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

• Correct answer
  A

  $3.5 \mathrm{~m} / \mathrm{s}^2$
• B

  $0.35 \mathrm{~m} / \mathrm{s}^2$
• C

  $2.5 \mathrm{~m} / \mathrm{s}^2$
• D

  $0.25 \mathrm{~m} / \mathrm{s}^2$

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
$\begin{aligned}<br/>& \mathrm{a}=\mathrm{r} \alpha \\ & \mathrm{a}=\frac{14}{4}=\frac{7}{2}=3.5 \mathrm{~m} / \mathrm{s}^2<br/>\end{aligned}$

Hence, the correct answer is option: 'A'.