Question

If a steel rod of a radius 10 mm and length 80 cm is streched by a force of 66 kN along its length, then the longitudinal stress on the rod is nearly

• A. $2.1 \times 10^2 \text{ N m}^{-2}$
• B. $2.1 \times 10^4 \text{ N m}^{-2}$
• C. $2.1 \times 10^5 \text{ N m}^{-2}$
• D. $2.1 \times 10^8 \text{ N m}^{-2}$

Hint:

When solving problems involving stress, strain, or other mechanical properties, always ensure that all physical quantities are converted to consistent SI units (meters, kilograms, seconds, Newtons) before performing calculations. Remember that stress is force per unit area, and the area calculation depends on the cross-sectional shape of the object. The length of the rod is not needed to calculate stress, only for strain or Young's modulus.

Answer 1

The Correct Option is D

Step 1: Identify the given information and the quantity to be calculated.
Given: \begin{itemize} \item Radius of the steel rod ($r$) = 10 mm. \item Length of the steel rod ($L$) = 80 cm (This is additional information not required to calculate stress). \item Force applied ($F$) = 66 kN. \end{itemize} We need to calculate the longitudinal stress ($\sigma$) on the rod. Step 2: Convert the given values to SI units.
Radius: $r = 10 \text{ mm} = 10 \times 10^{-3} \text{ m} = 0.01 \text{ m}$ Force: $F = 66 \text{ kN} = 66 \times 10^3 \text{ N}$ Step 3: Recall the formula for longitudinal stress.
Longitudinal stress ($\sigma$) is defined as the force ($F$) applied per unit cross-sectional area ($A$) of the object: $\sigma = \frac{F}{A}$ Step 4: Calculate the cross-sectional area of the rod.
Since the rod is circular, its cross-sectional area ($A$) is given by the formula for the area of a circle: $A = \pi r^2$ $A = \pi (0.01 \text{ m})^2$ $A = \pi \times 0.0001 \text{ m}^2$ $A = \pi \times 10^{-4} \text{ m}^2$ Using the approximation $\pi \approx 3.14159$: $A \approx 3.14159 \times 10^{-4} \text{ m}^2$ Step 5: Calculate the longitudinal stress.
Now, substitute the values of $F$ and $A$ into the stress formula: $\sigma = \frac{66 \times 10^3 \text{ N}}{3.14159 \times 10^{-4} \text{ m}^2}$ $\sigma = \frac{66}{3.14159} \times \frac{10^3}{10^{-4}} \text{ N/m}^2$ $\sigma \approx 21.008 \times 10^{(3 - (-4))} \text{ N/m}^2$ $\sigma \approx 21.008 \times 10^7 \text{ N/m}^2$ To express this in the form of the options, we adjust the decimal: $\sigma \approx 2.1008 \times 10^8 \text{ N/m}^2$ Step 6: Compare the calculated stress with the given options.
The calculated stress is approximately $2.1 \times 10^8 \text{ N m}^{-2}$, which matches option (4). The final answer is $\boxed{\text{2.1 \times 10^8 N m^{-2}}}$.