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(B) The letter $A$ consists of two side arms $AE$ and $ED$ of length $20\,cm$ each,and a cross-piece $BC$ of length $10\,cm$. The cross-piece $BC$ div

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$26.7$

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(B) The letter $A$ consists of two side arms $AE$ and $ED$ of length $20\,cm$ each,and a cross-piece $BC$ of length $10\,cm$. The cross-piece $BC$ divides the side arms into segments $AB$,$BE$,$EC$,and $CD$.Given the apex angle is $60^\circ$,the triangle formed by the top part is equilateral. Thus,$BE = EC = BC = 10\,cm$.Since the total length of each side arm is $20\,cm$,we have $AB = AE - BE = 20 - 10 = 10\,cm$ and $CD = ED - EC = 20 - 10 = 10\,cm$.All segments $(AB, BE, EC, CD, BC)$ have a length of $10\,cm$. With a resistance of $1.0\,\Omega/cm$,the resistance of each segment is $R = 10\,cm 	imes 1.0\,\Omega/cm = 10\,\Omega$.The circuit can be simplified: $AB$ is in series with the parallel combination of the upper path $(BE + EC = 10 + 10 = 20\,\Omega)$ and the cross-piece $(BC = 10\,\Omega)$,which is then in series with $CD$.The equivalent resistance of the parallel part is $R_p = \frac{20 	imes 10}{20 + 10} = \frac{200}{30} = 6.67\,\Omega$.The total resistance between $A$ and $D$ is $R_{eq} = R_{AB} + R_p + R_{CD} = 10 + 6.67 + 10 = 26.67\,\Omega \approx 26.7\,\Omega$.

$A$ letter $A$ is constructed of a uniform wire with resistance $1.0\,\Omega/cm$. The sides of the letter are $20\,cm$ and the cross-piece in the middle is $10\,cm$ long. The apex angle is $60^\circ$. The resistance between the ends $A$ and $D$ is ............. $\Omega$.

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