JEE MAIN - Physics (2025 - 3rd April Morning Shift - No. 11)
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
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$$ 2.5 \mathrm{~m} / \mathrm{s}^2 $$
$$ 3.5 \mathrm{~m} / \mathrm{s}^2 $$
$$ 0.25 \mathrm{~m} / \mathrm{s}^2 $$
$$ 0.35 \mathrm{~m} / \mathrm{s}^2 $$
Explanation
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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
$$\begin{aligned} & \mathrm{a}=\mathrm{r} \alpha \\ & \mathrm{a}=\frac{14}{4}=\frac{7}{2}=3.5 \mathrm{~m} / \mathrm{s}^2 \end{aligned}$$
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