The elastic limit of a metal is $\frac{400}{\pi} \mathrm{MPa}$. If a rod of this metal is to support a $484…

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
  1. $2.2 \mathrm{~mm}$
  2. $1.2 \mathrm{~mm}$
  3. $2 \mathrm{~mm}$
  4. $1.6 \mathrm{~mm}$

Solution

$\begin{aligned} & \text { } \mathrm{P}=\frac{\mathrm{F}}{\mathrm{A}} \Rightarrow \frac{400}{\pi} \times 10^6=\frac{484}{\pi \frac{\mathrm{d}^2}{4}} \Rightarrow \mathrm{d}^2=\frac{484 \times 4 \times \pi}{400 \pi \times 10^6} \\ & \Rightarrow \mathrm{d}^2=4.84 \times 10^{-6} \\ & \Rightarrow \mathrm{d}=2.2 \times 10^{-3} \mathrm{~m}\end{aligned}$

Asked in: AP EAMCET 2022 (06 Jul Shift 1)

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