Question 103 of 280

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

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

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Answer: (A) $3.5 \mathrm{~m} / \mathrm{s}^2$

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