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: $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

Read the following two statements below carefully and state, with reasons, if it is true or false.

$(a)$ The Young’s modulus of rubber is greater than that of steel;

$(b)$ The stretching of a coil is determined by its shear modulus.

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 $\frac{{YA}}{L}$ ]

The elongation of a wire on the surface of the earth is $10^{-4} \; m$. The same wire of same dimensions is elongated by $6 \times 10^{-5} \; m$ on another planet. The acceleration due to gravity on the planet will be $\dots \; ms ^{-2}$. (Take acceleration due to gravity on the surface of earth $=10 \; m / s ^{-2}$ )

A piece of copper having a rectangular cross-section of $15.2 \;mm \times 19.1 \;mm$ is pulled in tenston with $44,500\; N$ force, productng only elastic deformation. Calculate the resulting strain?