AP EAMCET · PHYSICS · Mechanical Properties of Solids
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 \(=70 \mathrm{GPa}\) )
- A \(0.81 \times 10^{-4} \mathrm{~m}\)
- B \(2 \times 10^{-4} \mathrm{~m}\)
- C \(0.2 \times 10^{-4} \mathrm{~m}\)
- D \(1.81 \times 10^{-4} \mathrm{~m}\)
Answer & Solution
Correct Answer
(D) \(1.81 \times 10^{-4} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
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, \(y=70 \times 10^{19} \mathrm{pa}\)…
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