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A cylindrical rod made of aluminum has length 1 meter and diameter of $10 \mathrm{~cm}$. The rod is subjected to a tensile force of $100 \mathrm{kN}$. The elongation in the rod is
(Young's modulus of aluminum $=70 \mathrm{GPa}$ )
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(Young's modulus of aluminum $=70 \mathrm{GPa}$ )
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Verified Answer
The correct answer is:
$1.81 \times 10^{-4} \mathrm{~m}$
Length of aluminum rod, $1=1 \mathrm{~m}$
Diameter, $\mathrm{d}=1 \mathrm{~m}$
Radius, $\mathrm{r}=\frac{0.1}{2}$
Tensile force, $\mathrm{F}=100 \times 10^3 \mathrm{~N}$
Young's modulus of aluminum, $y=70 \times 10^{19} \mathrm{pa}$
$\begin{aligned}
& \mathrm{Y}=\frac{\text { stress }}{\text { strain }}=\frac{\frac{\mathrm{F}}{\mathrm{A}}}{\frac{\Delta \mathrm{l}}{\mathrm{l}}}=\frac{\mathrm{F}}{\pi \mathrm{r}^2} \times \frac{1}{\Delta \mathrm{l}} \\
& \Delta \mathrm{l}=\frac{\mathrm{Fl}}{\pi \mathrm{r}^2 \mathrm{y}} \\
& =\frac{100 \times 10^3 \times 1}{3.14 \times(0.05)^2 \times 70 \times 10^9} \\
& =1.81 \times 10^{-4} \mathrm{~m} .
\end{aligned}$
Diameter, $\mathrm{d}=1 \mathrm{~m}$
Radius, $\mathrm{r}=\frac{0.1}{2}$
Tensile force, $\mathrm{F}=100 \times 10^3 \mathrm{~N}$
Young's modulus of aluminum, $y=70 \times 10^{19} \mathrm{pa}$
$\begin{aligned}
& \mathrm{Y}=\frac{\text { stress }}{\text { strain }}=\frac{\frac{\mathrm{F}}{\mathrm{A}}}{\frac{\Delta \mathrm{l}}{\mathrm{l}}}=\frac{\mathrm{F}}{\pi \mathrm{r}^2} \times \frac{1}{\Delta \mathrm{l}} \\
& \Delta \mathrm{l}=\frac{\mathrm{Fl}}{\pi \mathrm{r}^2 \mathrm{y}} \\
& =\frac{100 \times 10^3 \times 1}{3.14 \times(0.05)^2 \times 70 \times 10^9} \\
& =1.81 \times 10^{-4} \mathrm{~m} .
\end{aligned}$
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