Q55 / 75 · JEE Main 2026 (06 April Shift 1) · Physics
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 $\pi = 3.14$)
- Correct answerA$2.56$
- B$3.89$
- C$4.32$
- D$5.16$
Solution
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: $2.56$
Hence, the correct answer is option: 'A'.