The mass and length of a wire are $M$ and $L$ respectively. The density of the material of the wire is $d$. On applying a force $F$ on the wire,the increase in length is $l$. Then,the Young's modulus of the material of the wire will be:

The Young's modulus of a steel wire of length $6\,m$ and cross-sectional area $3\,mm^2$ is $2 \times 10^{11}\,N/m^2$. The wire is suspended from its support on a given planet. $A$ block of mass $4\,kg$ is attached to the free end of the wire. The acceleration due to gravity on the planet is $\frac{1}{4}$ of its value on the Earth. The elongation of the wire is (Take $g$ on the Earth $= 10\,m/s^2$):

The Young's modulus of a wire of length $L$ and radius $r$ is $Y$. If the length is reduced to $\frac{L}{2}$ and radius is $\frac{r}{2}$,then the Young's modulus will be

The density of a rubber cord is $d$. $A$ thick rubber cord of length $L$ and cross-sectional area $A$ undergoes elongation under its own weight when suspended vertically. This elongation is proportional to:

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.

$A$ $5\, m$ long aluminium wire $(Y = 7 \times 10^{10}\, N/m^2)$ of diameter $3\, mm$ supports a $40\, kg$ mass. In order to have the same elongation in a copper wire $(Y = 12 \times 10^{10}\, N/m^2)$ of the same length under the same weight,the diameter should now be,in $mm$.