A) \[nR\]
B) \[\frac{R}{n}\]
C) \[\frac{R}{{{n}^{2}}}\]
D) \[\frac{R}{{{n}^{4}}}\]
Correct Answer: D
Solution :
Initial resistance \[{{R}_{1}}=R\] Initial radius of wire \[{{r}_{1}}=r\] Final radius of wire \[{{r}_{2}}=nr\] (given) Let \[{{l}_{1}}\] and \[{{A}_{1}}\] is the length and the cross-sectional area of the original wire and \[{{l}_{2}}\]and \[{{A}_{2}}\] is the length and cross-sectional area of the compressed wire, since, volume of the wire remains constant. Hence, \[{{l}_{1}}\pi r_{1}^{2}={{l}_{2}}\pi r_{2}^{2}\] \[{{l}_{1}}{{r}^{2}}={{l}_{2}}{{(nr)}^{2}}\] \[\frac{{{l}_{1}}}{{{l}_{2}}}={{n}^{2}}\] and \[\frac{{{A}_{1}}}{{{A}_{2}}}=\frac{\pi {{r}^{2}}}{\pi {{n}^{2}}{{r}^{2}}}=\frac{1}{{{n}^{2}}}\] The resistance is given by \[R=\rho \frac{l}{A}\] Hence \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}}\times \frac{{{A}_{2}}}{{{A}_{1}}}={{n}^{2}}\times {{n}^{2}}={{n}^{4}}\] or \[{{R}_{2}}=\frac{{{R}_{1}}}{{{n}^{4}}}=\frac{R}{{{n}^{4}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
