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

How much force is required to produce an increase of $0.2\%$ in the length of a brass wire of diameter $0.6\, mm$ (Young’s modulus for brass = $0.9 \times {10^{11}}N/{m^2}$)

Explain experimental determination of Young’s modulus.

Young's modulus is determined by the equation given by $\mathrm{Y}=49000 \frac{\mathrm{m}}{\ell} \frac{\text { dyne }}{\mathrm{cm}^2}$ where $\mathrm{M}$ is the mass and $\ell$ is the extension of wre used in the experiment. Now error in Young modules $(\mathrm{Y})$ is estimated by taking data from $M-\ell$ plot in graph paper. The smallest scale divisions are $5 \mathrm{~g}$ and $0.02$ $\mathrm{cm}$ along load axis and extension axis respectively. If the value of $M$ and $\ell$ are $500 \mathrm{~g}$ and $2 \mathrm{~cm}$ respectively then percentage error of $\mathrm{Y}$ is :

The Young's modulus of a steel wire of length $6\,m$ and cross-sectional area $3\,mm ^2$, is $2 \times 11^{11}\,N / m ^2$. The wire is suspended from its support on a given planet. A block of mass $4\,kg$ is attached to the free end of the wire. The acceleration due to gravity on the planet is $\frac{1}{4}$ of its value on the earth. The elongation of wire is  (Take $g$ on the earth $=10$ $\left.m / s ^2\right):$