A cylindrical rod made of aluminum has length 1 meter and diameter of $10 \mathrm{~cm}$. The rod is…

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 $=70 \mathrm{GPa}$ )
  1. $0.81 \times 10^{-4} \mathrm{~m}$
  2. $2 \times 10^{-4} \mathrm{~m}$
  3. $0.2 \times 10^{-4} \mathrm{~m}$
  4. $1.81 \times 10^{-4} \mathrm{~m}$

Solution

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, $y=70 \times 10^{19} \mathrm{pa}$ $\begin{aligned} & \mathrm{Y}=\frac{\text { stress }}{\text { strain }}=\frac{\frac{\mathrm{F}}{\mathrm{A}}}{\frac{\Delta \mathrm{l}}{\mathrm{l}}}=\frac{\mathrm{F}}{\pi \mathrm{r}^2} \times \frac{1}{\Delta \mathrm{l}} \\ & \Delta \mathrm{l}=\frac{\mathrm{Fl}}{\pi \mathrm{r}^2 \mathrm{y}} \\ & =\frac{100 \times 10^3 \times 1}{3.14 \times(0.05)^2 \times 70 \times 10^9} \\ & =1.81 \times 10^{-4} \mathrm{~m} . \end{aligned}$

Asked in: AP EAMCET 2023 (17 May Shift 1)

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