JEE Main 2024 — Rotational Motion Question with Solution

To find the angular velocity ($\omega$) of the wheel after $10$ seconds, we first need to understand the relationship between the force applied through the string, the torque produced by this force, and how this torque affects the wheel's angular acceleration ($\alpha$).

The torque ($\tau$) produced by the force ($F$) is given by the product of the force and the radius ($r$) of the wheel through which the force is applied:

$\tau =F\cdot r$

Given that $F=40\mathrm{N}$ and $r=10\mathrm{cm}=0.1\mathrm{m}$, the torque can be calculated as:

$\tau =40\cdot 0.1=4\mathrm{Nm}$

The torque is related to the angular acceleration ($\alpha$) and the moment of inertia ($I$) of the wheel by the equation:

$\tau =I\cdot \alpha$

Given that $I=0.40{\mathrm{kg}\cdot \mathrm{m}}^2$, we can rearrange the above formula to solve for $\alpha$:

$\alpha =\frac{\tau }{I}=\frac{4}{0.40}=10{rad/s}^2$

With the angular acceleration ($\alpha$), we can calculate the angular velocity ($\omega$) after a given time ($t$) using the formula:

$\omega ={\omega }_0+\alpha \cdot t$

Where ${\omega }_0$ is the initial angular velocity. Since the wheel starts from rest, ${\omega }_0=0$. Thus, for $t=10\mathrm{s}$:

$\omega =0+10\cdot 10=100rad/s$

Therefore, the angular velocity ($\omega$) of the wheel after $10$ seconds is $100rad/s$, so $x=100$.