A) \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{1}{4}\]
B) \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{1}{2}\]
C) \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{2}{3}\]
D) \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{3}{5}\]
Correct Answer: B
Solution :
Here, resistance of wire\[=R=12\,\Omega \] Equivalent resistance of lengths \[{{l}_{1}}\] and \[{{l}_{2}}=\frac{8}{3}\Omega \] Let, resistance of length \[{{l}_{1}}\]and \[{{l}_{2}}\]are \[{{R}_{1}}{{R}_{2}}\] respectively. \[\therefore \]\[{{R}_{eq}}=\frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}\] We know, \[{{R}_{1}}+{{R}_{2}}=R=12\,\Omega \] ?(i) So, \[\frac{8}{3}=\frac{{{R}_{1}}{{R}_{2}}}{12}\] or, \[{{R}_{1}}{{R}_{2}}=32\] ?(ii) From (i) and (ii) \[{{R}_{1}}=4\,\Omega \]and \[{{R}_{2}}=8\,\Omega \] We know, \[R=\rho \frac{l}{A}\] or \[R\propto l\] So, \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}}\]or \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{1}{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
