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 length of an iron wire is $L$ and area of cross-section is $A$. The increase in length is $l$ on applying the force $F$ on its two ends. Which of the statement is correct

The area of cross-section of a wire of length $1.1$ metre is $1$ $mm^2$. It is loaded with $1 \,kg.$ If Young's modulus of copper is $1.1 \times {10^{11}}\,N/{m^2}$, then the increase in length will be ......... $mm$ (If $g = 10\,m/{s^2})$

Two separate wires $A$ and $B$ are stretched by $2 \,mm$ and $4\, mm$ respectively, when they are subjected to a force of $2\, N$. Assume that both the wires are made up of same material and the radius of wire $B$ is 4 times that of the radius of wire $A$. The length of the wires $A$ and $B$ are in the ratio of $a : b$. Then $a / b$ can be expressed as $1 / x$ where $x$ is

If Young's modulus for a material is zero, then the state of material should be

The load versus elongation graph for four wires of same length and the same material is shown in figure. The thinnest wire is represented by line