A wire is suspended from the ceiling and stretched under the action of a weight $F$ suspended from its other end. The force exerted by the ceiling on its is equal and opposite to the weight.
A rod of length I and negligible mass is suspended at its two ends by two wires of steel (wire $A$ ) and aluminium (wire $B$ ) of equal lengths (figure). The cross-sectional areas of wires $A$ and $B$ are $1.0 \mathrm{~mm}^2$ and $2.0 \mathrm{~mm}^2$, respectively.
$\left(Y_{\mathrm{Al}}=70 \times 10^9 \mathrm{Nm}^{-2}\right.$ and $\left.Y_{\text {steel }}=200 \times 10^9 \mathrm{Nm}^{-2}\right)$
For an ideal liquid,
A copper and a steel wire of the same diameter are connected end to end. A deforming force $F$ is applied to this composite wire which causes a total elongation of 1 cm . The two wires will have
The Young's modulus for steel is much more than that for rubber. For the same longitudinal strain, which one will have greater tensile stress?
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),
$$\begin{array}{ll} & \frac{(\text { stress })_{\text {steel }}}{(\text { stress })_{\text {rubber }}}>1 \\ \Rightarrow \quad & (\text { stress })_{\text {steel }}>(\text { stress })_{\text {rubber }} \end{array}$$