JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर Stretching of Wire

If a conducting wire stretches, it's length increases, area of cross-section decreases so resistance increases but volume remain constant.

Suppose for a conducting wire before stretching it's length \[={{l}_{1}},\] area of cross-section \[={{A}_{1}},\] radius \[={{r}_{1}},\] diameter \[={{d}_{1}},\] and resistance \[{{R}_{1}}=\rho \frac{{{l}_{1}}}{{{A}_{1}}}\]

After stretching length \[={{l}_{2}},\] area of cross-section \[={{A}_{2}},\] radius \[={{r}_{2}},\] diameter \[={{d}_{2}}\] and resistance \[={{R}_{2}}=\rho \frac{{{l}_{2}}}{{{A}_{2}}}\]

Ratio of resistances before and after stretching                

\[\frac{{{R}_{\mathbf{1}}}}{{{R}_{\mathbf{2}}}}=\frac{{{l}_{\mathbf{1}}}}{{{l}_{\mathbf{2}}}}\times \frac{{{A}_{\mathbf{2}}}}{{{A}_{\mathbf{1}}}}={{\left( \frac{{{l}_{\mathbf{1}}}}{{{l}_{\mathbf{2}}}} \right)}^{\mathbf{2}}}={{\left( \frac{{{A}_{\mathbf{2}}}}{{{A}_{\mathbf{1}}}} \right)}^{\mathbf{2}}}={{\left( \frac{{{r}_{\mathbf{2}}}}{{{r}_{\mathbf{1}}}} \right)}^{\mathbf{4}}}={{\left( \frac{{{d}_{\mathbf{2}}}}{{{d}_{\mathbf{1}}}} \right)}^{\mathbf{4}}}\]

(1) If length is given then \[R\propto {{l}^{2}}\,\Rightarrow \,\frac{{{R}_{1}}}{{{R}_{2}}}={{\left( \frac{{{l}_{1}}}{{{l}_{2}}} \right)}^{2}}\]

(2) If radius is given then \[R\propto \frac{1}{{{r}^{4}}}\,\Rightarrow \,\frac{{{R}_{1}}}{{{R}_{2}}}={{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{4}}\]