A lift of mass 1600 kg is supported by thick iron wire. If the maximum stress which the wire can withstand is 4 x 10^8 N/m^2 and its radius is 4 mm, then maximum acceleration the lift can take is _______ m/s^2. (take g = 10 m/s^2 and π = 3.14)

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q30 – JEE-MAIN 2026

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Home Physics Mechanical Properties of SolidsQuestion #291

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} {N/m}^{2}

and its radius is

4 mm

, then maximum acceleration the lift can take is _______

{m/s}^{2}

. (take

g = 10 {m/s}^{2}

and

π = 3.14

)

Clue 1 of 3: Understanding Stress

Stress is defined as the force per unit area acting on a material.

🧠 Full Solution Path

Step 1: Calculate the cross-sectional area of the wire✦ Active

The radius of the wire is $r = 4 mm = 4 \times 10^{- 3} m$ . The cross-sectional area $A$ of the circular wire is given by the formula:

A = π r^{2} = 3.14 \times (4 \times 10^{- 3} m)^{2} = 3.14 \times 16 \times 10^{- 6} m^{2} = 50.24 \times 10^{- 6} m^{2}

Step 2: Determine the maximum tension the wire can withstand○ Expand

The maximum stress $σ_{m a x}$ the wire can withstand is given as $4 \times 10^{8} {N/m}^{2}$ . The maximum force (tension) $F_{m a x}$ the wire can support is calculated using the stress formula:

F_{m a x} = σ_{m a x} \times A = (4 \times 10^{8} {N/m}^{2}) \times (50.24 \times 10^{- 6} m^{2}) = 200.96 \times 10^{2} N = 20096 N

Step 3: Calculate the maximum acceleration of the lift○ Expand

When the lift accelerates upwards with maximum acceleration $a_{m a x}$ , the net force acting on it is $F_{m a x} - m g$ . According to Newton's second law, $F_{n e t} = m a_{m a x}$ . The mass of the lift is $m = 1600 kg$ and $g = 10 {m/s}^{2}$ .

F_{m a x} - m g = m a_{m a x}

a_{m a x} = \frac{F_{m a x} - m g}{m} = \frac{20096 N - (1600 kg \times 10 {m/s}^{2})}{1600 kg}

a_{m a x} = \frac{20096 - 16000}{1600} = \frac{4096}{1600} = 2.56 {m/s}^{2}

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Physics Question 30 – JEE-MAIN 2026

Clue 1 of 3: Understanding Stress

🧠 Full Solution Path