Two wires of equal length and equal cross sectional areas are suspended as shown in the figure. Their Young's modulii are $Y_1$ and $Y_2$, respectively. The equivalent Young's modulus is

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Given, two wires have same length and equal cross-sectional area.

i.e., $l_1=l_2=l$ and $A_1=A_2=A$ Suppose, $m$ is the mass of the hanging load. From figure, $2 T=m g$ i.e., $T=\frac{m g}{2}$ Hence, stress on each wire $=\frac{T}{A}=\frac{m g}{2 A}$ $\therefore \quad Y_1=\frac{\text { stress }}{\text { strain }}=\frac{\frac{m g}{2 A}}{\frac{\Delta l}{l}}=\frac{m g l}{2 A \Delta l}$...(i) and $Y_2=\frac{\text { stress }}{\text { strain }}=\frac{\frac{m g}{2 A}}{\frac{\Delta l}{l}}=\frac{m g l}{2 A \Delta l}$...(ii) If $Y$ be the equivalent Young's modulus of the combination, then, $\quad Y=\frac{m g l}{A \Delta l}$...(iii) From Eqs. (i), (ii) and (iii), we get $Y=\frac{Y_1+Y_2}{2}$

Two wires of equal length and equal cross sectional areas are suspended as shown in the figure. Their…

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