$A$ mild steel wire of length $2l$ meter and cross-sectional area $A \; m^2$ is fixed horizontally between two pillars. $A$ small mass $m \; kg$ is suspended from the midpoint of the wire. If the extension in the wire is within the elastic limit,then the depression $x$ at the midpoint of the wire will be:

An area of cross-section of a rubber string is $2 \, cm^2$. Its length is doubled when stretched with a linear force of $2 \times 10^5 \, dynes$. The Young's modulus of the rubber in $dyne/cm^2$ will be:

Two wires of the same length and same cross-sectional area are suspended as shown in the figure. Their Young's moduli are $Y_1$ and $Y_2$. What is their equivalent Young's modulus?

In nature,the failure of structural members usually results from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling. In cases of tall cylindrical structures like trees,the torque is caused by their own weight bending the structure,such that the vertical line through the centre of gravity does not fall within the base. The elastic torque caused by this bending about the central axis of the tree is given by $\frac{Y\pi r^4}{4R}$,where $Y$ 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.

$A$ steel wire is stretched with a definite load. If the Young's modulus of the wire is $Y$,how can the value of $Y$ be decreased?

Two wires $A$ and $B$ are of the same material. Their lengths are in the ratio $1: 2$ and diameters are in the ratio $2: 1$. When stretched by forces $F_{A}$ and $F_{B}$ respectively,they get equal increase in their lengths. Then the ratio $F_{A} / F_{B}$ is