Give the relation between shear modulus and Young’s modulus.

A string of area of cross-section $4\,mm ^{2}$ and length $0.5$ is connected with a rigid body of mass $2\,kg$. The body is rotated in a vertical circular path of radius $0.5\,m$. The body acquires a speed of $5\,m / s$ at the bottom of the circular path. Strain produced in the string when the body is at the bottom of the circle is $\ldots . . \times 10^{-5}$. (Use Young's modulus $10^{11}\,N / m ^{2}$ and $g =10\,m / s ^{2}$ )

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

An aluminum rod (Young's modulus $ = 7 \times {10^9}\,N/{m^2})$ has a breaking strain of $0.2\%$. The minimum cross-sectional area of the rod in order to support a load of ${10^4}$Newton's is

A compressive force, $F$ is applied at the two ends of a long thin steel rod. It is heated, simultaneously, such that its temperature increases by $\Delta T$. The net change in its length is zero. Let $l$ be the length of the rod, $A$ its area of cross- section, $Y$ its Young's modulus, and $\alpha $ its coefficient of linear expansion. Then, $F$ is equal to

If in case $A$, elongation in wire of length $L$ is $l$, then for same wire elongation in case $B$ will be ......