Work Energy And Power Question 63

Question: A particle of mass m is moving in a horizontal circle of radius r under a centripetal force equal to K/r2 , where K is a constant. The total energy of the particle is [IIT 1977]

Options:

A) K2r

B) K2r

C) Kr

D) Kr

Show Answer

Answer:

Correct Answer: B

Solution:

Here mv2r=Kr2

K.E. =12mv2=K2r

U=rF.dr=r(Kr2)dr=Kr

Total energy E=K.E.+P.E.=K2rKr=K2r



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