A) \[4R\]
B) \[2R\]
C) R
D) none of these
Correct Answer: A
Solution :
Initial Resistance \[{{R}_{1}}=R\] Initial length \[{{l}_{1}}=l\] Final lenght \[{{l}_{2}}=2l\] The volume of wire is given by \[\pi r_{1}^{2}\times l\] . Since the wire of volume remains same after stretch therefore, \[\pi r_{1}^{2}\times l=\pi r_{2}^{2}\times 2l\] \[\Rightarrow \] \[{{r}_{2}}=\frac{{{r}_{1}}}{{{r}_{2}}}\] The resistance of wire is given by \[R=\rho \times \frac{l}{\Delta }=\rho \times \frac{l}{\pi {{r}^{2}}}\] \[\Rightarrow \]\[R\propto \frac{l}{{{r}^{2}}}\] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{l}_{1}}/r_{1}^{2}}{{{l}_{2}}/r_{2}^{2}}=\frac{l}{2l}=\frac{{{\left( {{r}_{1}}/\sqrt{2} \right)}^{2}}}{{{\left( {{r}_{1}} \right)}^{2}}}\] \[\Rightarrow \] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}\] \[\Rightarrow \] \[{{R}_{2}}=4{{R}_{1}}=4R\]
You need to login to perform this action.
You will be redirected in
3 sec
