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Circular Motion question

2024 · 1 Feb · Shift 1 · Q67
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  5. /2024 · 1 Feb · Shift 1 · Q67

Circular Motion question

2024 · 1 Feb · Shift 1 · Q67

JEE MainPhysicsCircular MotionMCQ+4 / −1
A ball of mass 0.5 kg0.5 \mathrm{~kg}0.5 kg is attached to a string of length 50 cm50 \mathrm{~cm}50 cm. The ball is rotated on a horizontal circular path about its vertical axis. The maximum tension that the string can bear is 400 N400 \mathrm{~N}400 N. The maximum possible value of angular velocity of the ball in rad/s\mathrm{rad} / \mathrm{s}rad/s is, :
  1. A
    1600
  2. B
    20
  3. C
    40
  4. D
    1000
View written solutionFree

Correct answer: C

  1. Identify the physical situation

The ball is moving in a horizontal circle while attached to a string and rotating about a vertical axis. The string provides the centripetal force.

Given:

  • Mass: m=0.5 kgm = 0.5\,\text{kg}m=0.5kg
  • Length of string: r=50 cm=0.5 mr = 50\,\text{cm} = 0.5\,\text{m}r=50cm=0.5m
  • Maximum tension: Tmax⁡=400 NT_{\max} = 400\,\text{N}Tmax​=400N

We need the maximum angular velocity ω\omegaω.

  1. Use centripetal force relation

For circular motion,

T=mω2rT = m\omega^2 rT=mω2r

At the maximum possible angular velocity, the tension reaches its maximum value:

400=(0.5)ω2(0.5)400 = (0.5)\omega^2(0.5)400=(0.5)ω2(0.5)
  1. Solve for ω\omegaω
400=0.25ω2400 = 0.25\omega^2400=0.25ω2 ω2=4000.25=1600\omega^2 = \frac{400}{0.25} = 1600ω2=0.25400​=1600 ω=1600=40 rad/s\omega = \sqrt{1600} = 40\,\text{rad/s}ω=1600​=40rad/s
  1. Match with options
  • A: 160016001600 ❌
  • B: 202020 ❌
  • C: 404040 ✅
  • D: 100010001000 ❌

So the correct option is C.

  1. Comparison with stored answer

Stored correct answer: C

My derived answer: C

They agree.

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