Question

A steel rod, 2 m in length, 40 mm in diameter, is subjected to a pull of 70 kN as shown in the figure



To what length should the bar be bored centrally from one end so that total extension will increase by $20$% under the same force (the bore diameter is $25mm$ and $E$ is $2\times {10}^{5}N/m{m}^2$)?

• A. 0.46 m
• B. 0.55 m
• C. 0.87 m
• D. 0.62 m

Answer 1

The correct option is D 0.62 m



$\Delta l=\frac{Pl}{AE}$
Initially when no boring.
$\Delta l_1=\frac{70\times {10}^3\times 2\times {10}^3}{\pi /4\times {40}^2\times 2\times {10}^5}$
After boring,
$\Delta l_2=\frac{70\times {10}^3\times X\times {10}^3}{\pi /4\times ({40}^2-{25}^2)\times 2\times {10}^5}+\frac{70\times {10}^3\times (2-X)\times {10}^3}{\pi /4\times {40}^2\times 2\times {10}^5}$
Given,
$\Delta l_2=1.2\Delta l_1$
$\Rightarrow \frac{350x}{\pi /4\times 65\times 15}+\frac{350(2-x)}{\pi /4\times 1600}=\frac{1.2\times 400}{\pi \times 1600}$
$\Rightarrow \frac{x}{3\times 13}+\frac{2-x}{64}=\frac{1.2\times 2}{64}$
$\Rightarrow \frac{64x+78-39x}{39\times 64}=\frac{2.4}{6.4}$
$\Rightarrow 25x=39\times 2.5-7.8$
$\Rightarrow x=0.624m$