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)
$30\frac{{mg}}{{AY}}\ cm + 40\ cm$
$20\frac{{mg}}{{AY}}\ cm + 40\ cm$
$50\frac{{mg}}{{AY}}\ cm + 40\ cm$
$40\frac{{mg}}{{AY}}\ cm + 40\ cm$
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..... $mm$
The maximum elongation of a steel wire of $1 \mathrm{~m}$ length if the elastic limit of steel and its Young's modulus, respectively, are $8 \times 10^8 \mathrm{~N} \mathrm{~m}^{-2}$ and $2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$, is:
A $14.5\; kg$ mass, fastened to the end of a steel wire of unstretched length $1.0 \;m ,$ is whirled in a vertical circle with an angular velocity of $2\;rev/s$ at the bottom of the circle. The cross-sectional area of the wire is $0.065 \;cm ^{2} .$ Calculate the elongation of the wire when the mass is at the lowest point of its path.
The ratio of diameters of two wires of same material is $n : 1$. The length of wires are $4\, m$ each. On applying the same load, the increase in length of thin wire will be
An aluminum rod (Young's modulus $ = 7 \times {10^9}\,N/{m^2})$ has a breaking strain of $0.2\%$. The minimum cross-sectional area of the rod in order to support a load of ${10^4}$Newton's is