The Young's modulus of a steel wire of length 6 m and cross-sectional area $$3 \text{ mm}^2$$, is $$2 \times 10^{11} \text{ N/m}^2$$. The wire is suspended from its support on a given planet. A block of mass 4 kg is attached to the free end of the wire. The acceleration due to gravity on the planet is $$\frac{1}{4}$$ of its value on the earth. The elongation of wire is (Take $$g$$ on the earth $$= 10 \text{ m/s}^2$$):

We are given a steel wire characterized by length $$L = 6$$ m, cross-sectional area $$A = 3 \text{ mm}^2 = 3 \times 10^{-6} \text{ m}^2$$, Young’s modulus $$Y = 2 \times 10^{11} \text{ N/m}^2$$, and mass $$m = 4$$ kg.

Since the gravitational acceleration on the planet is one-fourth of that on Earth, we have $$g' = \frac{g}{4} = \frac{10}{4} = 2.5 \text{ m/s}^2$$, and therefore the force acting on the wire due to the hanging mass is $$F = mg' = 4 \times 2.5 = 10 \text{ N}$$.

By definition of Young’s modulus, the elongation $$\Delta L$$ of the wire under tension is given by $$\Delta L = \frac{FL}{AY}$$, so substituting $$F = 10$$ N, $$L = 6$$ m, $$A = 3 \times 10^{-6}$$ m^2, and $$Y = 2 \times 10^{11}$$ N/m^2 yields $$\Delta L = \frac{10 \times 6}{3 \times 10^{-6} \times 2 \times 10^{11}} = \frac{60}{6 \times 10^{5}} = 1 \times 10^{-4} \text{ m} = 0.1 \text{ mm}$$.

Therefore, the elongation of the wire is 0.1 mm, which corresponds to Option C.