The Young's modulus of a steel wire of length $6 \mathrm{~m}$ and cross-sectional area $3 \mathrm{~mm}^{2}$, is $2 \times 10^{11}~\mathrm{N} / \mathrm{m}^{2}$. The wire is suspended from its support on a given planet. A block of mass $4 \mathrm{~kg}$ 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 $=10 \mathrm{~m} / \mathrm{s}^{2}$) :

The elongation of the wire can be calculated using the formula for stress and strain. The stress in the wire is given by: <br/><br/>$\sigma = \frac{mg}{A}$ <br/><br/>where m is the mass of the block (4 kg), g is the acceleration due to gravity on the planet (1/4 of its value on the earth, or 2.5 m/s²), and A is the cross-sectional area of the wire (3 mm²). <br/><br/>The strain in the wire is given by: <br/><br/>$\epsilon = \frac{\Delta L}{L}$ <br/><br/>where ΔL is the elongation of the wire and L is the original length of the wire (6 m). <br/><br/>Using Hooke's law, which states that stress is proportional to strain, we can find the elongation of the wire: <br/><br/>$\sigma = Y\epsilon$ <br/><br/>where Y is the Young's modulus of the wire (2 $\times$ 10¹¹ N/m²). <br/><br/>Combining the above equations, we can find the elongation of the wire: <br/><br/>$$\epsilon = \frac{\sigma}{Y} = \frac{mg}{A Y} = \frac{4 \times 2.5}{3 \times 10^{-6} \times 2 \times 10^{11}} = \frac{5}{3 \times 10^{-6} \times 10^{11}} = \frac{5}{3 \times 10^{5}}$$ <br/><br/>$\therefore$ $\frac{\Delta L}{L}$ $= \frac{5}{3 \times 10^{5}}$ <br/><br/>$\Rightarrow$ $\Delta L = {{5 \times 6} \over {3 \times {{10}^5}}}$ = ${1 \over {{{10}^4}}}$ = 0.1 mm <br/><br/>So, the elongation of the wire is 0.1 mm.