The Young's modulus of a steel wire of length $$6$$ m and cross-sectional area $$3$$ mm$$^2$$, is $$2 \times 10^{11}$$ 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 $$= 10$$ m/s$$^2$$):

We are given: length $$L = 6$$ m, cross-sectional area $$A = 3 \text{ mm}^2 = 3 \times 10^{-6} \text{ m}^2$$, Young's modulus $$Y = 2 \times 10^{11} \text{ N/m}^2$$, mass $$m = 4$$ kg, and gravity on the planet $$g' = g/4 = 10/4 = 2.5 \text{ m/s}^2$$.

The force on the wire is $$F = mg' = 4 \times 2.5 = 10$$ N. Using the elongation formula $$\Delta L = \frac{FL}{AY}$$, we get $$\Delta L = \frac{10 \times 6}{3 \times 10^{-6} \times 2 \times 10^{11}} = \frac{60}{6 \times 10^{5}} = 1 \times 10^{-4} \text{ m} = 0.1 \text{ mm}$$.

The correct answer is Option C: $$0.1$$ mm.