A) 0.02
B) 0.2
C) 3
D) 2
Correct Answer: A
Solution :
Since \[I\text{ }=\text{ }n\text{ }e\text{ }A\text{ }Vd\] \[1.5\text{ }=\text{ }9\text{ }\times \text{ }{{10}^{28}}\times \text{ }1.6\text{ }\times \text{ }{{10}^{-19}}\,\times \text{ }5\text{ }\times \text{ }l{{0}^{-6}}\,V\] \[\Rightarrow \,\,\,V\,\,=\,\,\frac{1.5}{9\,\,\times \,\,1.6\,\,\times \,\,5\,\,\times \,\,{{10}^{3}}}\,\,m/s\] \[V\,\,\,=\,\,\,\frac{0.3}{9\,\,\times \,\,1.6}\,\,mm/s\] \[V\,\,\,=\,\,\frac{1}{48}\,mm/s\] \[=\text{ }0.02\,\,mm/s\] Option is correct.You need to login to perform this action.
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