The interatomic distance for a metal is $3 \times {10^{ - 10}}\,m$. If the interatomic force constant is $3.6 \times {10^{ - 9}}\,N/{{\buildrel _{\circ} \over {\mathrm{A}}}}$, then the Young's modulus in $N/{m^2}$ will be

The dimensions of four wires of the same material are given below. In which wire the increase in length will be maximum when the same tension is applied

Young's modulus of rubber is ${10^4}\,N/{m^2}$ and area of cross-section is $2\,c{m^2}$. If force of $2 \times {10^5}$ dynes is applied along its length, then its initial length $l$ becomes

Each of three blocks $P$, $Q$ and $R$ shown in figure has a mass of $3 \mathrm{~kg}$. Each of the wire $A$ and $B$ has cross-sectional area $0.005 \mathrm{~cm}^2$ and Young's modulus $2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$. Neglecting friction, the longitudinal strain on wire $B$ is____________ $\times 10^{-4}$. $\left(\right.$ Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

In an experiment, brass and steel wires of length $1\,m$ each with areas of cross section $1\,mm^2$ are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress requires to produced a new elongation of $0.2\,mm$ is [Given, the Young’s Modulus for steel and brass are respectively $120\times 10^9\,N/m^2$ and $60\times 10^9\,N/m^2$ ]

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.