The mass and length of a wire are $M$ and $L$ respectively. The density of the material of the wire is $d$. On applying the force $F$ on the wire, the increase in length is $l$, then the Young's modulus of the material of the wire will be
$\frac{{Fdl}}{{Ml}}$
$\frac{{FL}}{{Mdl}}$
$\frac{{FMl}}{{dl}}$
$\frac{{Fd{L^2}}}{{Ml}}$
A mild steel wire of length $1.0 \;m$ and cross-sectional area $0.50 \times 10^{-2} \;cm ^{2}$ is stretched, well within its elastic limit, horizontally between two pillars. A mass of $100 \;g$ is suspended from the mid-point of the wire. Calculate the depression at the midpoint.
Longitudinal stress of $1\,kg/m{m^2}$ is applied on a wire. The percentage increase in length is $(Y = {10^{11}}\,N/{m^2})$
Figure shows the strain-stress curve for a given material. What are $(a)$ Young’s modulus and $(b)$ approximate yield strength for this material?
If in case $A$, elongation in wire of length $L$ is $l$, then for same wire elongation in case $B$ will be ......
A $100\,m$ long wire having cross-sectional area $6.25 \times 10^{-4}\,m ^2$ and Young's modulus is $10^{10}\,Nm ^{-2}$ is subjected to a load of $250\,N$, then the elongation in the wire will be :