Resistance of a wire is $8 \Omega$. It is drawn in such a way that it experiences a longitudinal strain of…

Resistance of a wire is $8 \Omega$. It is drawn in such a way that it experiences a longitudinal strain of $400 \%$. The final resistance of the wire is
  1. $100 \Omega$
  2. $200 \Omega$
  3. $300 \Omega$
  4. $400 \Omega$

Solution

$\mathrm{R}_1=8 \Omega, \varepsilon=\frac{\delta 1}{\mathrm{l}_1}=4 \Rightarrow \delta \mathrm{l}=4 \mathrm{l}_1$ $\therefore \quad l_2=l_1+81=l_1+4 l_1=5 l_1$
Now, resistance of a wire is given by $\begin{aligned} & \frac{\mathrm{R}_2}{\mathrm{R}_1}=\left(\frac{\mathrm{l}_2}{\mathrm{l}_1}\right)^2=\left(\frac{5 \mathrm{l}_1}{\mathrm{l}_1}\right)^2=25 \\ & \therefore \quad \mathrm{R}_2=25 \mathrm{R}_1=25 \times 8=200 \Omega \end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 2)

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