Exam

JEE Main

Year

2025

Date

3 April

Shift

Morning Shift

Subject

Physics

Chapter

Rotational Motion

JEE Main 2025 · 3 April · Morning Shift · Physics · Rotational Motion

A force of 49 N acts tangentially at the highest point of a sphere...

A force of 49 N acts tangentially at the highest point of a sphere (solid) of mass 20 kg , kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is

Step-by-step solution

Torque about bottom point
$$\begin{aligned}<br/>& \mathrm{F} \times 2 \mathrm{r}=\mathrm{I} \alpha \\ & 49 \times 2 \mathrm{r}=\frac{7}{5} \mathrm{mr}^2 \alpha \\ & 14=4 \mathrm{r} \alpha<br/>\end{aligned}$$
As sphere rolls without slipping
$$\begin{aligned}<br/>& \mathrm{a}=\mathrm{r} \alpha \\ & \mathrm{a}=\frac{14}{4}=\frac{7}{2}=3.5 \mathrm{~m} / \mathrm{s}^2<br/>\end{aligned}$$