JEE Mains · Physics · STD 11 - Newtons laws of motion
- JEE Mains 2026
- Medium
- MCQ
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 \(\pi = 3.14\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(2.56\)
Step-by-step Solution
Detailed explanation
Maximum stress \(\sigma_{max} = 4 \times 10^8\) N/m\(^2\)
Radius of wire \(r = 4 \times 10^{-3}\) m
Mass of lift \(m = 1600\) kg
Maximum tension the wire can withstand is given by:
\(T_{max} = \sigma_{max} \times \pi r^2\)
\(T_{max} = 4 \times 10^8 \times 3.14 \times (4 \times 10^{-3})^2\)
\(T_{max} = 4 \times 10^8 \times 3.14 \times 16 \times 10^{-6} = 20096\) N
When the lift accelerates upwards with maximum acceleration \(a\), the tension in the wire is:
\(T_{max} = m(g + a)\)
\(20096 = 1600(10 + a)\)
\(10 + a = \dfrac{20096}{1600} = 12.56\)
\(a = 12.56 - 10 = 2.56\) m/s\(^2\)
Answer: \(2.56\)
Radius of wire \(r = 4 \times 10^{-3}\) m
Mass of lift \(m = 1600\) kg
Maximum tension the wire can withstand is given by:
\(T_{max} = \sigma_{max} \times \pi r^2\)
\(T_{max} = 4 \times 10^8 \times 3.14 \times (4 \times 10^{-3})^2\)
\(T_{max} = 4 \times 10^8 \times 3.14 \times 16 \times 10^{-6} = 20096\) N
When the lift accelerates upwards with maximum acceleration \(a\), the tension in the wire is:
\(T_{max} = m(g + a)\)
\(20096 = 1600(10 + a)\)
\(10 + a = \dfrac{20096}{1600} = 12.56\)
\(a = 12.56 - 10 = 2.56\) m/s\(^2\)
Answer: \(2.56\)
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