Mechanical Properties of Solids

JEE Main level, single correct, medium

10 of 118

Question 10 of 118

A lift of mass 16001600 kg is supported by thick iron wire. If the maximum stress which the wire can withstand is 4×1084 \times 10^8 N/m2^2 and its radius is 44 mm, then maximum acceleration the lift can take is _______ m/s2^2. (take g=10g = 10 m/s2^2 and π=3.14\pi = 3.14)

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Answer: (A) 2.562.56

Maximum stress σmax=4×108\sigma_{max} = 4 \times 10^8 N/m2^2

Radius of wire r=4×10−3r = 4 \times 10^{-3} m

Mass of lift m=1600m = 1600 kg

Maximum tension the wire can withstand is given by:

Tmax=σmax×πr2T_{max} = \sigma_{max} \times \pi r^2

Tmax=4×108×3.14×(4×10−3)2T_{max} = 4 \times 10^8 \times 3.14 \times (4 \times 10^{-3})^2

Tmax=4×108×3.14×16×10−6=20096T_{max} = 4 \times 10^8 \times 3.14 \times 16 \times 10^{-6} = 20096 N

When the lift accelerates upwards with maximum acceleration aa, the tension in the wire is:

Tmax=m(g+a)T_{max} = m(g + a)

20096=1600(10+a)20096 = 1600(10 + a)

10+a=200961600=12.5610 + a = \dfrac{20096}{1600} = 12.56

a=12.56−10=2.56a = 12.56 - 10 = 2.56 m/s2^2

Answer: 2.562.56

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