Two wires P and Q are made of the same material and have the same cross-sectional area - AQA - A-Level Physics - Question 28 - 2022 - Paper 1

Using Hooke's Law, extension can be represented as: $\text{Extension} = \frac{\text{Force} \cdot \text{Length}}{\text{Area} \cdot \text{Young's Modulus}}$ For wire P, the extension is (\frac{F \cdot L}{A \cdot E} = x). For wire Q, substituting its parameters yields: $\text{Extension of Q} = \frac{2F \cdot 2L}{A \cdot E} = \frac{4F \cdot L}{A \cdot E} = 4x$ So this statement is also incorrect.

Strain is defined as the ratio of extension to original length: $\text{Strain} = \frac{\text{Extension}}{\text{Original Length}}$ Strain in P: $\text{Strain}_P = \frac{x}{L}$ Strain in Q: $\text{Strain}_Q = \frac{4x}{2L} = \frac{2x}{L}$ Thus, strain in Q is indeed double that in P, making this statement correct.

As established, the stress in P is ( \frac{F}{A} ) and the stress in Q is ( \frac{2F}{A} ). Thus, stress in P is indeed half of that in Q, confirming this statement is correct.