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Circular Motion question

2023 · 31 Jan · Shift 2 · Q61
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  5. /2023 · 31 Jan · Shift 2 · Q61

Circular Motion question

2023 · 31 Jan · Shift 2 · Q61

JEE MainPhysicsCircular MotionMCQ+4 / −1
A stone of mass 1 kg1 \mathrm{~kg}1 kg is tied to end of a massless string of length 1 m1 \mathrm{~m}1 m. If the breaking tension of the string is 400 N400 \mathrm{~N}400 N, then maximum linear velocity, the stone can have without breaking the string, while rotating in horizontal plane, is :
  1. A
    20 ms−120 \mathrm{~ms}^{-1}20 ms−1
  2. B
    40 ms−140 \mathrm{~ms}^{-1}40 ms−1
  3. C
    400 ms−1400 \mathrm{~ms}^{-1}400 ms−1
  4. D
    10 ms−110 \mathrm{~ms}^{-1}10 ms−1
View written solutionFree

Correct answer: A

  1. Given data

    • Mass of stone: m=1 kgm = 1\,\text{kg}m=1kg
    • Length of string (radius): r=1 mr = 1\,\text{m}r=1m
    • Maximum tension before breaking: Tmax⁡=400 NT_{\max} = 400\,\text{N}Tmax​=400N

  2. Condition for horizontal circular motion

    While rotating in a horizontal plane, the string tension provides the required centripetal force. Thus, T=mv2rT = \frac{mv^2}{r}T=rmv2​

    For the maximum possible speed without breaking, Tmax⁡=mvmax⁡2rT_{\max} = \frac{mv_{\max}^2}{r}Tmax​=rmvmax2​​

  3. Substitute the values 400=(1) vmax⁡21400 = \frac{(1)\,v_{\max}^2}{1}400=1(1)vmax2​​

    So, vmax⁡2=400v_{\max}^2 = 400vmax2​=400 vmax⁡=20 m/sv_{\max} = 20\,\text{m/s}vmax​=20m/s

  4. Match with options

    • A: 20 ms−120\,\text{ms}^{-1}20ms−1 ✅
    • B: 40 ms−140\,\text{ms}^{-1}40ms−1
    • C: 400 ms−1400\,\text{ms}^{-1}400ms−1
    • D: 10 ms−110\,\text{ms}^{-1}10ms−1

Therefore, the correct option is A.

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