Question
Figure represents the extension ( $\Delta l$ ) of a wire of length 1 meter, suspended from the ceiling of the room at one end with load W connected to the other end. If the cross- sectional area of the wire is $10^{- 5}{\text{ }m}^{2}$ then the Young's modulus of the wire is
$\_\_\_\_$ $N/m^{2}$.
Options
Solution
Sol. [Key Concept: Young's modulus $Y = \frac{slope\ of\ W\ vs\ \Delta\ l \times \ L}{A}$; read the slope $\frac{W}{\Delta\ l}{= 10}^{5}$ from the graph.]
$$\frac{\Delta\mathcal{l}}{w} = tan\theta = \frac{\Delta y}{\Delta x} = \frac{1 \times 10^{- 4}}{10} = 10^{- 5}$$
$$\begin{matrix} & \frac{w}{\Delta\mathcal{l}} = 10^{5} \\ & \ \Rightarrow y = \frac{F/A}{\Delta\mathcal{l}/\mathcal{l}} = \frac{F\mathcal{l}}{\text{ }A\Delta\mathcal{l}} = \left( \frac{w}{\Delta\mathcal{l}} \right)\frac{\mathcal{l}}{A} \\ & \ \Rightarrow y = \frac{10^{5} \times 1}{10^{- 5}} = 10^{10}\text{ }N/m^{2} \end{matrix}$$
Create a free account to view solution
View Solution Free