A) \[6\times {{10}^{5}}\,m{{s}^{-1}}\]
B) \[8\times {{10}^{5}}\,m{{s}^{-1}}\]
C) \[1.8\times {{10}^{6}}\,m{{s}^{-1}}\]
D) \[1.8\times {{10}^{5}}\,m{{s}^{-1}}\]
Correct Answer: C
Solution :
\[\frac{1}{2}mv_{\max }^{2}=e{{V}_{0}}\Rightarrow {{v}_{\max }}=\sqrt{2\left( \frac{e}{m} \right){{V}_{0}}}\] \[=\sqrt{2\times 1.8\times {{10}^{11}}\times 9}=1.8\times {{10}^{6}}m/s.\]
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