A sample of wire has a Young modulus $E$ - AQA - A-Level Physics - Question 24 - 2017 - Paper 1

The area for a circular wire is (A = \frac{\pi d^2}{4}). For the first sample with diameter (d_1), the area is (A_1 = \frac{\pi d_1^2}{4}). For the second sample with half the diameter (d_2 = \frac{1}{2}d_1), the area becomes:

$A_2 = \frac{\pi \left( \frac{1}{2}d_1 \right)^2}{4} = \frac{\pi d_1^2}{16} = \frac{1}{4}A_1$

Next, the second sample's length is three times the first, so (L_2 = 3L_1).

The stress for the first sample is:

$\text{Stress}_1 = \frac{F}{A_1}$

And for the second sample:

$\text{Stress}_2 = \frac{F}{A_2} = \frac{F}{\left(\frac{1}{4}A_1\right)} = 4 \cdot \frac{F}{A_1} = 4 \cdot \text{Stress}_1$

The strain in the first sample is given by (\frac{\Delta L_1}{L_1}) while in the second it is (\frac{\Delta L_2}{L_2}), where (\Delta L_2 = \Delta L_1) remains the same.

Thus, considering both samples together:

$E_2 = \frac{\text{Stress}_2}{\text{Strain}_2} = \frac{4 \cdot \text{Stress}_1}{\frac{\Delta L_1}{3L_1}} = 12 \cdot \frac{\text{Stress}_1}{\frac{\Delta L_1}{L_1}} = 12E$

Therefore, the Young modulus of the second sample of wire in terms of $E$ is (12E).