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A particle of mass m is moving in a hori...

A particle of mass m is moving in a horizontal circle of radius r, under a centripetal force equal to `-(k//r^2)` where k is constant. The total energy of the particle is

A

`(K)/(2r)`

B

`-(K)/(2r)`

C

`-(K)/(r )`

D

`(K)/(r )`

Text Solution

Verified by Experts

The correct Answer is:
B
Here `(mv^2)/(r )=(K)/(r^2)` `KE=(1)/(2)mv^2=(K)/(2r)`
`U=-int_(infty)^(r )Fdr=-int_(infty)^(r )(-(K)/(r^2))dr=-(K)/(r )`
Total energy `E=KE+PE=(K)/(2r)-(K)/(r )=-(K)/(2r)`
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