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AIIMS 201827 May
Exam
AIIMS
Year
2018
Date
27 May
Subject
Physics
Chapter
Rotational Motion

AIIMS 2018 · 27 May · Physics · Rotational Motion

An elevator starts going up with an acceleration 2 ~m / s^2 . If...

An elevator starts going up with an acceleration 2 m/s22 \mathrm{~m} / \mathrm{s}^2. If radius of the wheel attached to the elevator is 0.1 m0.1 \mathrm{~m}, then find out number of revolutions in 10 s10 \mathrm{~s}.

Options

  1. A160160
  2. B170170
  3. C180180
  4. D190190

Correct answer

A. 160160

Step-by-step solution

θ=2πn=12(α)t2\theta=2 \pi n=\frac{1}{2}(\alpha) t^2 n=12(ar)t22πn=\frac{1}{2}\left(\frac{a}{r}\right) \frac{t^2}{2 \pi} Here, a=2 m/s2,r=0.1 m,t=10 s,n=a=2 \mathrm{~m} / \mathrm{s}^2, r=0.1 \mathrm{~m}, t=10 \mathrm{~s}, n= ? ∴n=12(20.1)1026.28≈160\therefore n=\frac{1}{2}\left(\frac{2}{0.1}\right) \frac{10^2}{6.28} \approx 160
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