Question

33.A conductor of resistance 3ohm is stretched uniformly till its length is doubled.The wire now is bent in the form of an equiletral triangle.The effective resistance between the ends of any side of the triangle in ohms is

Answer 1

If you double the length of the conductor, then its cross-sectional area must become half of its original value, since $\rm volume = area \times length$ is constant. 
Therefore, we can say that $R'=\rho\frac{2l}{A/2} = 4 \rho \frac{l}{A}=4R_0=4 \times 3 \Omega=12 \Omega$
Now your wire is bent such that each side has a resistance 4 ohms, since the resistance is divided uniformly. 
Now if you make a diagram, you'll see that between vertices, there is an arrangement of two 4 ohms in series connected in parallel with another 4 ohm resistor. The net resistance becomes

8*4/(8+4) = 8/3 ohms.