A) R
B) R/4
C) 3R/4
D) R/2
Correct Answer: B
Solution :
(b) R/4 The resistance of rod before reformation \[{{R}_{1}}=R=\frac{\rho {{l}_{1}}}{\pi r_{1}^{2}}\] \[\left[ \because \,\,R=\frac{\rho l}{A}=\frac{\rho l}{\pi {{r}^{2}}} \right]\] Now, the rod is reformed such that \[{{l}_{2}}=\frac{{{l}_{1}}}{2}\] \[\therefore \pi r_{1}^{2}{{l}_{1}}=\pi r_{2}^{2}{{l}_{2}}\] (\[\because \] Volume remains constant) or \[\frac{r_{1}^{2}}{r_{2}^{2}}=\frac{{{l}_{2}}}{{{l}_{1}}}\] (1) Now, the resistance of the rod after reformation \[{{R}_{2}}=\frac{\rho {{l}_{2}}}{\pi r_{2}^{2}}\] \[\therefore \frac{{{R}_{1}}}{{{R}_{2}}}={\frac{\rho {{l}_{1}}}{\pi r_{1}^{2}}}/{\frac{\rho {{l}_{2}}}{\pi r_{2}^{2}}=\frac{{{l}_{1}}}{{{l}_{2}}}\times \frac{r_{2}^{2}}{r_{1}^{2}}}\;\] Or \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}}\times \frac{{{l}_{1}}}{{{l}_{2}}}={{\left( \frac{{{l}_{1}}}{{{l}_{2}}} \right)}^{2}}={{(2)}^{2}}\](using eq. (1)) \[\therefore {{R}_{2}}=\frac{R}{4}\]
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