Q136 / 150 · MHT CET 2021 (20 Sep Shift 2) · Physics
A disc of radius $0.4 \mathrm{~m}$ and mass $1 \mathrm{~kg}$ rotates about an axis passing through its center and perpendicular to its plane. The angular acceleration of the disc is $10 \mathrm{rad} / \mathrm{s}^2$. The tangential force applied to the rim of the disc is
- A$4 \mathrm{~N}$
- B$1 \mathrm{~N}$
- Correct answerC$2 \mathrm{~N}$
- D$8 \mathrm{~N}$
Solution
$\mathrm{R}=0.4 \mathrm{~m}, \mathrm{M}=1 \mathrm{~kg}, \alpha=10 \mathrm{rad} / \mathrm{s}^2$
Moment of inertia, $\mathrm{I}=\frac{\mathrm{MR}^2}{2}=\frac{1 \times(0.4)^2}{2}=0.08 \mathrm{~kg} \mathrm{~m}^2$
Torque, $\tau=\mathrm{RF}=\mathrm{I} \alpha$
$\therefore \mathrm{F}=\frac{\mathrm{I} \alpha}{\mathrm{R}}=\frac{0.08 \times 10}{0.4}=2 \mathrm{~N}$
Hence, the correct answer is option: 'C'.