Q 11-06-256JEE MainJEE Main 2025 (3 Apr, Shift 1)Medium
A force of $49\ \text{N}$ acts tangentially at the highest point of a solid sphere of mass $20\ \text{kg}$, kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is:
Answer: (A) $3.5\ \text{m/s}^2$
Take torques about the contact point P, which is the instantaneous axis when rolling; friction acts at P and gives no torque there.
$I_P = \tfrac25mr^2 + mr^2 = \tfrac75mr^2$, and the force at the top is $2r$ from P:
$$F(2r) = \tfrac75mr^2\alpha \Rightarrow a = r\alpha = \frac{10F}{7m} = \frac{10\times49}{7\times20} = 3.5\ \text{m/s}^2$$
Solution by Sreeraj P, M.Sc Physics