A) 10m/s
B) 20m/s
C) 60m/s
D) 90m/s
Correct Answer: D
Solution :
[d] \[P{{E}_{B}}=\frac{K(1\mu C)\,(1mC)}{1}\] \[K{{E}_{B}}=0\] When it is at A \[P{{E}_{A}}=K\frac{(1\mu C)(1mC)}{10}\] \[K{{E}_{A}}=\frac{1}{2}m{{u}^{2}}(let)\] Conservation of energy \[{{K}_{A}}+{{P}_{A}}={{K}_{B}}+{{P}_{B}}\] \[\frac{1}{2}m{{u}^{2}}+\frac{9\times {{10}^{9}}\times {{10}^{-6}}\times {{10}^{-3}}}{10}=\frac{9\times {{10}^{9}}\times {{10}^{-6}}\times {{10}^{-3}}}{1}\]\[\frac{2\times {{10}^{-3}}}{2}{{u}^{2}}+0.9=9\] \[{{u}_{2}}=8100\] \[u=90m/s\]You need to login to perform this action.
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