Questions
A particle of given mass is moving in a circular path of radius ‘r’ with kinetic energy K. If radius of circular path is reduced to r/2 keeping the angular momentum constant, the final kinetic energy of the particle will be
detailed solution
Correct option is D
mvr=L ⇒ v=L/mr∴ K=12 mv2 = 12 m Lmr2 = L22mr2So, for constant angular momentum, K ∝ 1r2∴ K'K=r2r/22=4 ⇒ K'=4K.Talk to our academic expert!
Similar Questions
A particle of mass m = 5 units is moving with a uniform speed , units in the XOY plane along the line y = x + 4. The magnitude of the angular momentum of the Particle about the origin is
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
free study material
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests
