Question

A $12\,cm$ wire is given a shape of a right angled triangle $ABC$ having sides $3\,cm$, $4\, cm$ and $5 \,cm$ as shown in the $B$ figure.

The resistance between two ends $(AB, BC, CA)$ of the respective sides are measured one by one by a multi-meter. The resistances will be in the ratio

• A. 9 : 16 : 25
• B. 27 : 32 : 35
• C. 21 : 24 : 25
• D. 3 : 4 : 5

Answer 1

Answer: (B)

Let $\rho$ and $A$ be resistivity and area of cross-section of the wire respectively.
The wire is bent in the form of right triangle as shown in adjacent figure.

Resistance of side $AB$ is $R_{1}=\frac{3 \rho}{A}$
Resistance of side $BC$ is $R_{2}=\frac{4 \rho}{A}$
Resistance of side $AC$ is $R_{3}=\frac{5 \rho}{A}$
$\therefore$ The resistance between the ends $A$ and $B$ is
$R_{A B}=\frac{R_{1}\left(R_{2}+R_{3}\right)}{R_{1}+R_{2}+R_{3}}=\frac{\frac{3 \rho}{A}\left(\frac{4 \rho}{A}+\frac{5 \rho}{A}\right)}{\frac{3 \rho}{A}+\left(\frac{4 \rho}{A}+\frac{5 \rho}{A}\right)}=\frac{27}{12} \frac{\rho}{A}$
The resistance between the ends $B$ and $C$ is
$R_{B C}=\frac{R_{2}\left(R_{1}+R_{3}\right)}{R_{2}+R_{1}+R_{3}}=\frac{\frac{4 \rho}{A}\left(\frac{3 \rho}{A}+\frac{5 \rho}{A}\right)}{\frac{4 p}{A}+\frac{3 p}{A}+\frac{5 p}{A}}=\frac{32}{12} \frac{p}{A}$
The resistance between the ends $A$ and $C$ is
$R_{A C}=\frac{R_{3}\left(R_{1}+R_{2}\right)}{R_{3}+R_{1}+R_{2}}$
$=\frac{\frac{5 \rho}{A}\left(\frac{3 \rho}{A}+\frac{4 \rho}{A}\right)}{\frac{5 p}{A}+\frac{3 \rho}{A}+\frac{4 \rho}{A}}=\frac{35}{12} \frac{\rho}{A}$
$\therefore R_{A B}: R_{B C}: R_{A C}$
$=\frac{27}{12}: \frac{32}{12}: \frac{35}{12}=27: 32: 35$