AP EAMCET · PHYSICS · Mechanical Properties of Solids
The elastic limit of a metal is \(\frac{400}{\pi} \mathrm{MPa}\). If a rod of this metal is to support a \(484 \mathrm{~N}\) load without exceeding its elastic limit, the minimum diameter of the rod is
- A \(2.2 \mathrm{~mm}\)
- B \(1.2 \mathrm{~mm}\)
- C \(2 \mathrm{~mm}\)
- D \(1.6 \mathrm{~mm}\)
Answer & Solution
Correct Answer
(A) \(2.2 \mathrm{~mm}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { } \mathrm{P}=\frac{\mathrm{F}}{\mathrm{A}} \Rightarrow \frac{400}{\pi} \times 10^6=\frac{484}{\pi \frac{\mathrm{d}^2}{4}} \Rightarrow \mathrm{d}^2=\frac{484 \times 4 \times \pi}{400 \pi \times 10^6} \\ & \Rightarrow \mathrm{d}^2=4.84 \times 10^{-6} \\ &…
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