Two wires of the same material and length but the cross-sectional area in the ratio 1 ∶ 2 are used to suspend the same load. The extensions in them will be in the ratio -

Concept:

Stress (σ):

The tensile force per unit cross-sectional area of the rod is called stress.

\(\sigma =\frac{Force~(F)}{Area~(A)}\)

Strain (ϵ):

The change in length per unit length is called strain.

\(\epsilon=\frac{Change~in~length~(Δ l)}{Original~length~(l)}\)

The ratio of stress and strain is called Young’s modulus of elasticity (E).

\(E=\frac{stress}{strain}=\frac{\frac{F}{A}}{\frac{Δ l}{l}}=\frac{F~l}{A~Δ l}\)

∴ change in length (Δl) = \(\frac{Fl}{EA}\)

Where F is the force applied, A is the cross-sectional area, Δl is change in length and l is the original length.

Calculation:

It is given that the material of both wire is the same, so the Young’s modulus of elasticity will be the same for both the wires.

Second wire is double in cross section as compared to first wire.

⇒ Area of first wire = A, ⇒ Area of second wire = 2A

They are of the same length (l) and the same load (F) is applied on both the wires.

The extension or change in length is calculated as:

∴ extension in first wire (Δl₁) = \(\frac{Fl}{EA}\)

∴ extension in second wire (Δl₂) = \(\frac{Fl}{E (2A)}=\frac{Fl}{2E A}\)

So the ratio of extension of both the wire is given as 

\(⇒\frac{\Delta l_1}{\Delta l_2}=\frac{\frac{Fl}{E A}}{\frac{Fl}{2E A}}=\frac{2}{1}\)

So the wire with half of the cross-section area compared to the second wire will extend 2 times for the same load.

Extension or change in length is found in 2 : 1 ratio.