Question 162 of 380

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:

A$50.0 \Omega$B$10 \Omega$C$36.7 \Omega$D$26.7 \Omega$

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Answer: (D) $26.7 \Omega$

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
$$\begin{aligned} \mathrm{R}_{\mathrm{BC}} & =\frac{20 \times 10}{10+2 \times 10}+20-10+20-10 \\ & =\frac{80}{3}=26.7 \Omega \end{aligned}$$