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
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Overall changes in volume and radii of a uniform cylindrical steel wire are $0.2 \%$ and $0.002 \%$ respectively when subjected to some suitable force. Longitudinal tensile stress acting on the wire is ($Y = 2.0 × 10^{11} Nm^{-2}$)
A wire of length $L$ and radius $r$ is clamped rigidly at one end. When the other end of the wire is pulled by a force $f$, its length increases by $l$. Another wire of same material of length $2 L$ and radius $2 r$ is pulled by a force $2 f$. Then the increase in its length will be
If average depth of an ocean is $4000 \mathrm{~m}$ and the bulk modulus of water is $2 \times 10^9 \mathrm{Nm}^{-2}$, then fractional compression $\frac{\Delta V}{V}$ of water at the bottom of ocean is $\alpha \times 10^{-2}$. The value of $\alpha$ is ___________(Given, $\mathrm{g}=10 \mathrm{~ms}^{-2}, \rho=1000 \mathrm{~kg} \mathrm{~m}^{-3}$ )
Consider two wires of same material having their ratio of radii to be $2: 1$. If these two wires are stretched by equal force, then the ratio of stress produced in them is
The breaking stress of aluminium is $7.5 \times 10^7 \,Nm ^{-2}$. The greatest length of aluminium wire that can hang vertically without breaking is ......... $\times 10^3 \,m$ (Density of aluminium is $2.7 \times 10^3 \,kg m ^{-3}$ )
In the given figure, two elastic rods $A$ & $B$ are rigidly joined to end supports. $A$ small mass $‘m’$ is moving with velocity $v$ between the rods. All collisions are assumed to be elastic & the surface is given to be frictionless. The time period of small mass $‘m’$ will be : [$A=$ area of cross section, $Y =$ Young’s modulus, $L=$ length of each rod ; here, an elastic rod may be treated as a spring of spring constant $\frac{{YA}}{L}$ ]