Question

Overall changes in volume and radii of a uniform cylindrical steel wire are 0.2% and 0.002% respectively when subjected to some suitable force. Longitudinal tensile stress acting on the wire is :- (Y = 2.0 × 10¹¹ Nm⁻²)

• A. 3.2 × 10⁹ Nm⁻²
• B. 3.2 × 10⁷ Nm⁻²
• C. 3.6 × 10⁹ Nm⁻²
• D. 3.9 × 10⁸ Nm⁻²

Answer 1

Answer: (D)

3.9 × 10⁸ Nm⁻²

Answer 2

Let the original length of the wire be $L$, and the original radius be $r$. The original volume is $V = \pi r^2 L$. When subjected to a tensile force, the length increases by $\Delta L$ and the radius decreases by $\Delta r$. The longitudinal strain is $\epsilon_L = \frac{\Delta L}{L}$. The radial strain is $\epsilon_r = \frac{\Delta r}{r}$. For tensile stress, the radius decreases, so $\epsilon_r$ is negative.

The volume change can be related to the strains. Taking the natural logarithm of the volume formula: $\ln V = \ln(\pi r^2 L) = \ln \pi + 2 \ln r + \ln L$ Differentiating with respect to small changes: $\frac{dV}{V} = 2 \frac{dr}{r} + \frac{dL}{L}$ For small strains, this can be written as: $\frac{\Delta V}{V} \approx 2 \epsilon_r + \epsilon_L$

The problem states that the overall changes in volume and radii are 0.2% and 0.002% respectively. It is most consistent to interpret the longitudinal strain $\epsilon_L$ as $0.2\%$ and the radial strain $\epsilon_r$ as $-0.002\%$.

If $\epsilon_L = 0.2\% = 0.002$ and $\epsilon_r = -0.002\% = -0.00002$: The implied volume change is $\frac{\Delta V}{V} = \epsilon_L + 2\epsilon_r = 0.002 + 2(-0.00002) = 0.002 - 0.00004 = 0.00196$, which is $0.196\%$. This is very close to the stated $0.2\%$.

The longitudinal tensile stress ($\sigma$) is given by Young's modulus ($Y$) and longitudinal strain ($\epsilon_L$): $\sigma = Y \times \epsilon_L$ Given $Y = 2.0 \times 10^{11} \text{ Nm}^{-2}$ and $\epsilon_L = 0.002$: $\sigma = (2.0 \times 10^{11} \text{ Nm}^{-2}) \times (0.002)$ $\sigma = 4.0 \times 10^8 \text{ Nm}^{-2}$

This calculated value ($4.0 \times 10^8 \text{ Nm}^{-2}$) is closest to option (D) $3.9 \times 10^8 \text{ Nm}^{-2}$. Let's check if option (D) is consistent: If $\sigma = 3.9 \times 10^8 \text{ Nm}^{-2}$, then $\epsilon_L = \frac{\sigma}{Y} = \frac{3.9 \times 10^8}{2.0 \times 10^{11}} = 1.95 \times 10^{-3} = 0.00195$. This is $0.195\%$, very close to $0.2\%$. With $\epsilon_r = -0.00002$, the volume change would be $\frac{\Delta V}{V} = 0.00195 + 2(-0.00002) = 0.00195 - 0.00004 = 0.00191$, or $0.191\%$, which is also close to $0.2\%$. Therefore, option (D) is the most appropriate answer.