JEE Physics 2025 Question with Solution

Given: Side length of cube = $0.1 \, \text{m}$, mass = $0.01 \, \text{kg}$, density of water = $1000 \, \text{kg/m}^3$, and $g = 10 \, \text{m/s}^2$.

Find: The value of $y$ in $T = y\pi \times 10^{-2} \, \text{s}$.

For a floating body displaced vertically by a small distance $x$, the change in buoyant force provides the restoring force:

$$F_b = -\rho g A x$$

where $A$ is the cross-sectional area.

For the cube,

$$A = 0.1 \times 0.1 = 0.01 \, \text{m}^2$$

So the effective spring constant is

$$k = \rho g A = 1000 \times 10 \times 0.01 = 100 \, \text{N/m}$$

Using the time period formula for SHM,

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

Substituting the values,

$$T = 2\pi \sqrt{\frac{0.01}{100}}$$ $$T = 2\pi \sqrt{0.0001} = 2\pi \times 0.01 = 0.02\pi \, \text{s}$$

Now compare with

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

Thus,

$$y = 2$$

Therefore, the correct option is A.

Given: The cube floats in water and executes small vertical oscillations.

Find: The coefficient $y$.

The restoring force comes from the additional displaced volume when the cube is pushed down by $x$. Hence,

$$F = -\rho g A x$$

This is of the form $F = -kx$, so

$$k = \rho g A$$

Now,

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

Therefore,

$$k = 1000 \times 10 \times 0.01 = 100 \, \text{N/m}$$

The cube mass is

$$m = 10 \, \text{g} = 0.01 \, \text{kg}$$

So,

$$T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{0.01}{100}} = 0.02\pi \, \text{s}$$

Hence,

$$y = 2$$

The second provided approach contains an inconsistency: it introduces an effective mass of displaced fluid and obtains $0.2\pi$ s, but then still concludes $y = 2$. The correct working is the first approach, which gives $0.02\pi$ s. Therefore, the correct option remains A.