JEE Main 2025 — Rotational Motion Question with Solution

To determine the angular velocity of the wheel after the cord is unwound by 1 meter, we start by calculating the work done by the force applied on the cord.

Work Done by Force (W_F):

$W_F=20\text{N}\times 1\text{m}=20\text{J}$

Relationship with Kinetic Energy:

The work done (W_F) is converted into the change in kinetic energy of the wheel. Therefore, we have:

$\Delta \text{KE}=20\text{J}=\frac{1}{2}I{\omega }^2$

where $I$ is the moment of inertia of the wheel, and $\omega$ is the angular velocity.

Determine the Moment of Inertia (I):

Since the wheel has mass $M=10\text{kg}$ and radius $R=10\text{cm}=0.1\text{m}$, and knowing it's a solid disk:

$I=M{R}^2=10\times 0.1^2=0.1\text{kg}\cdot {\text{m}}^2$

Solve for Angular Velocity ($\omega$):

Apply the equation for kinetic energy:

$20=\frac{1}{2}\times 0.1\times {\omega }^2$

Solving for $\omega$, we get:

$20=0.05{\omega }^2$

${\omega }^2=\frac{20}{0.05}=400$

$\omega =\sqrt{400}=20\text{rad/s}$

Therefore, the angular velocity of the wheel after the cord is unwound by 1 meter is $20\text{rad/s}$.