Steel and copper wires of same length are stretched by the same weight one after the other. Young's modulus of steel and copper are $2 \times {10^{11}}\,N/{m^2}$ and $1.2 \times {10^{11}}\,N/{m^2}$. The ratio of increase in length

If the temperature of a wire of length $2 \,m$ and area of cross-section $1 \,cm ^2$ is increased from $0^{\circ} C$ to $80^{\circ} C$ and is not allowed to increase in length, then force required for it is …………$N$ $\left\{Y=10^{10} \,N / m ^2, \alpha=10^{\left.-6 /{ }^{\circ} C \right\}}\right.$

The temperature of a wire of length $1$ metre and area of cross-section $1\,c{m^2}$ is increased from $0°C$ to $100°C$. If the rod is not allowed to increase in length, the force required will be $(\alpha = {10^{ – 5}}/^\circ C$ and $Y = {10^{11}}\,N/{m^2})$

Young's modules of material of a wire of length ' $L$ ' and cross-sectional area $A$ is $Y$. If the length of the wire is doubled and cross-sectional area is halved then Young's $modules$ will be :

In nature the failure of structural members usually result from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus, the vertical through the centre of gravity does not fall withinthe  base. The elastic torque caused because of this bending about the central axis of the tree is given by $\frac{{Y\pi {r^4}}}{{4R}}$ $Y$ is the Young’s modulus, $r$ is the radius of the trunk and $R$ is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.