A) 4R
B) 8R
C) 16K
D) 2R
Correct Answer: C
Solution :
Resistance of a wire of length\[l,\]area A and specific resistance p is given by \[R=\rho \frac{l}{A}\] Also, Volume (V)\[=Length\,(l)\times Area(A)\] where\[A=\pi {{r}^{2}}\](r is radius) When the wire is stretched its volume (V) remains constant. Hence, \[R=\frac{\rho V}{{{\pi }^{2}}{{r}^{4}}}\] ?.. (i) When radius is halved \[R=\frac{\rho V}{{{\pi }^{2}}{{\left( \frac{r}{2} \right)}^{4}}}\] \[\therefore \] \[\frac{R}{R}=\frac{16\rho V}{{{\pi }^{2}}{{r}^{4}}}\times \frac{{{\pi }^{2}}{{r}^{4}}}{\rho V}=16\] \[\Rightarrow \] \[{{y}^{2}}=\frac{{{a}^{2}}}{2}=\frac{{{4}^{2}}}{2}=8\] \[\Rightarrow \] \[y=2\sqrt{2}\,cm\] Hence, new resistance increases to sixteen times its original value.
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