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Question

A particle of mass m and charge q is fastened to one end of a string of length l. The other end of the 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 in then switched on. Then:
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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 particle 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 options are
A the speed of the particle when it reaches B is qElm.
B the tension in the string when particle reaches at B is 2qE.
Here the work done(W) by the electric field is the kinetic energy(KE) of the mass m.
W=Fdx=qE(OBOC)=qE(llcos60)=12qEl
KE=12mv2
therefore, 12qEl=12mv2...(1)
v=qElm
If T is the tension in the string, qE=Tmv2l....(2)
using (1), (2) become: qE=TqET=2qE
ans: (B),(C)
118696_112162_ans.png

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