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Q.

A charge particle ‘q’ is shot towards another charged particle ‘Q’ which is fixed, with a speed ‘v’. It approaches ‘Q’ upto a closest distance r and then returns. If q were given a speed of ‘5v’ the closest distances of approach would be

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a

r/5

b

2r

c

r

d

r/25

answer is D.

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Detailed Solution

12mv2=kQqr ⇒12m(5v)2=kqQr' ⇒r'=r25 

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