JEE Main Mechanical Properties of Solids: related Physics practice
Revise Mechanical Properties of Solids with the original worked examples below. The practice bank below covers broader Physics topics; a dedicated Mechanical Properties of Solids lesson is not available yet.
Stress describes force per area; strain describes relative deformation. In the linear elastic region, their ratio is an elastic modulus. Check that the load and geometry suit the selected modulus before using Hooke’s law.
Key ideas and when to use them
Tensile stress: $\sigma=F/A$.
Longitudinal strain: $\epsilon=\Delta L/L$.
Young’s modulus: $Y=\sigma/\epsilon=FL/(A\Delta L)$ in the linear elastic region.
Common mistakes
Strain has no unit; stress and elastic modulus have units of pascals.
Use the cross-sectional area, not the wire’s curved surface area.
Three worked examples
Example 1: A 200 N tensile load acts on a cross-section of 2 × 10⁻⁴ m². Find the stress.
Divide force by cross-sectional area.
$\sigma=200/(2\times10^{-4})$.
1 × 10⁶ Pa
Example 2: A 2 m wire extends by 0.20 mm. Find its longitudinal strain.
Convert the extension to $0.20\times10^{-3}$ m.
$\epsilon=(0.20\times10^{-3})/2$.
1 × 10⁻⁴
Example 3: A 1 m wire has area 10⁻⁶ m² and Young’s modulus 2 × 10¹¹ Pa. Find its extension under 20 N, within its linear elastic region.
alleryStart learning →