A) \[\frac{R}{2}\]
B) \[\frac{R}{4}\]
C) \[\frac{R}{8}\]
D) \[\frac{R}{16}\]
Correct Answer: D
Solution :
We have, \[R=\frac{\rho l}{A}\] ?(i) Since, \[m=V\times \rho =Al\rho \] \[A=\frac{m}{l\rho }\Rightarrow l=\frac{m}{A\rho }\] ?(ii) From Eqs. (i) and (ii) \[R=\frac{\rho M}{A\rho A}=R\propto \frac{1}{{{A}^{2}}}\] Hence, \[\frac{{{R}_{2}}}{{{R}_{1}}}={{\left( \frac{{{A}_{1}}}{{{A}_{2}}} \right)}^{2}}=\frac{\pi r_{1}^{2}}{\pi r_{2}^{2}}={{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{4}}={{\left( \frac{{{r}_{1}}}{2{{r}_{1}}} \right)}^{4}}=\frac{1}{16}\] So, \[{{R}_{2}}=\frac{{{R}_{1}}}{16}=\frac{R}{16}\](since\[{{R}_{1}}=R\])
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