 cm and mass 10 g , is…

We can determine the time period of the oscillations by considering that when the cube is depressed by a small displacement, the additional buoyant force provided by the displaced water acts as a restoring force. Here’s a step‐by‐step explanation:

Define the given values:

Side length of the cube, $L = 10\text{ cm} = 0.1\text{ m}$ .

Mass of the cube, $m = 10\text{ g} = 0.01\text{ kg}$ .

Density of water, $\rho = 10^3\ \text{kg/m}^3$ .

Acceleration due to gravity, $g = 10\ \text{m/s}^2$ .

Cross-sectional area of the cube (face area),

$A = L^2 = (0.1)^2 = 0.01\text{ m}^2.$

When the cube is depressed by a small distance $x$ , the additional volume of water displaced is

$\Delta V = A\,x.$

Thus, the additional buoyant force is:

$F_b = \rho\,g\,\Delta V = \rho\,g\,A\,x.$

This force acts in the upward (restoring) direction. Notice that the force is proportional to the displacement $x$ , which is the hallmark of simple harmonic motion (SHM).

The effective spring constant, $k$ , associated with this SHM is:

$k = \rho\,g\,A.$

The period of oscillation for SHM is given by:

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

which, upon substituting for $k$ , becomes:

$T = 2\pi \sqrt{\frac{m}{\rho\,g\,A}}.$

Substitute the given values into the expression:

$T = 2\pi \sqrt{\frac{0.01}{1000 \times 10 \times 0.01}}.$

Calculate the denominator:

$1000 \times 10 \times 0.01 = 100,$

so,

$T = 2\pi \sqrt{\frac{0.01}{100}} = 2\pi \sqrt{0.0001}.$

Since,

$\sqrt{0.0001} = 0.01,$

we have:

$T = 2\pi \times 0.01 = 0.02\pi\ \text{s}.$

The problem states that the time period is given by:

$y \pi \times 10^{-2}\ \text{s}.$

Writing our result in the same form:

$0.02\pi\ \text{s} = 2\pi \times 10^{-2}\ \text{s}.$

We see that $y = 2.$

Therefore, the correct answer is:

Option A: 2.