A) 22.8 cm/s
B) 228 cm/s
C) 16.8 m/s
D) 168 m/s
Correct Answer: A
Solution :
Given : Potential at A, V= 600 volt mass of the ball \[m=1\,gm={{10}^{-3}}kg\]charge on the ball \[q={{10}^{-8}}C\]velocity at\[B,{{\upsilon }_{B}}=20cm/s\] Now from the relation \[\upsilon _{A}^{2}-\upsilon _{B}^{2}=\frac{2qV}{m}=\frac{2\times {{10}^{-8}}\times 600}{{{10}^{-3}}}=0.012\] Thus \[\upsilon _{A}^{2}=0.012+0.04=0.052\] \[{{\upsilon }_{A}}=0.22.8cm/\sec .\] \[{{\upsilon }_{A}}=22.8\,cm/\sec .\]
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