Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A particle of mass \(m\) moves with constant speed along a circular path of radius \(r\) under the action of force \(F\). Its speed is
PhysicsMotion In Two DimensionsAIIMSAIIMS 2008
Options:
  • A \(\sqrt{\frac{F r}{m}}\)
  • B \(\sqrt{\frac{F}{r}}\)
  • C \(\sqrt{F m r}\)
  • D \(\sqrt{\frac{F}{m r}}\)
Solution:
2771 Upvotes Verified Answer
The correct answer is: \(\sqrt{\frac{F r}{m}}\)
Centripetal force $(F)=\frac{m v^2}{r} \therefore v=\sqrt{\frac{F r}{m}}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.