In steel,the Young's modulus and the strain at the breaking point are $2 \times 10^{11} \, N/m^2$ and $0.15$ respectively. The stress at the breaking point for steel is therefore:

The force required to stretch a wire of cross-section $1 \ cm^{2}$ to double its length will be ........ $\times 10^{7} \ N$. (Given Young's modulus of the wire $= 2 \times 10^{11} \ N/m^{2}$)

$A$ copper wire and an aluminium wire have lengths in the ratio $5: 2$, diameters in the ratio $4: 3$ and forces applied in the ratio $4: 5$. Find the ratio of increase in length of the copper wire to that of the aluminium wire. (Given: $Y_{Cu} = 1.1 \times 10^{11} \text{ Nm}^{-2}$, $Y_{Al} = 0.7 \times 10^{11} \text{ Nm}^{-2}$)

When a force of $4 \, N$ is applied to a wire,its length is $a \, m$. When a force of $5 \, N$ is applied,its length is $b \, m$. What will be its length when a force of $9 \, N$ is applied?

The area of a cross-section of a steel wire is $0.1 \, cm^2$ and Young's modulus of steel is $2 \times 10^{11} \, N \, m^{-2}$. The force required to stretch it by $0.1 \%$ of its original length is ......... $N$.

$A$ compressive force, $F$ is applied at the two ends of a long thin steel rod. It is heated, simultaneously, such that its temperature increases by $\Delta T$. The net change in its length is zero. Let $l$ be the length of the rod, $A$ its area of cross-section, $Y$ its Young's modulus, and $\alpha$ its coefficient of linear expansion. Then, $F$ is equal to