Question

A uniform steel rod of length $1m$ and area of cross section $20c{m}^2$ is hanging from a fixed support. Find the increase in the length of the rod. $(Y_{steel}=2.0\times {10}^{11}N{m}^{-2},{\rho }_{steel}=7.85\times {10}^{3}Kg{m}^{-3})$

• A. $1.923\times {10}^{-5}cm$
• B. $2.923\times {10}^{-5}cm$
• C. $\mathrm{1..123}\times {10}^{-5}cm$
• D. $3.123\times {10}^{-5}cm$

Answer 1

The correct option is A $1.923\times {10}^{-5}cm$

(strain)e = $\frac{stress}{youngsmodulus}.$

$stress=\frac{force}{Area}.$

$strain=\frac{△L}{L}=\frac{changeinlength}{originallength}$

$\therefore Force=mg$

$\therefore \frac{△L}{L}=\frac{mg}{AE}.$

$(L=\frac{p}{2})$

$\therefore △L=\frac{mgL}{2AE}.$

M = density $\times$ volume

$=S\times A\times L$

$=△L=\frac{SALgL}{2AE}.$

$=\frac{S{L}^{2}g}{2E}.=\frac{7850\times 1^2\times 9.81}{2\times 2\times {10}^{11}}$

$=1.92\times {10}^{-5}cm.$

Self wt deflection, short cut formula

$sl=\frac{y{l}^2}{2E}(y=sg)$