Question

The elastic limit of steel is 8 × 10⁸ N m⁻² and its Young modulus 2 × 10¹¹ N m⁻². Find the maximum elongation of a half-metre steel wire that can be given without exceeding the elastic limit.

 

Answer 1

Given : 

\[\text{ Elastic limit of steel } \frac{F}{A} = 8 \times {10}^5 \text{ N/ m}^2 \]
\[\text{ Young's modulus of steel Y }= 2 \times {10}^{11} \text{ N/ m}^2 \]
\[\text{ Length of steel wire L } = \frac{1}{2}\text{ m = 0 . 5 m }\]

The elastic limit of steel indicates the maximum pressure that steel can bear.
Let the maximum elongation of steel wire be \[∆ L\] .

\[Y = \frac{F}{A} \frac{L}{∆ L}\]
\[ \Rightarrow ∆ L = \frac{FL}{AY}\]
\[ \Rightarrow ∆ L = \frac{8 \times {10}^5 \times \left( 0 . 5 \right)}{2 \times {10}^{11}}\]
\[ = 2 \times {10}^{- 3} \text{ m = 2 mm }\]

Hence, the required elongation of steel wire is 2 mm.