Question

One end of the steel rod is clamped to the roof and the other end is attached to a mass of 1000 kg as shown in the figure. The length of the rod is 50 cm and its cross-sectional area is 1000 mm$^2$. The change in the length of the rod due to the weight of the mass is \underline{\hspace{2cm}} .

• A. 0.025 mm
• B. 0.10 mm
• C. 0.050 mm
• D. 0.075 mm

Hint:

Young's modulus formula: $Y = \frac{FL}{A\Delta L}$. Rearrange to find $\Delta L$.
Ensure all units are consistent (SI units are recommended: N, m, m$^2$, Nm$^{-2}$).
Convert final answer to the units required by the options (mm in this case).
$1 \text{ mm} = 10^{-3} \text{ m}$, so $1 \text{ mm}^2 = (10^{-3} \text{ m})^2 = 10^{-6} \text{ m}^2$.

Answer 1

The Correct Option is A

Young's modulus ($Y$) is defined as the ratio of stress ($\sigma$) to strain ($\epsilon$): $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}$ where $F$ is the applied force, $A$ is the cross-sectional area, $L$ is the original length, and $\Delta L$ is the change in length. We need to find $\Delta L$: $\Delta L = \frac{FL}{AY}$. Given values: Mass $m_{load} = 1000 \text{ kg}$. Acceleration due to gravity $g = 10 \text{ ms}^{-2}$. Force $F = m_{load}g = 1000 \text{ kg} \times 10 \text{ ms}^{-2} = 10000 \text{ N}$. Original length $L = 50 \text{ cm} = 0.5 \text{ m}$. Cross-sectional area $A = 1000 \text{ mm}^2 = 1000 \times (10^{-3} \text{ m})^2 = 1000 \times 10^{-6} \text{ m}^2 = 10^{-3} \text{ m}^2$. Young's modulus for steel $Y = 2 \times 10^{11} \text{ Nm}^{-2}$. Substitute these values into the formula for $\Delta L$: $\Delta L = \frac{(10000 \text{ N}) \times (0.5 \text{ m})}{(10^{-3} \text{ m}^2) \times (2 \times 10^{11} \text{ Nm}^{-2})}$. $\Delta L = \frac{5000}{2 \times 10^{11-3}} \text{ m} = \frac{5000}{2 \times 10^8} \text{ m}$. $\Delta L = \frac{2500}{10^8} \text{ m} = 2500 \times 10^{-8} \text{ m} = 25 \times 10^{-6} \text{ m}$. To convert to millimeters (mm), since $1 \text{ m} = 1000 \text{ mm}$: $\Delta L = 25 \times 10^{-6} \text{ m} \times \frac{1000 \text{ mm}}{1 \text{ m}} = 25 \times 10^{-3} \text{ mm} = 0.025 \text{ mm}$. \[ \boxed{0.025 \text{ mm}} \]