A letter ${ }^{\prime} \mathrm{A}^{\prime}$ is constructed of a uniform wire with resistance $1.0 \Omega$…
- $50.0 \Omega$
- $10 \Omega$
- $36.7 \Omega$
- $26.7 \Omega$
Solution

For $\mathrm{ADE} \frac{1}{\mathrm{R}^{\prime}}=\frac{1}{2 \mathrm{x}}+\frac{1}{10}$ or $\quad R^{\prime}=\frac{20 x}{10+2 x}$ $ \mathrm{R}_{\mathrm{BC}}=\frac{20 \mathrm{x}}{10+2 \mathrm{x}}+20-\mathrm{x}+20-\mathrm{x} $ or $\frac{20 \mathrm{x}}{10+2 \mathrm{x}}+40=2 \mathrm{x}$ Solving we get $ \mathrm{x}=10 \Omega $ Putting the value of $x=10 \Omega$ in equation (i) We get $ \begin{aligned} \mathrm{R}_{\mathrm{BC}} & =\frac{20 \times 10}{10+2 \times 10}+20-10+20-10 \\ & =\frac{80}{3}=26.7 \Omega \end{aligned} $
Asked in: JEE Main 2013 (09 Apr Online)