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A letter ${ }^{\prime} \mathrm{A}^{\prime}$ is constructed of a uniform wire with resistance $1.0 \Omega$ per $\mathrm{cm}$, The sides of the letter are $20 \mathrm{~cm}$ and the cross piece in the middle is $10 \mathrm{~cm}$ long. The apex angle is 60 . The resistance between the ends of the legs is close to:
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Verified Answer
The correct answer is:
$26.7 \Omega$
$26.7 \Omega$
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
$$
\begin{aligned}
\mathrm{R}_{\mathrm{BC}} & =\frac{20 \times 10}{10+2 \times 10}+20-10+20-10 \\
& =\frac{80}{3}=26.7 \Omega
\end{aligned}
$$
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