A) \[\frac{256R}{81}\]
B) \[\frac{81R}{256}\]
C) \[\frac{16R}{9}\]
D) \[\frac{9R}{16}\]
Correct Answer: A
Solution :
The resistance of a wire of length\[l,\]area of cross-section A and specific resistance p is. \[R=\rho \frac{l}{A}\] Also \[volume=length\times area\] \[=lA=constant\] \[{{l}_{1}}{{A}_{1}}={{l}_{2}}{{A}_{2}}\] \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{{{A}_{2}}}{{{A}_{1}}}=\frac{\pi \left( \frac{3r}{4} \right)}{\pi {{r}^{2}}}=\frac{9}{16}\] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}}\times \frac{{{A}_{2}}}{{{A}_{1}}}\] \[=\frac{9}{16}\times \frac{9}{16}=\frac{81}{256}\] \[{{R}_{2}}=\frac{256{{R}_{1}}}{81}=\frac{256R}{81}\]
You need to login to perform this action.
You will be redirected in
3 sec
