Published by:
CGP EDU Academic Team
Published on: September 12, 2026
A ball of mass 1 g and charge $10^{-8} \, \mathrm{C}$ moves from a point A. where potential is 600 volt to the point B where potential is zero. Velocity of the ball at the point B is 20 cm/s. The velocity of the ball at the point A will be
Text Solution
Verified by ExpertsThe correct answer is:
A
By using $\frac{1}{2} m (v_1^2 - v_2^2) = QV$
⇒ ⇒ $\frac{1}{2} \times 10^{-3} \{ v_1^2 - (0.2)^2 \} = 10^{-8} (600 - 0)$
⇒ ⇒ $v_1 = 22.8 \text{ cm/s}$
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