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Question

A simple pendulum with a bob of mass m=1 kg, charge q=5μC and string length l=1 m is given a horizontal velocity u in a uniform electric field E=2×106 V/m at its bottom most point A as shown in figure. It is given that the speed u is such that the string slacks at point C. Find the speed u.
[Take g=10 m/s2]


A
37 m/s
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B
5 m/s
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C
31 m/s
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D
7 m/s
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Solution

The correct option is C 31 m/s
Given:
m=1 kg;q=5μC=5×106 C,
l=1 m;E=2×106 V/m


Let the speed of the bob at point C is v.

Now apply work-energy theorem between points A and C.

From work energy theorem,

12m(v2u2)=mgl(1+sin60)+qElcos60

12×1×(v2u2)=[1×10×1(1+sin60)]+[5×106×2×106×1×cos60]

u2v2=27.3 .........(1)

Further, at point C, tension in the string will be zero.

Hence,

mv2l=mgsin60qEcos60

1×v21=1×10×sin605×106×2×106×cos60

v2=3.7 ...........(2)

From eqs. (1) and (2)

u=31 m/s

Hence, option (c) is the correct answer.

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