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, the extension ( \Delta L ) can be calculated as:

$\Delta L = \frac{FL}{AY}$

For wire P: $\Delta L_P = \frac{F \cdot L}{A \cdot Y} = x$

For wire Q, substituting the lengths and forces: $\Delta L_Q = \frac{2F \cdot 2L}{A \cdot Y} = \frac{4FL}{AY} = 4x$

Therefore, the extension of Q is not 2x, making this statement incorrect.

Strain is defined as the ratio of extension to original length:

$\text{Strain} = \frac{\Delta L}{L}$

For wire P: $\text{Strain}_P = \frac{x}{L}$

For wire Q: $\text{Strain}_Q = \frac{4x}{2L} = \frac{2x}{L}$

This shows that the strain of Q is indeed double the strain of P, therefore this statement is correct.

From our earlier calculations:

• Stress in P: ( \text{Stress}_P = \frac{F}{A} )
• Stress in Q: ( \text{Stress}_Q = \frac{2F}{A} )

Thus, ( \text{Stress}_P = \frac{1}{2} \text{Stress}_Q ), indicating that the stress for P is half that of Q. This statement is correct.