    Question

# 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=OB−OC=l−l2=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=12mv2−12m(0)2 ⇒v=√qElm Now, at B let tension be T, So, T−qE=mv2R ⇒T−qE=m(qElm)l ∴T=2qE Hence, options (b) and (c) are the correct answers.  Suggest Corrections  0      Explore more