Longitudinal stress of $1\,kg/m{m^2}$ is applied on a wire. The percentage increase in length is $(Y = {10^{11}}\,N/{m^2})$

Two wires are made of the same material and have the same volume. The first wire has cross-sectional area $A$ and the second wire has cross-sectional area $3A$. If the length of the first wire is increased by $\Delta l$ on applying a force $F$, how much force is needed to stretch the second wire by the same amount?

The force required to stretch a wire of crosssection $1 cm ^{2}$ to double its length will be ........ $ \times 10^{7}\,N$

(Given Yong's modulus of the wire $=2 \times 10^{11}\,N / m ^{2}$ )

The length of wire, when $M_1$ is hung from it, is $I_1$ and is $I_2$ with both $M_1$ and $M_2$ hanging. The natural length of wire is ........

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 :

A steel wire of diameter $0.5 mm$ and Young's modulus $2 \times 10^{11} N m ^{-2}$ carries a load of mass $M$. The length of the wire with the load is $1.0 m$. A vernier scale with $10$ divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count $1.0 mm$, is attached. The $10$ divisions of the vernier scale correspond to $9$ divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by $1.2 kg$, the vernier scale division which coincides with a main scale division is. . . . Take $g =10 m s ^{-2}$ and $\pi=3.2$.