The interatomic distance for a metal is $3 \times {10^{ - 10}}\,m$. If the interatomic force constant is $3.6 \times {10^{ - 9}}\,N/{{\buildrel _{\circ} \over {\mathrm{A}}}}$, then the Young's modulus in $N/{m^2}$ will be
$1.2 \times {10^{11}}$
$4.2 \times {10^{11}}$
$10.8 \times {10^{ - 19}}$
$2.4 \times {10^{10}}$
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 following four wires are made of same material. Which one will have the largest elongation when subjected to the same tension ?
Density of rubber is $d$. $ A$ thick rubber cord of length $L$ and cross-section area $A$ undergoes elongation under its own weight on suspending it. This elongation is proportional to
The elongation of a wire on the surface of the earth is $10^{-4} \; m$. The same wire of same dimensions is elongated by $6 \times 10^{-5} \; m$ on another planet. The acceleration due to gravity on the planet will be $\dots \; ms ^{-2}$. (Take acceleration due to gravity on the surface of earth $=10 \; m / s ^{-2}$ )
The Young's modulus of steel is twice that of brass. Two wires of same length and of same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights added to the steel and brass wires must be in the ratio of