Physics - Current Electricity Question with Solution | TestHub

Resistance of side $AB=R_1=\frac{3\rho }{A}$ Resistance of side $BC=R_2=\frac{4\rho }{A}$ Resistance of side $\mathrm{AC}={\mathrm{P}}_3=\frac{5\rho }{\mathrm{A}}$ (Where $A=$ Area of cross-section and $\rho =$ resistivity of wire)
$\therefore$ The resistance between $\mathrm{A}$ and $\mathrm{B}$ is $$\begin{array}{rl}& {\mathrm{R}}_{\mathrm{AB}}=\frac{R_1\left(R_2+R_3\right)}{R_1+R_2+R_3}=\frac{\frac{3\rho }{A}\left(\frac{4\rho }{A}+\frac{5\rho }{A}\right)}{\frac{3\rho }{A}+\frac{4\rho }{A}+\frac{5\rho }{A}}\\ & =\frac{27\rho }{12A}\\ & \text{ Similarly }{R}_{BC}=\frac{R_2\left(R_1+R_3\right)}{R_1+R_2+R_3}\\ & \frac{\frac{4\rho }{A}\left(\frac{3\rho }{A}+\frac{5\rho }{A}\right)}{\frac{3\rho }{A}+\frac{4\rho }{A}+\frac{5\rho }{A}}=\frac{32}{12}\frac{\rho }{A}\\ & \text{ And }{R}_{AC}=\frac{R_3\left(R_1+R_2\right)}{R_1+R_2+R_3}\\ & =\frac{\frac{5\rho }{A}\left(\frac{3\rho }{A}+\frac{3\rho }{A}\right)}{\frac{5\rho }{A}+\frac{3\rho }{A}+\frac{4\rho }{A}}=\frac{35}{12}\frac{\rho }{A}\\ & \therefore R_{AB}:R_{BC}:R_{AC}=\frac{27}{12}:\frac{32}{12}:\frac{35}{12}\\ & =27:32:35\\ & \end{array}$$