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Question

A small ball of mass m and charge +q tied with a string of length l is rotating in a vertical circle under gravity and a uniform horizontal electric field E as shown. The tension in the string will be minimum for:
1028368_f56225baa2404af79399f87e1eeeefe7.PNG

A
θ=tan1(qEmg)
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B
θ=π
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C
θ=πtan1(qEmg)
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D
θ=π+tan1(qEmg)
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Solution

The correct option is A θ=tan1(qEmg)

mass =m
charge =q
length =l
Electric field =E
Tension in the string will be minimum f0. Let, the velocity at bottom most point is V0. Then an angle θ let the velocity of particle is V.
TmgcosθqFsinθ=mV2l(1)
Using energy conservation,
12mV212mV20=mgl(1cosθ)+qElsinθ
mV2l=mV20l=2mg(1cosθ)+2qEsinθ
Putting in equation (1) we get,
T=mgcosθ+qEsinθ+mV20l2mg(1cosθ)+2qEsinθ
T=3mgcosθ+3qEsinθ+mV20l2mg
For T to be minimum
dTdθ=0 d2Tdθ2<0
Thus, dTdθ=3mgsinθ+3qEcosθ=0
tanθ=qEmg
θ=tan1(qEmg) or, π+tan1(qEmg)

1241184_1028368_ans_0425a430e97e485b934eda938e4af5d0.jpg

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