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 solid disc of radius \( 20 \mathrm{~cm} \) and mass \( 10 \mathrm{~kg} \) is rotating with an angular velocity of \( 600 \mathrm{rpm} \), about an axis normal to its circular plane and passing through its centre of mass. The retarding torque required to bring the disc at rest in \( 10 \mathrm{~s} \) is _____
PhysicsRotational MotionJEE Main
Options:
  • A \(4 \pi \times 10^{-2}\)
  • B \(2 \pi\)
  • C \(4 \pi \times 10^{-1}\)
  • D \(2 \pi \times 10^{-2}\)
Solution:
2070 Upvotes Verified Answer
The correct answer is: \(4 \pi \times 10^{-2}\)

τ=ΔLΔt=Iωf-ωiΔt

τ=mR22×[0-ω]Δt

=10×20×10-222×600×π30×10

=0.4π=4π×10-2

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.