TS EAMCET · Physics · Mechanical Properties of Solids
A steel rod has a radius of \(10 \mathrm{~mm}\) and a length of \(1 \mathrm{~m}\). A \(80 \mathrm{kN}\) force stretches it along its length. If the Young's modulus of the rod is \(2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\), then the change in length is
- A \(\frac{2}{\pi} \mathrm{mm}\)
- B \(\frac{4}{\pi} \mathrm{mm}\)
- C \(\frac{3}{\pi} \mathrm{mm}\)
- D \(1 \mathrm{~mm}\)
Answer & Solution
Correct Answer
(B) \(\frac{4}{\pi} \mathrm{mm}\)
Step-by-step Solution
Detailed explanation
We know that, Young's modulus is given by \(Y=\frac{\text { Stress }}{\text { Strain }}=\frac{F / a}{\Delta l / l}\) \(\ldots\) (i) or \(\quad \Delta l=\frac{F}{a} \times \frac{l}{Y}\) \(\ldots\) (ii) Given, \(\quad F=80 \mathrm{kN}=80 \times 10^3 \mathrm{~N}\)…
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