Question

A conductor of resistance $3\Omega$ is stretched uniformly till its length is doubled. The wire is now bent in the form of an equilateral triangle. The effective resistance between the ends of any side of the triangle in $\Omega$

• A. $\frac{9}{2}$
• B. $\frac{8}{3}$
• C. $2$
• D. $1$

Answer 1

The correct option is B $\frac{8}{3}$

If we double the length of the conductor$,$ then its cross$-$sectional area must become half of its original value (since$,$ $volume=area\times length$ is constant).

Therefore$,$ we can say that after stretching the new resistance is $R^{\prime }=\rho \frac{2l}{A/2}=4\rho \frac{1}{A}=4R_0=4\times 3\Omega =12\Omega$

Now$,$ the wire is bent such that each side has a resistance $4\Omega ,$ (since the resistance is divided uniformly).

Now from the diagram, we can see that between two vertices$.$ there is an arrangement of two $4\Omega$ resistors in series connected in parallel with another $4\Omega$ resistor$.$

The net resistance becomes$:-$ $\frac{8\times 4}{8+4}=\frac{8}{3}\Omega$

Hence,

option $\left(B\right)$ is correct answer.