$A$ $5\, m$ 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 $mm$.

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?

Two wires $A$ and $B$ of the same cross-section are connected end to end. When the same tension is applied to both wires,the elongation in wire $B$ is twice the elongation in wire $A$. If $L_A$ and $L_B$ are the initial lengths of the wires $A$ and $B$ respectively,then (Young's modulus of material of wire $A = 2 \times 10^{11} \ Nm^{-2}$ and Young's modulus of material of wire $B = 1.1 \times 10^{11} \ Nm^{-2}$):

$A$ wire of cross-sectional area $A$,modulus of elasticity $2 \times 10^{11} \text{ N m}^{-2}$,and length $2L = 2 \text{ m}$ is stretched between two vertical rigid supports. When a mass of $2 \text{ kg}$ is suspended at the middle,it sags from its original position,making an angle $\theta = \frac{1}{100} \text{ radian}$ with the horizontal at the points of support. The value of $A$ is . . . . . . $\times 10^{-4} \text{ m}^2$. (Given: $g = 10 \text{ m/s}^2$)

Which one is more elastic,steel or plastic? Why?

$A$ 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 $\pi = \frac{22}{7}$)