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 $60\times 10^9\,N/m^2$ ]
$1.8\times 10^6\,N/m^2$
$0.2\times 10^6\,N/m^2$
$1.2\times 10^6\,N/m^2$
None of these
The length of an elastic string is a metre when the longitudinal tension is $4\, N$ and $b$ metre when the longitudinal tension is $5\, N$. The length of the string in metre when the longitudinal tension is $9\, N$ is
A wire extends by $1 mm$ when a force is applied. Double the force is applied to another wire of same material and length but half the radius of cross-section. The elongation of the wire in mm will be ........
The mean distance between the atoms of iron is $3 \times {10^{ - 10}}m$ and interatomic force constant for iron is $7\,N\,/m$The Young’s modulus of elasticity for iron is
A wire of cross sectional area $A$, modulus of elasticity $2 \times 10^{11} \mathrm{Nm}^{-2}$ and length $2 \mathrm{~m}$ is stretched between two vertical rigid supports. When a mass of $2 \mathrm{~kg}$ is suspended at the middle it sags lower from its original position making angle $\theta=\frac{1}{100}$ radian on the points of support. The value of $A$ is. . . . . . $\times 10^{-4} \mathrm{~m}^2$ (consider $\mathrm{x}<\mathrm{L}$ ).
(given: $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )
In nature the failure of structural members usually result from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus, the vertical through the centre of gravity does not fall withinthe base. The elastic torque caused because of this bending about the central axis of the tree is given by $\frac{{Y\pi {r^4}}}{{4R}}$ $Y$ is the Young’s modulus, $r$ is the radius of the trunk and $R$ is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.