A) \[{{10}^{19}}\]
B) \[{{10}^{21}}\]
C) \[1.6\times {{10}^{19}}\]
D) \[1.4\times {{10}^{20}}\]
Correct Answer: B
Solution :
Power, \[P=\frac{{{V}^{2}}}{R}\] \[R=\frac{{{V}^{2}}}{P}\] \[=\frac{{{(60)}^{2}}}{160}=22.5\Omega \] Now, according to Ohms law \[V=IR\] \[\therefore \] \[I=\frac{60}{22.5}\] \[\Rightarrow \] \[I=2.6A\] Here, \[t=60s\] As \[I=\frac{ne}{t}\] \[\Rightarrow \]\[n=\frac{I\times t}{e}\] \[=\frac{2.6\times 60}{1.6\times {{10}^{-19}}}\] \[\approx {{10}^{21}}\] (The charge of electron \[e=1.6\times {{10}^{-19}}C\])
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