Q: A lift is tied with thick iron wires and its mass is 1000 kg. the minimum diameter of wire if the maximum acceleration of lift is 1.2 ms^{-2} and the maximum safe stress is 1.4 × 10^{8} N m^{-2} is : (g = 9.8 ms^{-2} )

(a) 0.00141 m

(b) 0.00282 m

(c) 0.005 m

(d) 0.01 m

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Ans: (d)

Sol: T = m (g + a)

T = 1000(9.8 + 1.2) = 11000 N

$ Stress = \frac{T}{A} = \frac{T}{\pi r^2}$

$1.4 \times 10^8 = \frac{11000}{\pi r^2} $

$1.4 \times 10^8 = \frac{11000}{\pi (D/2)^2} $

$D^2 = \frac{4 \times 11000}{\pi \times 1.4 \times 10^8}$

$D = \sqrt{\frac{4 \times 11000}{(22/7) \times 1.4 \times 10^8}}$

D = 0.01 m