Two wires X and Y have the same extension for the same load - AQA - A-Level Physics - Question 26 - 2021 - Paper 1

To solve this, we first analyze the conditions given:

1. Volume and Mass Consistency: Since wires X and Y have the same mass and length, their volumes must also be equivalent. The volume of wire X can be expressed as:

   $$$$

ho imes L}{E} \quad (1) $$

Where:

• $A_X$ is the cross-sectional area of wire X, given by: $A_X = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4} \quad (2)$

Therefore, substituting (2) into (1):

$V_X = \frac{\pi d^2}{4} \times L$

Meanwhile, for wire Y with diameter 2d:

$A_Y = \pi \left(\frac{2d}{2}\right)^2 = \pi d^2 \quad (3)$

Thus:

$V_Y = \pi d^2 \times L$

2. Density Calculation: The density of wire Y ($\rho_Y$) can be found as follows:

   Maintaining the same mass for both wires, we equate their volumes:

   $\rho_Y = \frac{\rho}{4} \quad (4)$

3. Young's Modulus: The Young modulus of wire Y ($E_Y$) can be determined given that both wires are subjected to the same extension under the same load:

   Using the relation for Young modulus:

   $E = \frac{FL}{A \Delta L}$

   We can assert:

   $E_Y = 4E \quad (5)$

So, the final results for density and Young's modulus are:

• Density of wire Y: $\frac{\rho}{4}$
• Young's modulus of wire Y: $4E$.