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A particle is moving in a circular path ...

A particle is moving in a circular path of radius a under the action of an attractive potential `U=-(k)/(2r^(2))`. Its total energy is :

A

Zero

B

` - (3)/(2) (k)/( a^(2))`

C

` - (k)/( 4 a^(2))`

D

`(k)/( 2 a^(2))`

Text Solution

Verified by Experts

The correct Answer is:
A
Since ` U = (-k)/(2 r ^(2))` . . . (i)
` F = ( - dU)/( dr) = (- k)/( r^(3))`
Now `(k)/( r^(3)) = (mv^(2))/( r)` (force provides centripetal force)
`:. KE = (1)/(2) mv^(2) = (k)/(2 r ^(2))` . . . (ii)
Total energy = KE + PE = 0 [By eq. (i) and (ii) ]
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