A) \[16R\]
B) \[3R\]
C) \[2R\]
D) \[R\]
Correct Answer: A
Solution :
Here: Initial resistance\[{{R}_{1}}=R\] Initial radius\[{{r}_{1}}=r\], final radius\[{{r}_{2}}=\frac{r}{2}\] The volume of the wire remain constant even when radius is reduced. It means \[{{A}_{1}}{{l}_{1}}={{A}_{2}}{{l}_{2}}\] or \[\pi {{r}_{1}}^{2}{{l}_{1}}=\pi {{r}_{2}}^{2}{{l}_{2}}\] \[\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)=\frac{{{l}_{1}}}{{{l}_{2}}}\] or \[{{l}_{2}}=4{{l}_{1}}\] Also, resistance of wire is \[R=\frac{\rho l}{A}\propto \frac{l}{A}\] Hence, \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{l}_{1}}}{{{A}_{1}}}\times \frac{{{A}_{2}}}{{{l}_{2}}}\] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{l}_{1}}}{\pi r_{1}^{2}}\times \frac{\pi r_{2}^{2}}{{{l}_{2}}}\] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{l}_{1}}}{2{{l}_{2}}}\times \frac{{{l}_{1}}}{{{l}_{2}}}\] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{l}_{1}}}{4{{l}_{1}}}\times \frac{{{l}_{1}}}{4{{l}_{1}}}=\frac{1}{4}\] So, new resistance is given by, \[{{R}_{2}}=16{{R}_{1}}=16R\]
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