Question

A mild steel bar, circular in cross-section, tapers from $40mm$ diameter to $20mm$ diameter over its length of $800mm$. It is subjected to an axial pull of $20kN$. $E=2\times {10}^{5}N/m{m}^2$. The increase in the length of the rod will be

• A. $\frac{1}{10\pi }mm$
• B. $\frac{2}{5\pi }mm$
• C. $\frac{4}{5\pi }mm$
• D. $\frac{1}{5\pi }mm$

Answer 1

The correct option is B $\frac{2}{5\pi }mm$

$\Delta =\frac{4PL}{\pi d_1{d}_{2}E}$
$=\frac{4\times 20\times {10}^3\times 800}{\pi \times 20\times 40\times 2\times {10}^5}=\frac{2}{5\pi }$