$A$ uniform plank of Young's modulus $Y$ is moved over a smooth horizontal surface by a constant horizontal force $F$. The area of cross-section of the plank is $A$. The compressive strain on the plank in the direction of the force is

The maximum elongation of a steel wire of $1 \,m$ length if the elastic limit of steel and its Young's modulus,respectively,are $8 \times 10^8 \,N \,m^{-2}$ and $2 \times 10^{11} \,N \,m^{-2}$,is: (in $\,mm$)

$5 \,m$ 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 be (in $mm$):

$A$ metal rod of length $L$ and cross-sectional area $A$ is heated through $T^{\circ} C$. What is the force required to prevent the expansion of the rod lengthwise? $[Y=$ Young's modulus of the material of rod,$\alpha=$ coefficient of linear expansion $]$

$A$ 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: $Y_{Cu} = 1.1 \times 10^{11} \text{ Nm}^{-2}$,$Y_{Al} = 0.7 \times 10^{11} \text{ Nm}^{-2}$)

$A$ steel wire of length $3.2 \, m$ $(Y_{S} = 2.0 \times 10^{11} \, N/m^{2})$ and a copper wire of length $4.4 \, m$ $(Y_{C} = 1.1 \times 10^{11} \, N/m^{2})$,both of radius $1.4 \, mm$,are connected end to end. When stretched by a load,the net elongation is found to be $1.4 \, mm$. The load applied,in Newtons,is. (Given $\pi = \frac{22}{7}$)