A $12 \mathrm{~cm}$ wire is given a shape of a right angled triangle $A B C$ having sides $3 \mathrm{~cm}$, $4 \mathrm{~cm}$ and $5 \mathrm{~cm}$ as shown in the figure. The resistance between two ends (AB, $B C, C A)$ of the respective sides are measured one by one by a multi-meter. The resistance will be in the ratio:

• A

  $9: 16: 25$
• Correct answer
  B

  $27: 32: 35$
• C

  $21: 24: 25$
• D

  $3: 4: 5$

Solution

Resistance of side $A B=R_1=\frac{3 \rho}{A}$ Resistance of side $B C=R_2=\frac{4 \rho}{A}$ 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 $$ \begin{aligned} & \mathrm{R}_{\mathrm{AB}}=\frac{R_1\left(R_2+R_3\right)}{R_1+R_2+R_3}=\frac{\frac{3 \rho}{A}\left(\frac{4 \rho}{A}+\frac{5 \rho}{A}\right)}{\frac{3 \rho}{A}+\frac{4 \rho}{A}+\frac{5 \rho}{A}} \\ & =\frac{27 \rho}{12 A} \\ & \text { Similarly } R_{B C}=\frac{R_2\left(R_1+R_3\right)}{R_1+R_2+R_3} \\ & \frac{\frac{4 \rho}{A}\left(\frac{3 \rho}{A}+\frac{5 \rho}{A}\right)}{\frac{3 \rho}{A}+\frac{4 \rho}{A}+\frac{5 \rho}{A}}=\frac{32}{12} \frac{\rho}{A} \\ & \text { And } R_{A C}=\frac{R_3\left(R_1+R_2\right)}{R_1+R_2+R_3} \\ & =\frac{\frac{5 \rho}{A}\left(\frac{3 \rho}{A}+\frac{3 \rho}{A}\right)}{\frac{5 \rho}{A}+\frac{3 \rho}{A}+\frac{4 \rho}{A}}=\frac{35}{12} \frac{\rho}{A} \\ & \therefore R_{A B}: R_{B C}: R_{A C}=\frac{27}{12}: \frac{32}{12}: \frac{35}{12} \\ & =27: 32: 35 \\ & \end{aligned} $$

Hence, the correct answer is option: 'B'.