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6. system of particles and rotational motion — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics6. system of particles and rotational motion1 Mark
MCQ
A solid cylinder of mass $2\; kg$ and radius $4 \;\mathrm{cm}$ is rotating about its axis at the rate of $3\; \mathrm{rpm}$. The torque required to stop after $2 \pi$ revolutions is
  • ✓
    $2 \times 10^{-6} \; \mathrm{Nm}$
  • B
    $2 \times 10^{-3} \; \mathrm{Nm}$
  • C
     $12 \times 10^{-4}\; \mathrm{Nm}$
  • D
    $2 \times 10^{6} \; \mathrm{Nm}$
✓

Answer

Correct option: A.
$2 \times 10^{-6} \; \mathrm{Nm}$
a
$\theta= 2 \pi$ radian

$\omega_{0}=3 \mathrm{rpm} \Rightarrow \frac{2 \pi}{60}(3) \frac{\mathrm{rad}}{\mathrm{sec}}$

$\omega^{2}=\omega_{0}^{2}-2 \alpha \theta$

$0=\left(\frac{3 \times 2 \pi}{60}\right)^{2}-2 \alpha\left(4 \pi^{2}\right)$

$\therefore \alpha=\frac{1}{800} \mathrm{rad} / \mathrm{s}^{2}$

$\tau=\frac{m R^{2}}{2} \alpha=\frac{2}{2} \times\left(\frac{4}{100}\right)^{2} \times \frac{1}{800}=2 \times 10^{-6} \mathrm{Nm}$

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