A) zero
B) 2 R
C) 8 R
D) 16 R
Correct Answer: D
Solution :
For a wire of length \[l,\]area A and specific resistance \[\rho ,\] resistance (R) is given by \[R=\rho \frac{l}{A}\] ?(i) Also volume = length x area which remains constant on stretching the wire, hence \[{{l}_{1}}{{A}_{1}}={{l}_{2}}{{A}_{2}}\] If \[{{r}_{1}}\] and \[{{r}_{2}}\] are radii of wire then, \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{{{A}_{2}}}{{{A}_{1}}}\] \[=\frac{\pi r_{2}^{2}}{\pi r_{1}^{2}}\] \[=\left( \frac{0.5\times r}{r} \right)=\frac{1}{4}\] Using Eq. (i), we have \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}}\times \frac{{{A}_{2}}}{{{A}_{1}}}\] \[=\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}\] \[\Rightarrow \] \[{{R}_{2}}=16{{R}_{1}}=16R\]
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