JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
The centre of a wheel rolling on a plane surface moves with a speed \(v_{0} .\) A particle on the rim of the wheel at the same level as the centre will be moving at a speed \(\sqrt{x}\, v_{0}\). Then the value of \(x\) is ...... .
- A \(9\)
- B \(2\)
- C \(4\)
- D \(81\)
Answer & Solution
Correct Answer
(B) \(2\)
Step-by-step Solution
Detailed explanation
For no slipping \({v}_{0}=\omega {R}\) Now \(v_{A}=v_{B}=\sqrt{v_{0}^{2}+(\omega R)^{2}}\) \(=\sqrt{2} v_{0}\) \(x=2\)
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