Related Formula
Young's modulus Y$Y$ is defined as:
Y = StressStrain = (F / A)/(Δ L / L) Δ L = (F L)/(A Y)$$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta L / L} \implies \Delta L = \frac{F L}{A Y}$$
Area of cross-section of wire with diameter d$d$ is:
A = (π d²)/(4) Δ L = (4 F L)/(π d² Y)$$A = \frac{\pi d^2}{4} \implies \Delta L = \frac{4 F L}{\pi d^2 Y}$$Core Logic
Since both wires are made of the same material (
$
Core Logic
Since both wires are made of the same material ($
Y_A = Y_B
) and stretched with the same force ($) and stretched with the same force ($F_A = F_B
):$):
$Δ L ∝ (L)/(d²)$\Delta L \propto \frac{L}{d^2}$Set up the ratio for wires A and B:
$
Set up the ratio for wires A and B:
$(Δ LA)/(Δ LB) = ( (LA)/(LB) ) × ( (dB)/(dA) )²$\frac{\Delta L_A}{\Delta L_B} = \left( \frac{L_A}{L_B} \right) \times \left( \frac{d_B}{d_A} \right)^2$Step 1: Substitute Given Ratios
Substitute the ratios
$
Step 1: Substitute Given Ratios
Substitute the ratios $
\frac{L_A}{L_B} = \frac{1}{3}
and$ and $\frac{d_A}{d_B} = 2 \implies \frac{d_B}{d_A} = \frac{1}{2}
:$:
$(Δ LA)/(Δ LB) = ( (1)/(3) ) × ( (1)/(2) )² = (1)/(3) × (1)/(4) = (1)/(12)$\frac{\Delta L_A}{\Delta L_B} = \left( \frac{1}{3} \right) \times \left( \frac{1}{2} \right)^2 = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}$Therefore, the ratio of their elongations is
$
Therefore, the ratio of their elongations is $
1:12
. Pattern Recognition
Sees: Same material, same stretching force. Shortcut: Elongation scales directly with length and inversely with the square of the diameter (radius). Thus, ratio is
$.
Pattern Recognition
Sees: Same material, same stretching force.
Shortcut: Elongation scales directly with length and inversely with the square of the diameter (radius). Thus, ratio is $
(1/3) / (2^2) = 1/12$.
Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
