A wire of length $L$ and radius $r$ is clamped rigidly at one end. When the other end of the wire is pulled by a force $f$ its length increases by $l$. Another wire of the same material of length $2L$ and radius $2r$ is pulled by a force $2f$. Then find the increase in length of this wire.
$\mathrm{Y}=\frac{\mathrm{FL}}{\mathrm{Al}}=\frac{f \mathrm{~L}}{\pi r^{2} l}$
$\therefore l=\frac{f \mathrm{~L}}{\pi r^{2} \mathrm{Y}}$
Now for increase in length for second wire,
$l^{\prime}=\frac{(2 f)(2 \mathrm{~L})}{\pi(2 r)^{2} l}$
$=\frac{4 f L}{4 \pi r^{2} l}$
$l^{\prime}=l$
Two identical solid balls, one of ivory and the other of wet-clay are dropped from the same height on the floor. Which one will rise to a greater height after striking the floor and why ?
A load $W$ produces an extension of $1mm$ in a thread of radius $r.$ Now if the load is made $4W$ and radius is made $2r$ all other things remaining same, the extension will become..... $mm$
In steel, the Young's modulus and the strain at the breaking point are $2 \times {10^{11}}\,N{m^{ - 2}}$ and $0.15$ respectively. The stress at the breaking point for steel is therefore
A meter scale of mass $m$ , Young modulus $Y$ and cross section area $A$ is hanged vertically from ceiling at zero mark. Then separation between $30\ cm$ and $70\ cm$ mark will be :-( $\frac{{mg}}{{AY}}$ is dimensionless)
On increasing the length by $0.5\, mm$ in a steel wire of length $2\, m$ and area of cross-section $2\,m{m^2}$, the force required is $[Y$ for steel$ = 2.2 \times {10^{11}}\,N/{m^2}]$