Which of the following cases will have the greatest strain energy (F is the stretching force, A is the area of cross section and s is the strain)

The correct option is: F = 15 N, A = 2 cm², s = 10⁻³

To find which case has the greatest strain energy, we start with the formula for the strain energy ($U$) stored in a stretched wire:

$U=\frac{1}{2}\times \text{Stress}\times \text{Strain}\times \text{Volume}$

We know the definitions of stress and volume:

$\text{Stress}=\frac{F}{A}$

$\text{Volume}=A\times L$

where:
• $F$ is the stretching force,
• $A$ is the area of cross-section,
• $s$ is the strain, and
• $L$ is the original length of the wire.

Substituting these definitions back into the strain energy formula:

$U=\frac{1}{2}\times \left(\frac{F}{A}\right)\times s\times (A\times L)$

Simplifying the expression, the cross-sectional area $A$ cancels out:

$U=\frac{1}{2}\times F\times s\times L$

Assuming the length $L$ of the wire is constant across all cases, the strain energy is directly proportional to the product of the stretching force ($F$) and the strain ($s$):

$U\propto F\times s$

Now, let's calculate the product $F\times s$ for each given option to determine which one is the largest:

Case 1: F = 10 N, A = 1 cm², s = 10⁻³
$F\times s=10\times {10}^{-3}=0.010\text{N}$

Case 2: F = 15 N, A = 2 cm², s = 10⁻³
$F\times s=15\times {10}^{-3}=0.015\text{N}$

Case 3: F = 10 N, A = 1/2 cm², s = 10⁻⁴
$F\times s=10\times {10}^{-4}=0.001\text{N}$

Case 4: F = 5 N, A = 3 cm², s = 10⁻³
$F\times s=5\times {10}^{-3}=0.005\text{N}$

Comparing the values, Case 2 ($F\times s=0.015$) has the greatest value, and therefore it will have the greatest strain energy.