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2. Electric Potential and Capacitance
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A ball of mass $1\, g$ and charge ${10^{ - 8}}\,C$ 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
A
$22.8\, cm/s$
B
$228\, cm/s$
C
$16.8\, m/s$
D
$168\, m/s$
Solution
(a) By using $\frac{1}{2}m(v_1^2 – v_2^2) = QV$
$==>$ $\frac{1}{2} \times {10^{ – 3}}\{ v_1^2 – {(0.2)^2}\} = {10^{ – 8}}(600 – 0)$
$==>$ ${v_1} = 22.8\,cm/s$
Standard 12
Physics
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