A lift of mass 1600 kg is supported by thick iron wire. If the maximum stress which the wire can withstand is 4 × 10⁸ N/m² and its radius is 4 mm, then maximum acceleration the lift can take is m/s². (take g = 10 m/s² and π = 3.14) (a) 2.56 (b) 3.89 (c) 4.32 (d) 5.16

Answer: (a). Maximum tension T = S π r² = 4 × 10⁸ × 3.14 × (4 × 10⁻³)² = 20096 N. For upward acceleration: T - mg = ma ⇒ a = T/m - g a = (20096)/(1600) - 10 = 12.56 - 10 = 2.56 m/s² So the answer is (a) 2.56.

A lift of mass 1600 kg is supported by thick iron wire. If the… | JEE Main 2026 (6 Apr, Shift 1)

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Thermal Stress & Breaking Stress

A lift of mass 1600 kg is supported by thick iron wire. If the maximum stress which the wire can withstand is $4 \times 10^{8}\text{ N/m}^2$ and its radius is 4 mm, then maximum acceleration the lift can take is $\underline{\qquad}\ \text{m/s}^2$ .

(take $\text{g} = 10\text{ m/s}^2$ and $\pi = 3.14$ )

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Answer: (a)

Maximum tension $T = S \pi r^2 = 4 \times 10^8 \times 3.14 \times (4 \times 10^{-3})^2 = 20096\text{ N}$ .
For upward acceleration: $T - mg = ma \Rightarrow a = \dfrac{T}{m} - g$
$a = \dfrac{20096}{1600} - 10 = 12.56 - 10 = 2.56\text{ m/s}^2$
So the answer is (a) 2.56.

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