A) \[8.1\times {{10}^{4}}s\]
B) \[2.7\times {{10}^{4}}s\]
C) \[9\times {{10}^{3}}s\]
D) \[3\times {{10}^{3}}s\]
Correct Answer: B
Solution :
Here, Number density of free electrons, \[n=8.5\times {{10}^{28}}{{m}^{-3}}\] Area of cross-section of a wire, \[A=2.0\times {{10}^{-6}}{{m}^{2}}\] Length of the wire \[l=3.0\,m\] Current, \[l=3.0\,A\] The drift velocity of an electron is \[{{v}_{d}}=\frac{I}{ne\,A}\] (i) The time taken by the electron to drift from one end to other end of the wire is : \[t=\frac{l}{{{v}_{d}}}=\frac{l\,ne\,A}{I}\] (using (i)) \[=\frac{\left( 3.0\,m \right)\,\left( 8.5\times {{10}^{28}}{{m}^{-3}} \right)\left( 1.6\times {{10}^{-19}}C \right)\left( 2.0\times {{10}^{-6}}{{m}^{2}} \right)}{\left( 3.0\,\,A \right)}\]\[=2.7\times {{10}^{4}}\,s\]
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