Q84 / 180 · AIIMS 2018 (27 May) · Physics

An elevator starts going up with an acceleration $2 \mathrm{~m} / \mathrm{s}^2$. If radius of the wheel attached to the elevator is $0.1 \mathrm{~m}$, then find out number of revolutions in $10 \mathrm{~s}$.

  • Correct answer
    A
    $160$
  • B
    $170$
  • C
    $180$
  • D
    $190$

Solution

$\theta=2 \pi n=\frac{1}{2}(\alpha) t^2$ $n=\frac{1}{2}\left(\frac{a}{r}\right) \frac{t^2}{2 \pi}$ Here, $a=2 \mathrm{~m} / \mathrm{s}^2, r=0.1 \mathrm{~m}, t=10 \mathrm{~s}, n=$ ? $\therefore n=\frac{1}{2}\left(\frac{2}{0.1}\right) \frac{10^2}{6.28} \approx 160$

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

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