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

We know,
$\mathrm{R}=\rho \frac{\ell }{\mathrm{A}}$
where$\rho$ is resistivity
$\mathcal{l}$ is length
A is area of cross-section
$\therefore$ For a wire, when bent to form the triangle
ABC, resistance of each side is proportional
to length$\mathrm{R}\propto \ell$ .
$\Rightarrow R_{AB}=3k,R_{BC}=4k,R_{CA}=5k$
Let k be a constant.
When resistance is measured between the ends using a multimeter: For ends A and B:
The sides AC and CB are in series and these sides (AC and BC) together are in parallel with AB.
$\begin{array}{l}\therefore R_{\text{eff }}=\frac{R_{AB}\times \left(R_{AC}+R_{CB}\right)}{R_{AB}+R_{AC}+R_{CB}}\\ =\frac{3k\times \left(9k\right)}{12k}\\ =\frac{27}{12}k\end{array}$
Similaly, for B and C:
$\begin{array}{l}{R}_{\text{eff }}=\frac{R_{BC}\times \left(R_{CA}+R_{AB}\right)}{R_{AB}+R_{AC}+R_{BC}}\\ =\frac{4k\times \left(8k\right)}{12k}=\frac{32}{12}k\end{array}$
Similaly, for C and A:
$\begin{array}{l}{R}_{\text{eff }}=\frac{R_{CA}\times \left(R_{AB}+R_{BC}\right)}{R_{AB}+R_{BC}+R_{CA}}\\ =\frac{5k\times \left(7k\right)}{12k}=\frac{35}{12}k\end{array}$
$\therefore$Ratio of these resistances is 27 : 32 : 35.