Discussion Forum

A conducting wire has length L₁ and diameter d₁. After stretching the same wire length becomes L₂ and diameter d₂ . The ratio  of resistances before and after stretching is 

Answer:

Explanation:

Resistance of wire is given by R= $\rho\frac{l}{A}$

 where, l is the length of conductor and A is the cross-sectional area.According to the question,

  $R_{1}=\rho\frac{l_{1}}{A_{1}}$  ............(i)

 in the second condition,

 $R_{2}=\rho\frac{l_{2}}{A_{2}}$  ............(ii)

 On dividing  Eq.(i) by Eq.(ii) , we get

 $\frac{R_{1}}{R_{2}}=\frac{l_{1}}{A_{1}}\times\frac{A_{2}}{l_{2}}$

$\frac{R_{1}}{R_{2}}=\frac{A_{2}}{A_{1}}\times\frac{l_{1}}{l_{2}}$

$\frac{R_{1}}{R_{2}}=\frac{\pi(\frac{d_{2}}{2})^{2}}{\pi(\frac{d_{1}}{2})^{2}}\times\frac{l_{1}}{l_{2}}$

$\frac{R_{1}}{R_{2}}=\frac{({d_{2}})^{2}}{({d_{1}})^{2}}\times\frac{l_{1}}{l_{2}}$............(iii)

 Volume of wire  remains constant, Hence , V₁ = V₂

            A₁l₁ = A₂l₂

 $\Rightarrow$   $ \pi \left(\frac{d_{1}}{2}\right)^{2}l_{1}=\pi \left(\frac{d_{2}}{2}\right)^{2}l_{2}$

$\Rightarrow$   $  \frac{\pi d_{1}}{4}^{2}l_{1}=\frac{\pi d_{2}}{4}^{2}l_{2}$

$\frac{l_{1}}{l_{2}}=\frac{(d_{2})^{2}}{(d_{1})^{2}}$    ...............(iv)

Substituting the value of Eq.(iv) in Eq.(iii) , we get

$\frac{R_{1}}{R_{2}}=\frac{(d_{2})^{2}}{(d_{1})^{2}}\times\frac{(d_{2})^{2}}{(d_{1})^{2}}$

 $\frac{R_{1}}{R_{2}}=\frac{(d_{2})^{4}}{(d_{1})^{4}}$