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Two rods of equal length and equal cross-sectional area are joined end to end to form a composite rod and stretched by a force F applied at its ends. Given l1 = l2 = 1 m, A = 1 m², F = 2 N, Young's modulus y1 = 2*10¹¹ N/m² and y2 = 3*10¹¹ N/m², find the total (net) elongation of the composite rod. If the elongation is x * 10⁻¹¹ m, what is x (rounded to two decimals)?

1. 1.00
2. 1.67
3. 2.40
4. 0.67

Correct answer: 1.67

Solution

For a rod, elongation = F*l/(A*y). Since the rods are in series, the same force F acts on each and the total elongation is the sum. With l = 1 m, A = 1 m², F = 2 N: first rod gives 2/(2*10¹¹) = 1.0*10⁻¹¹ m, second gives 2/(3*10¹¹) = 0.667*10⁻¹¹ m, summing to about 1.67*10⁻¹¹ m.

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