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Q.

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

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a

K/2

b

2K

c

K/4

d

4K

answer is D.

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Detailed Solution

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.
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