Question

A $12cm$ wire is given a shape of a right angled triangle $ABC$ having sides $3cm$, $4cm$ and $5cm$ 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
$R_{AB}=\frac{R_1\left(R_2+R_3\right)}{R_1+R_2+R_3}=\frac{\frac{3\rho }{A}\left(\frac{4\rho }{A}+\frac{5\rho }{A}\right)}{\frac{3\rho }{A}+\left(\frac{4\rho }{A}+\frac{5\rho }{A}\right)}=\frac{27}{12}\frac{\rho }{A}$
The resistance between the ends $B$ and $C$ is
$R_{BC}=\frac{R_2\left(R_1+R_3\right)}{R_2+R_1+R_3}=\frac{\frac{4\rho }{A}\left(\frac{3\rho }{A}+\frac{5\rho }{A}\right)}{\frac{4p}{A}+\frac{3p}{A}+\frac{5p}{A}}=\frac{32}{12}\frac{p}{A}$
The resistance between the ends $A$ and $C$ is
$R_{AC}=\frac{R_3\left(R_1+R_2\right)}{R_3+R_1+R_2}$
$=\frac{\frac{5\rho }{A}\left(\frac{3\rho }{A}+\frac{4\rho }{A}\right)}{\frac{5p}{A}+\frac{3\rho }{A}+\frac{4\rho }{A}}=\frac{35}{12}\frac{\rho }{A}$
$\therefore R_{AB}:R_{BC}:R_{AC}$
$=\frac{27}{12}:\frac{32}{12}:\frac{35}{12}=27:32:35$