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Question

# A thin fixed ring of radius 1 m has a positive charge 1×10−5C uniformly distributed over it. A particle of mass 0.9gm having a negative charge of 1×10−6C is placed on the axis at a distance of 1cm from the centre of the ring. Assuming that the oscillations has small amplitude, the time period of oscillations is:

A
0.23s
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B
0.39s
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C
0.49s
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D
0.63s
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Solution

## The correct option is D 0.63sWe know E on the axis of the ring is:E=kqx(r2+x2)3/2For small amplitudes r>>x:⇒E=kqxr3We can see that force is always applied in the opposite direction of displacement i.e:F=−kqxr3(1×10−6)C⇒a=−kqx(1×10−6)mr3We know for a particle in S. H. M: a=−ω2x∴ω2=kq(1×10−6)mr3=9×109×1×10−5×1×10−60⋅9×10−3×13=100⇒ω=10Now, T=2πw=2×3⋅1410=0.63s

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