A letter $'A'$ is constructed of a uniform wire with resistance $1.0\,\Omega $ per $cm$ , The sides of the letter are $20\, cm$ and the cross piece in the middle is $10\, cm$ long. The apex angle is $60$ . The resistance between the ends of the legs is close to ................ $\Omega$
JEE MAIN 2013, Diffcult
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For $ADE$ $\frac{1}{\mathrm{R}^{\prime}}=\frac{1}{2 \mathrm{x}}+\frac{1}{10}$
or $\quad \mathrm{R}^{\prime}=\frac{20 \mathrm{x}}{10+2 \mathrm{x}}$
$\mathrm{R}_{\mathrm{BC}}=\frac{20 \mathrm{x}}{10+2 \mathrm{x}}+20-\mathrm{x}+20-\mathrm{x}$ ....$(i)$
or $\frac{20 x}{10+2 x}+40=2 x$
Solving we get
$x=10\, \Omega$
Putting the value of $x=10\, \Omega$ in equation $( i )$ We get
$ R_{B C} =\frac{20 \times 10}{10+2 \times 10}+20-10+20-10 $
$=\frac{80}{3}=26.7\, \Omega $
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