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A letter'A' 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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The correct answer is:
$26.7 \Omega$
For ADE $\frac{1}{R^{\prime}}=\frac{1}{2 x}+\frac{1}{10} \quad$ or $\quad R^{\prime}=\frac{20 x}{10+2 x}$
$\mathrm{R}_{\mathrm{BC}}=\frac{20 \mathrm{x}}{10+2 \mathrm{x}}+20-\mathrm{x}+20-\mathrm{x} \quad \ldots$(i)
or $\quad \frac{20 \mathrm{x}}{10+2 \mathrm{x}}+40=2 \mathrm{x}$
Solving we get $\mathrm{x}=10 \Omega$
Putting the value of $x=10 \Omega$ in equation (i) We get $\mathrm{R}_{\mathrm{BC}}=\frac{20 \times 10}{10+2 \times 10}+20-10+20-10$
$=\frac{80}{3}=26.7 \Omega$
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