Question 1 of 13intermediate🔧 ApplyNumerical3 marks

A steel wire of length 4.7 m and cross-sectional area 3.0×103.0 \times 10⁻⁵ m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0×104.0 \times 10⁻⁵ m² under a given load. What is the ratio of the Young's modulus of steel to that of copper?

Correct Answer

The ratio of Young's modulus of steel to that of copper is 1.79 (or approximately 1.8).

Exercise: EXERCISES | Q: 8.1
For More Understanding

Explanation

This question tests the understanding of Young's modulus and its relationship with elongation. From the context, the formula for elongation is ΔL = (F/A)L/Y = FL/(AY). When two wires stretch by the same amount under the same load, we can equate their elongations and solve for the ratio of Young's moduli. The key insight is that equal elongation under equal load means the product A×Y/L must be the same for both wires.

Solution Steps

  1. Step 1: For a wire under tension, the elongation is given by ΔL = FL/(AY), where F is the applied force (load), L is length, A is cross-sectional area, and Y is Young's modulus.

  2. Step 2: Since both wires stretch by the same amount under the same load, we have: ΔL_s = ΔL_c, which gives: (F L_s)/(A_s Y_s) = (F L_c)/(A_c Y_c)

  3. Step 3: Canceling F from both sides and rearranging: Y_s/Y_c = (L_s × A_c)/(A_s × L_c)

  4. Step 4: Substituting the given values: Y_s/Y_c = (4.7×4.04.7 \times 4.0 × 10⁻⁵)/(3.0×103.0 \times 10⁻⁵ × 3.5)

  5. Step 5: Simplifying: Y_s/Y_c = (4.7×4.04.7 \times 4.0)/(3.0×3.53.0 \times 3.5) = 18.8/10.5 = 1.79

Next