The modulus of elasticity is dimensionally equivalent to

The area of a cross-section of steel wire is $0.1\,\,cm^2$ and Young's modulus of steel is $2\,\times \,10^{11}\,\,N\,\,m^{-2}.$  The force required to stretch by $0.1\%$ of its length is ……… $N$.

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 diameter $0.25 \;cm ,$ one made of steel and the other made of brass are loaded as shown in Figure. The unloaded length of steel wire is $1.5 \;m$ and that of brass wire is $1.0 \;m .$ Compute the elongations of the steel and the brass wires.

A mild steel wire of length $1.0 \;m$ and cross-sectional area $0.50 \times 10^{-2} \;cm ^{2}$ is stretched, well within its elastic limit, horizontally between two pillars. A mass of $100 \;g$ is suspended from the mid-point of the wire. Calculate the depression at the midpoint.