Two steel wires of same length $L$ but radii $r$ and $2r$ are connected together end to end and tied to a wall as shown. The force $F$ stretches the combination by $10 \ mm$. How far does the junction point $A$ move (in $mm$)?

The dimensional formula for Young's modulus is

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

$A$ copper wire and a steel wire of the same diameter and length are connected end to end and a force is applied,which stretches their combined length by $1 \ cm$. The two wires will have

Two wires of equal length and cross-section are suspended as shown. Their Young's moduli are $Y_1$ and $Y_2$ respectively. The equivalent Young's modulus will be

$A$ steel rod of length $1\,m$ and cross-sectional area $10^{-4}\,m^2$ is heated from $0^{\circ}C$ to $200^{\circ}C$ without being allowed to extend or bend. The compressive force produced in the rod is $........\times 10^4\,N$. (Given: Young's modulus of steel $Y = 2 \times 10^{11}\,N/m^2$,coefficient of linear expansion $\alpha = 10^{-5}\,K^{-1}$)