Two wires $‘A’$ and $‘B’$ of the same material have radii in the ratio $2 : 1$ and lengths in the ratio $4 : 1$. The ratio of the normal forces required to produce the same change in the lengths of these two wires is

A steel wire of length $3.2 m \left( Y _{ S }=2.0 \times 10^{11}\,Nm ^{-2}\right)$ and a copper wire of length $4.4\,M$ $\left( Y _{ C }=1.1 \times 10^{11}\,Nm ^{-2}\right)$, 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 Newton, will be. (Given $\pi=\frac{22}{7}$)

If Young's modulus of iron is $2 \times {10^{11}}\,N/{m^2}$ and the interatomic spacing between two molecules is $3 \times {10^{ - 10}}$metre, the interatomic force constant is ......... $N/m$

Two similar wires under the same load yield elongation of $0.1$ $mm$ and $0.05$ $mm$ respectively. If the area of cross- section of the first wire is $4m{m^2},$ then the area of cross section of the second wire is..... $mm^2$

A rod of length $1.05\; m$ having negligible mass is supported at its ends by two wires of steel (wire $A$) and aluminium (wire $B$) of equal lengths as shown in Figure. The cross-sectional areas of wires $A$ and $B$ are $1.0\; mm ^{2}$ and $2.0\; mm ^{2}$. respectively. At what point along the rod should a mass $m$ be suspended in order to produce $(a)$ equal stresses and $(b)$ equal strains in both steel and alumintum wires.

The Young's modulus of a wire of length $L$ and radius $r$ is $Y$ $N/m^2$. If the length and radius are reduced to $L/2$ and $r/2,$ then its Young's modulus will be