Question

A round steel bar of overall length $40cm$ consists of two equal portions of $20cm$ each having diameters of $10cm$ and $8cm$ respectively. Take $E$ as $2\times {10}^{6}kg/c{m}^2$. If the rod is subjected to a tensile load of $10$ tonnes, the elongation in cm will be given by

• A. $\frac{1}{10\pi }(\frac{1}{25}+\frac{1}{16})$
• B. $\frac{2}{10\pi }(\frac{1}{25}+\frac{1}{16})$
• C. $\frac{3}{10\pi }(\frac{1}{25}+\frac{1}{16})$
• D. $\frac{4}{10\pi }(\frac{1}{25}+\frac{1}{16})$

Answer 1

The correct option is A $\frac{1}{10\pi }(\frac{1}{25}+\frac{1}{16})$

Elongation $=\frac{PL}{E{A}_1}+\frac{PL}{E{A}_2}$
$=\frac{PL}{E}\times \frac{4}{\pi }(\frac{1}{d_1^2}+\frac{1}{d_2^2})$
$=\frac{10\times 1000\times 20}{2\times {10}^6}\times \frac{4}{\pi }(\frac{1}{100}+\frac{1}{64})$
$=\frac{1}{10\pi }(\frac{1}{25}+\frac{1}{16})$