Question

Two circular mild steel bars $A$ and $B$ of equal lengths $l$ have diameters $d_A=2cm$ and $d_B=3cm$. Each is subjected to a tensile load of magnitude $P$. The ratio of the elongations of the bars $l_A/l_B$ is

• A. $\frac{2}{3}$
• B. $\frac{3}{4}$
• C. $\frac{4}{9}$
• D. $\frac{9}{4}$

Answer 1

The correct option is D $\frac{9}{4}$

The elongation of a bar subjected to a tensile load $P$ is given by
$\Delta =\frac{PL}{AE}$
Now for bar $A$,
${\Delta }_A=\frac{PL}{\frac{\pi d_A^2}{4}\times E}$
Similarly for bar $B$,
${\Delta }_B=\frac{PL}{\frac{\pi d_B^2}{4}\times E}$
$\therefore \frac{{\Delta }_A}{{\Delta }_B}=\frac{d_B^2}{d_A^2}$
$\therefore \frac{{\Delta }_A}{{\Delta }_B}=\frac{(3{)}^2}{(2{)}^2}=\frac{9}{4}$