KCET 2025 Physics Paper

The relationship between the torque, moment of inertia, and angular acceleration for each wheel can be established with the following equations:For the smaller wheel:$T_1=I_1 \alpha_1$ where $T_1$ is the torque applied to the smaller wheel, $I_1=4 \mathrm{kg} \cdot \mathrm{m}^2$ is its moment of inertia, and $\alpha_1$ is the angular acceleration.The angular acceleration ratio between the two wheels is given by:$\alpha_2 r_2=\alpha_1 r_1$ where $\alpha_2$ is the angular acceleration of the larger wheel, $r_1=0.2 \mathrm{m}$, and $r_2=0.3 \mathrm{m}$.Given that a torque of 10 Nm is applied to the smaller wheel:$10=\alpha_1 \times 4$ Solving for $\alpha_1$ :$\alpha_1=\;\frac{10}{4}=\;\frac{5}{2} \mathrm{rad} / \mathrm{s}^2$Using the relation for the angular accelerations:$\;\alpha_2 \times 0.3 = \left(\;\frac{5}{2}\right) \times 0.2$$\;\alpha_2=\;\frac{5}{3} \mathrm{rad} / \mathrm{s}^2$ Finally, for the larger wheel, the torque $T_2$ is given by:$T_2=I_2 \alpha_2$where $I_2=20 \mathrm{kg} \cdot \mathrm{m}^2$. Plugging in the values:$T_2=20 \times \;\frac{5}{3}=\;\frac{100}{3} \mathrm{Nm}$ Thus, the angular accelerations and the torque can be summarized as follows:Angular acceleration of the smaller wheel $\alpha_1=\;\frac{5}{2} \mathrm{rad} / \mathrm{s}^2$Angular acceleration of the larger wheel $\alpha_2=\;\frac{5}{3} \mathrm{rad} / \mathrm{s}^2$Torque on the larger wheel $T_2=\;\frac{100}{3} \mathrm{Nm}$