Maximum Acceleration of Lift Supported by Iron Wire

To find the maximum acceleration that the lift can take, we need to calculate the maximum force (tension) the iron wire can support before reaching its breaking stress threshold.

1. Cross-sectional area of the wire: The cross-sectional area $A$ of the wire with radius $r = 4\text{ mm} = 4 \times 10^{-3}\text{ m}$ is given by: $A = \pi r^2$

Substituting the given values ($\pi = 3.14$): $A = 3.14 \times \left(4 \times 10^{-3}\text{ m}\right)^2$ $A = 3.14 \times 16 \times 10^{-6}\text{ m}^2 = 50.24 \times 10^{-6}\text{ m}^2$

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2. Maximum tension supported by the wire: The maximum allowable stress $\sigma_{\text{max}}$ is $4 \times 10^8\text{ N/m}^2$. The maximum tension $T_{\text{max}}$ that the wire can bear is: $T_{\text{max}} = \sigma_{\text{max}} \times A$ $T_{\text{max}} = \left(4 \times 10^8\text{ N/m}^2\right) \times \left(50.24 \times 10^{-6}\text{ m}^2\right)$ $T_{\text{max}} = 200.96 \times 10^2\text{ N} = 20096\text{ N}$

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3. Maximum acceleration of the lift: When the lift accelerates upwards with maximum acceleration $a_{\text{max}}$, the force balance equation is: $T_{\text{max}} = m(g + a_{\text{max}})$

Where:

• Mass of the lift, $m = 1600\text{ kg}$
• Acceleration due to gravity, $g = 10\text{ m/s}^2$

Substituting the values into the equation: $20096 = 1600(10 + a_{\text{max}})$ $10 + a_{\text{max}} = \frac{20096}{1600}$ $10 + a_{\text{max}} = 12.56$ $a_{\text{max}} = 12.56 - 10 = 2.56\text{ m/s}^2$

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Conclusion: The maximum acceleration the lift can take is $2.56\text{ m/s}^2$, which corresponds to option A.