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A particle of mass m and charge (+q) is fastened to one end of a light string of length l. The other end of string is fixed to the point O. The whole system lies on a frictionless horizontal plane. Initially, the mass is at rest at A. A uniform electric field in the direction shown is switched on Then:


A
The speed of the particle when it reaches B is 2qElm
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B
The speed of the particle when it reaches B is qElm
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C
The tension in the string when particles reaches at B is 2qE
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D
The tension in the string when the particle reaches at B is zero.
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Solution

The correct option is C The tension in the string when particles reaches at B is 2qE

From diagram,
OC=lcos60=l2

CB=OBOC=ll2=l2

Work done by electric force from A to B,

W=F×d=qE×CB=qEl2

Work done by tension is zero because it is perpendicular to the path taken.

So, using work energy theorem,
Net work done=change in kinetic energy

qEl2=12mv212m(0)2

v=qElm

Now, at B let tension be T,

So, TqE=mv2R

TqE=m(qElm)l

T=2qE

Hence, options (b) and (c) are the correct answers.

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