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(B) The total length of the wire corresponds to a resistance of $24\,\Omega$. Since the wire is bent into an equilateral triangle with a tail,the tota

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$10$

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(B) The total length of the wire corresponds to a resistance of $24\,\Omega$. Since the wire is bent into an equilateral triangle with a tail,the total perimeter of the triangle is $3 	imes 5\,cm = 15\,cm$,and the tail is $5\,cm$ long. Total length = $20\,cm$. Resistance per unit length = $\frac{24\,\Omega}{20\,cm} = 1.2\,\Omega/cm$.Each side of the triangle has a resistance of $5\,cm 	imes 1.2\,\Omega/cm = 6\,\Omega$. The tail has a resistance of $5\,cm 	imes 1.2\,\Omega/cm = 6\,\Omega$.The triangle part between the two nodes has two branches in parallel: one branch is a single side $(6\,\Omega)$ and the other branch consists of two sides in series $(6\,\Omega + 6\,\Omega = 12\,\Omega)$.The equivalent resistance of this parallel combination is $R_p = \frac{6 	imes 12}{6 + 12} = \frac{72}{18} = 4\,\Omega$.This $4\,\Omega$ is in series with the tail resistance of $6\,\Omega$.Therefore,the total effective resistance between $A$ and $B$ is $R_{eq} = 4\,\Omega + 6\,\Omega = 10\,\Omega$.

$A$ wire has a resistance of $24\,\Omega$ and is bent into the shape shown in the figure. The effective resistance between $A$ and $B$ is .............. $\Omega$.

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