A small ball of mass $2 \times 10^{- 3} k g$ having a charge of $1 μ c$ is suspended by a string of length $0.8 m .$ Another identical ball having the same charge is kept at the point of suspension. Determine the minimum horizontal velocity which should be imparted to the lower ball so that it can make complete revolution. $(g = 10 m / s^{2})$

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$5.86 m / s$

$7.58 m / s$

$4.68 m / s$

$8.68 m / s$

answer is B.

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Detailed Solution

At A:- $T_{A} - m g - F_{e} = \frac{m V_{A}^{2}}{l}$ ………. (1)
At B:- $T_{B} + m g - F_{e} = \frac{m V_{B}^{2}}{l}$ …..….. (2)
For just completing circle, $T_{B} = 0$
$V_{B}^{2} = \frac{(m g - F e) l}{m}$
From conservation of energy,
$\frac{1}{2} m V_{A}^{2} = \frac{1}{2} m V_{B}^{2} + m g (2 l)$

$\frac{1}{2} m V_{A}^{2} = \frac{1}{2} m \frac{(m g - F e)}{m} l + m g (2 l) V_{A}^{2} = g l - \frac{F_{e}}{m} l + 4 g l V_{A} = \sqrt{5 g l - \frac{F_{e}}{m} l} = \sqrt{5 (10) (0.8) - [\frac{9 \times 10^{9} \times 10^{- 12}}{{(0.8)}^{2}} \times \frac{0.8}{2 \times 10^{- 3}}]} = 5.86 m / s$