A) 3
B) 1/3
C) 8/9
D) 2
Correct Answer: B
Solution :
| [b] The resistance of a conductor material is given by \[R=\rho \frac{l}{A}\] which implies \[R\propto l\] and \[R\propto \frac{l}{A}.\]Also Ohm's law is \[i=\frac{V}{R}\]. |
| Since, voltage remains same in parallel. |
| \[\therefore \] \[i\propto \frac{1}{R}\] |
| \[\Rightarrow \] \[\frac{{{i}_{1}}}{{{i}_{2}}}=\frac{{{R}_{2}}}{{{R}_{1}}}\] |
| \[\frac{{{i}_{1}}}{{{i}_{2}}}=\frac{\rho {{l}_{2}}/{{A}_{2}}}{\rho {{l}_{1}}/{{A}_{1}}}\] \[\left( \because R=\frac{\rho l}{A} \right)\] |
| \[\Rightarrow \] \[\frac{{{i}_{1}}}{{{i}_{2}}}=\frac{{{l}_{2}}}{{{l}_{1}}}\times {{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}\] \[(\because A=\pi {{r}^{2}})\] |
| \[\Rightarrow \] \[\frac{{{i}_{1}}}{{{i}_{2}}}=\frac{3}{4}\times {{\left( \frac{2}{3} \right)}^{2}}\] |
| Hence, \[\frac{{{i}_{1}}}{{{i}_{2}}}=\frac{1}{3}\] |
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