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Circular Motion question

2024 · 5 Apr · Shift 2 · Q67
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  5. /2024 · 5 Apr · Shift 2 · Q67

Circular Motion question

2024 · 5 Apr · Shift 2 · Q67

JEE MainPhysicsCircular MotionMCQ+4 / −1
A man carrying a monkey on his shoulder does cycling smoothly on a circular track of radius 9 m9 \mathrm{~m}9 m and completes 120 resolutions in 3 minutes. The magnitude of centripetal acceleration of monkey is (in m/s2\mathrm{m} / \mathrm{s}^2m/s2 ) :
  1. A
    4π2 ms−24 \pi^2 \mathrm{~ms}^{-2}4π2 ms−2
  2. B
    16π2 ms−216 \pi^2 \mathrm{~ms}^{-2}16π2 ms−2
  3. C
    57600π2 ms−257600 \pi^2 \mathrm{~ms}^{-2}57600π2 ms−2
  4. D
    Zero
View written solutionFree

Correct answer: B

  1. Given data

    • Radius of circular track: r=9 mr = 9\,\text{m}r=9m
    • Number of revolutions: 120120120
    • Time taken: 3 min=180 s3\,\text{min} = 180\,\text{s}3min=180s

  2. Find frequency Frequency is revolutions per second: f=120180=23 Hzf = \frac{120}{180} = \frac{2}{3}\,\text{Hz}f=180120​=32​Hz

  3. Find angular speed ω=2πf=2π⋅23=4π3 rad/s\omega = 2\pi f = 2\pi \cdot \frac{2}{3} = \frac{4\pi}{3}\,\text{rad/s}ω=2πf=2π⋅32​=34π​rad/s

  4. Centripetal acceleration Magnitude of centripetal acceleration is ac=ω2ra_c = \omega^2 rac​=ω2r Substituting: ac=(4π3)2⋅9a_c = \left(\frac{4\pi}{3}\right)^2 \cdot 9ac​=(34π​)2⋅9 ac=16π29⋅9=16π2 m/s2a_c = \frac{16\pi^2}{9} \cdot 9 = 16\pi^2\,\text{m/s}^2ac​=916π2​⋅9=16π2m/s2

  5. Match with options 16π2 m/s216\pi^2\,\text{m/s}^216π2m/s2 corresponds to Option B.

Final Answer: 16π2 m/s2\boxed{16\pi^2\,\text{m/s}^2}16π2m/s2​

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