ExamBro
ExamBro
JEE Mains · Physics · STD 11 - System of particles and rotational motion
  • JEE Mains 2021
  • Medium
  • Numerical

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 ...... .

Pick an option to check your answer.
Verified Solution

Answer & Solution

Correct Answer

(B) \(2\)

Step-by-step Solution

Detailed explanation

For no slipping \({v}_{0}=\omega {R}\)

Now \(v_{A}=v_{B}=\sqrt{v_{0}^{2}+(\omega R)^{2}}\)

\(=\sqrt{2} v_{0}\)

\(x=2\)

Same subject
Explore more questions on app