A load of 10 kg is suspended by a metal wire 3 m long and having a cross-sectional area 4 mm^{2}. Find (a) the stress (b) the strain and (c) the elongation. Young modulus of the metal is 2.0 × 10^{11} N m^{−2}.

#### Solution

Given:

Mass of the load (m) = 10 kg

Length of wire (L) = 3 m

Area of cross-section of the wire (A) = 4 mm^{2} = 4.0 × 10^{−6} m^{2}

Young's modulus of the metal Y = 2.0 × 10^{11} N m^{−2}

(a) Stress = F/A

F = mg

= \[10 \times 10\] = 100 N (g = 10 m/s^{2})

\[\therefore \frac{F}{A} = \frac{100}{4 \times {10}^{- 6}}\]

\[ = 2 . 5 \times {10}^7 \text{ N/ m }^2\]

(b) Strain = \[\frac{\Delta L}{L}\]

Or,

\[\text{ Strain } = \frac{\text{ Stress }}{Y}\]

\[\text{ Strain }= \frac{2 . 5 \times {10}^7}{2 \times {10}^{11}}\]

\[ = 1 . 25 \times {10}^{- 4} \text{ N/ m}^2\]

(c) Let the elongation in the wire be \[∆ L\] .

\[\text { Strain} = \frac{\Delta L}{L}\]

\[ \Rightarrow \Delta L = \left( \text{ Strain } \right) \times L\]

\[ = 1 . 25 \times {10}^{- 4} \times 3\]

\[ = 3 . 75 \times {10}^{- 4} \text{ m}\]