Question 151 of 380

A metal wire of resistance $3 \Omega$ is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle ${60}^{o}$ at the center, the equivalent resistance between these two points will be:

A$\dfrac{5}{3} \Omega$B$\dfrac{12}{5} \Omega$C$\dfrac{7}{2} \Omega$D$\dfrac{5}{2} \Omega$

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Answer: (A) $\dfrac{5}{3} \Omega$

We know, $R=\dfrac{\rho l}{A}=\dfrac{\rho {l}^{2}}{Al}=\dfrac{\rho {l}^{2}}{Volume}$
$R\propto {l}^{2}$ , if the volume is constant.
$∴$ New resistance $=4R$

${R}_{1}=\dfrac{4R}{360}\times 300=\dfrac{10R}{3}$
${R}_{2}=\dfrac{4R}{360}\times 60=\dfrac{2R}{3}$
$∴$ Equivalent resistance ${R}_{eq}=\dfrac{{R}_{1}{R}_{2}}{{R}_{1}+{R}_{2}}$
$=\dfrac{\dfrac{10R}{3}\times \dfrac{2R}{3}}{\dfrac{10R}{3}+\dfrac{2R}{3}}=\dfrac{5R}{9}$
$=\dfrac{5}{3}\Omega$