A positive charged particle of charge q and mass m is suspended from a point O by a string of length $l$ . . In the entire space a uniform horizontal electric field E exists. The particle is drawn aside so that the string becomes vertical and then particle is projected horizontally with velocity v such that the particle starts to move along a circle with same constant speed v. Then choose the correct statements :

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Kinetic energy of mass is $\frac{q^{2} E^{2} l}{2 m g}$

Tension in string is $m g [1 + {(\frac{q E}{m g})}^{2}]$

Tension in string is $\sqrt{{(m g)}^{2} + {(q E)}^{2}}$

Kinetic energy of mass is $\frac{5}{2} m l \sqrt{g^{2} + \frac{q^{2} E^{2}}{m^{2}}}$

answer is B, C.

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Detailed Solution

$T \cos θ = m g_{e f f}$ ……….. (i)
$T \sin θ = \frac{m v^{2}}{l \sin θ}$ ……….. (ii)
By (i)
$T = \frac{m g_{e f f} \cos θ}{\cos^{2} θ} = \frac{m g}{\frac{g^{2}}{[g^{2} + {(\frac{q E}{m})}^{2}]}} = m g [1 + {(\frac{q E}{m g})}^{2}]$
By (i) & (ii)
$\frac{m g_{e f f}}{\cos θ} \sin θ = \frac{m v^{2}}{l \sin θ} \Rightarrow v = \sqrt{(g_{e f f} \sin θ) l \tan θ} = \frac{q E}{m} \sqrt{\frac{l}{g}}$
$∴ K . E = \frac{1}{2} m {(\frac{q E}{m})}^{2} \frac{l}{g} = \frac{q^{2} E^{2} l}{2 m g}$