Young’s modulus of perfectly rigid body material is

Four identical hollow cylindrical columns of mild steel support a big structure of mass $50 \times 10^{3} {kg}$, The inner and outer radii of each column are $50\; {cm}$ and $100 \;{cm}$ respectively. Assuming uniform local distribution, calculate the compression strain of each column. [Use $\left.{Y}=2.0 \times 10^{11} \;{Pa}, {g}=9.8\; {m} / {s}^{2}\right]$

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

The force required to stretch a steel wire of $1\,c{m^2}$ cross-section to $1.1$ times its length would be $(Y = 2 \times {10^{11}}\,N{m^{ - 2}})$

Two wires $A$ and $B$ are of same materials. Their lengths are in the ratio $1 : 2$ and diameters are in the ratio $2 : 1$ when stretched by force ${F_A}$ and ${F_B}$ respectively they get equal increase in their lengths. Then the ratio ${F_A}/{F_B}$ should be

The Young's modulus of a steel wire of length $6\,m$ and cross-sectional area $3\,mm ^2$, is $2 \times 11^{11}\,N / m ^2$. 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 wire is  (Take $g$ on the earth $=10$ $\left.m / s ^2\right):$