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

To find the density and Young modulus of wire Y, let's analyze the situation:

Given Data

• Diameter of wire X: $d$
• Density of wire X: $ho$
• Young modulus of wire X: $E$
• Diameter of wire Y: $2d$ (double the diameter of X)
• Wire Y has the same mass and length as wire X.

Step 1: Calculate the Cross-Sectional Areas

The cross-sectional area $A$ of a wire can be calculated using the formula for the area of a circle: A = rac{ ext{diameter}^2 imes ext{π}}{4}

• For wire X: A_X = rac{d^2 imes ext{π}}{4}
• For wire Y (diameter $2d$): A_Y = rac{(2d)^2 imes ext{π}}{4} = rac{4d^2 imes ext{π}}{4} = d^2 imes ext{π}

Step 2: Analyze the Mass and Length

As both wires have the same mass $m$ and length $L$, we can express the mass based on the volume (Volume = Area × Length): V_X = A_X imes L = rac{d^2 imes ext{π}}{4} imes L \ V_Y = A_Y imes L = d^2 imes ext{π} imes L

Given that $m_X = ho imes V_X$ and $m_Y = ho_Y imes V_Y$, since $m_X = m_Y$, we have:

ho imes V_X = ho_Y imes V_Y $$ This leads us to:

ho imes rac{d^2 imes ext{π}}{4} imes L = ho_Y imes d^2 imes ext{π} imes L $$

Step 3: Solve for Density of Wire Y

Cancelling common terms, we get:

ho imes rac{1}{4} = ho_Y \\ ext{Thus: } ho_Y = rac{ ho}{4} $$ ### Step 4: Young's Modulus Relationship The Young's modulus of a material is described by: $$ Y = rac{F imes L}{A imes ext{extension}} $$ Since both wires are subjected to the same load, have the same length and extension, we can write: $$ rac{E}{E_Y} = rac{A_Y}{A_X} $$ This simplifies to: $$ rac{E}{E_Y} = rac{d^2 imes ext{π}}{rac{d^2 imes ext{π}}{4}} = 4 $$ ### Step 5: Solve for Young's Modulus of Wire Y Thus: $$ E_Y = rac{E}{4} $$ ### Final Answers - Density of wire Y: $\rho_Y = \frac{\rho}{4}$ - Young modulus of wire Y: $E_Y = \frac{E}{4}$