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 chain of 125 links is 1.25 m long and has mass of 2 kg with the ends fastened together. It is set for rotating at 50 revolutions per second. The centripetal force on each link is

PhysicsLaws of MotionNEET
Options:
  • A 3.14 N
  • B 0.314 N
  • C 314 N
  • D None of these
Solution:
2038 Upvotes Verified Answer
The correct answer is: 314 N
circumference =2πr=1.25 m

so radius r=1.25m2π=0.199 m

Mass of each link is m =2 kg125 =0.016 kg

angular velocity is w=50 revolutions per second =50×2π rad s-1=314 rad s-1

The centripetal force on each link is f=m×r×w2=0.016×0.199×3142 N

f=314 N

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.