A conductor of resistance $3 \Omega$ is stretched uniformly till its length is doubled. The wire is now bent in the form of an equilateral triangle. The effective resistance between the ends of any side of the triangle in ohms is

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Aicharya · Accepted Answer

Resistance $R \propto l^2$

$ \begin{aligned} \frac{R_1}{R_2} & =\left(\frac{l_1}{l_2}\right)^2 \\ \frac{3}{R_2} & =\left(\frac{l}{2 l}\right)^2 \\ \Rightarrow \quad R_2 & =12 \Omega \end{aligned} $ Stretched wire is bent in the form of an equilateral triangle. $\therefore$ Resistance of each side $=\frac{12}{3}=4 \Omega$ $ R_1=R_2=R_3=4 \Omega $ Equivalent resistance of $R_1$ and $R_2$ (in series) $ \begin{aligned} R^{\prime} & =R_1+R_2 \\ & =4+4=8 \Omega \end{aligned} $ Now, $R^{\prime}$ and $R_3$ are in parallel so total resistance $ \begin{aligned} R & =\frac{R^{\prime} R_3}{R^{\prime}+R_3}=\frac{8 \times 4}{8+4} \\ & =\frac{32}{12}=\frac{8}{3} \Omega \end{aligned} $

A conductor of resistance $3 \Omega$ is stretched uniformly till its length is doubled. The wire is now bent…

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