To determine Young's modulus of a wire, the formula is $Y = \frac{F}{A}.\frac{L}{{\Delta L}}$ where $F/A$ is the stress and $L/\Delta L$ is the strain. The conversion factor to change $Y$ from $CGS$ to $MKS$ system is

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..... $mm^2$

The Young's modulus of steel is twice that of brass. Two wires of same length and of same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights added to the steel and brass wires must be in the ratio of

A wire of area of cross-section ${10^{ - 6}}{m^2}$ is increased in length by $0.1\%$. The tension produced is $1000 N$. The Young's modulus of wire is

An elastic material of Young's modulus $Y$ is subjected to a stress $S$. The elastic energy stored per unit volume of the material is

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 ......... $N/m$