Figure represents the extension ( l ) of a wire of length 1 meter, suspended from the ceiling of the room at one end with a load W connected to the other end. If the cross-sectional area of the wire is 10⁻⁵ m ^2 then the Young's modulus of the wire is __________ N/m ^2 .

Step-by-step solution

From the given graph, we can observe that for a load W = 20 N, the extension is l = 2 10⁻⁴ m. The formula for Young's modulus Y is: Y = Stress Strain = W / A l / L = W L A l Given: Length of the wire, L = 1 m Cross-sectional area, A = 10⁻⁵ m ^2 Substituting the values into the formula: Y = 20 1 10⁻⁵ 2 10⁻⁴ Y = 20 2 10⁻⁹ Y = 10 10^9 = 1.0 10¹⁰ N/m ^2 Answer: 1.0 10¹⁰