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Circular Motion question

2023 · 29 Jan · Shift 1 · Q58
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  5. /2023 · 29 Jan · Shift 1 · Q58

Circular Motion question

2023 · 29 Jan · Shift 1 · Q58

JEE MainPhysicsCircular MotionMCQ+4 / −1
A car is moving on a horizontal curved road with radius 50 m. The approximate maximum speed of car will be, if friction between tyres and road is 0.34. [take g = 10 ms −2^{-2}−2]
  1. A
    3.4 ms −1^{-1}−1
  2. B
    13 ms −1^{-1}−1
  3. C
    22.4 ms −1^{-1}−1
  4. D
    17 ms −1^{-1}−1
View written solutionFree

Correct answer: B

  1. Identify the force providing centripetal force

On a horizontal curved road, the frictional force provides the necessary centripetal force.

The maximum static friction is: fmax⁡=μmgf_{\max}=\mu mgfmax​=μmg

This must be equal to the required centripetal force at maximum speed: mv2r=μmg\frac{mv^2}{r}=\mu mgrmv2​=μmg

  1. Cancel common terms

v2r=μg\frac{v^2}{r}=\mu grv2​=μg

So, v=μgrv=\sqrt{\mu g r}v=μgr​

  1. Substitute the given values

Given:

  • μ=0.34\mu = 0.34μ=0.34
  • g=10 m s−2g = 10\,\text{m s}^{-2}g=10m s−2
  • r=50 mr = 50\,\text{m}r=50m

v=0.34×10×50v=\sqrt{0.34 \times 10 \times 50}v=0.34×10×50​

v=170v=\sqrt{170}v=170​

v≈13.0 m s−1v\approx 13.0\,\text{m s}^{-1}v≈13.0m s−1

  1. Match with the options

The closest option is:

  • B: 13 m s−113\,\text{m s}^{-1}13m s−1
  1. Comparison with stored answer

Stored correct answer: B

Our derived answer is also B, so they agree.

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