Question

The elastic limit of a metal is \[ \frac{400}{\pi}\text{ MPa}. \] If a rod of this metal is to support a \(484\text{ N}\) load without exceeding its elastic limit, the minimum diameter of the rod is

• A. \(2.2\text{ mm}\)
• B. \(1.2\text{ mm}\)
• C. \(2\text{ mm}\)
• D. \(1.6\text{ mm}\)

Hint:

For rods under load, \[ \text{Stress}=\frac{F}{A} \] and for a circular cross-section, \[ A=\frac{\pi d^2}{4}. \]

Answer 1

The Correct Option is A

Step 1: Use the stress formula.
Stress is \[ \text{Stress}=\frac{\text{Force}}{\text{Area}} \] At elastic limit, \[ \frac{F}{A}=\frac{400}{\pi}\times 10^6 \] Given: \[ F=484\text{ N} \] For a circular rod, \[ A=\frac{\pi d^2}{4} \] Therefore, \[ \frac{484}{\pi d^2/4} = \frac{400}{\pi}\times 10^6 \]

Step 2: Simplify the equation.
\[ \frac{484\times 4}{\pi d^2} = \frac{400\times 10^6}{\pi} \] Cancelling \(\pi\), \[ \frac{1936}{d^2}=400\times 10^6 \] Thus, \[ d^2=\frac{1936}{400\times 10^6} \] \[ d^2=4.84\times 10^{-6} \] \[ d=\sqrt{4.84\times 10^{-6}} \] \[ d=2.2\times 10^{-3}\text{ m} \] \[ d=2.2\text{ mm} \]

Step 3: Final conclusion.
Hence, the minimum diameter of the rod is \[ \boxed{2.2\text{ mm}} \]