QJEEM24Ph6b11: JEE Main 2024 Physics - Class 11 JEEM24Ph6b11 Solution: A string is wrapped around the rim of a wheel of moment of…

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

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

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 $t = 10 \, \mathrm{s}$ :

$\omega = 0 + 10 \cdot 10 = 100 \, \mathrm{rad/s}$

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