Young's modulus $(Y)=\frac{\text { Stress }}{\text { Longitudinal strain }}$

For same longitudinal strain, $\quad Y \propto$ stress

$$\therefore \quad \frac{Y_{\text {steel }}}{Y_{\text {rubber }}}=\frac{(\text { stress })_{\text {steel }}}{(\text { stress })_{\text {rubber }}} \quad \text{... (i)}$$

$$\begin{aligned} \text{But}\quad Y_{\text {steel }} & >Y_{\text {rubber }} \\ \frac{Y_{\text {steel }}}{Y_{\text {rubber }}} & >1 \end{aligned}$$

Therefore, from Eq. (i),

$$\begin{array}{ll} & \frac{(\text { stress })_{\text {steel }}}{(\text { stress })_{\text {rubber }}}>1 \\ \Rightarrow \quad & (\text { stress })_{\text {steel }}>(\text { stress })_{\text {rubber }} \end{array}$$

$$\text { Stress }=\frac{\text { Magnitude of internal reaction force }}{\text { Area of cross }- \text { section }}$$

Therefore, stress is a scalar quantity not a vector quantity.

Work done in stretching a wire is given by $W=\frac{1}{2} F \times \Delta l$

[where $F$ is applicd force and $\Delta l$ is extension in the wire]

As springs of steel and copper are equally stretched. Therefore, for same force $(F)$,

$$W \propto \Delta l\quad \text{... (i)}$$

$$\text { Young's modulus }(Y)=\frac{F}{A} \times \frac{l}{\Delta l} \Rightarrow \Delta l=\frac{F}{A} \times \frac{l}{Y}$$

As both springs are identical,

$$\Delta l \propto \frac{1}{Y}\quad \text{... (ii)}$$

From Eqs. (i) and (ii), we get

$$W \propto \frac{1}{y}$$

$$\begin{aligned} &\begin{array}{lll} \therefore & \frac{W_{\text {steel }}}{W_{\text {copper }}}=\frac{Y_{\text {copper }}}{Y_{\text {steel }}}< 1 & \left(\text { As, } Y_{\text {steel }}>Y_{\text {copper }}\right) \\ \Rightarrow & W_{\text {steel }}< W_{\text {copper }} \end{array}\\ &\text { Therefore, more work will be done for stretching copper spring. } \end{aligned}$$