The length of an elastic string is a metre when the longitudinal tension is $4\, N$ and $b$ metre when the longitudinal tension is $5\, N$. The length of the string in metre when the longitudinal tension is $9\, N$ is

Four identical hollow cylindrical columns of mild steel support a big structure of mass $50 \times 10^{3} {kg}$, The inner and outer radii of each column are $50\; {cm}$ and $100 \;{cm}$ respectively. Assuming uniform local distribution, calculate the compression strain of each column. [Use $\left.{Y}=2.0 \times 10^{11} \;{Pa}, {g}=9.8\; {m} / {s}^{2}\right]$

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 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 $=2 \times 10^{11}\,N / m ^{2}$ )

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 metallic wire is $l$. The tension in the wire is $T_1$ for length $l_1$ and tension in the wire is $T_2$ for length $l_2$. Find the original length.