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AIIMS 2014
Exam
AIIMS
Year
2014
Subject
Physics
Chapter
Motion in One Dimension

AIIMS 2014 · Physics · Motion in One Dimension

A steel wire with cross-section 3 ~cm^2 has elastic limit 2.4 ×...

A steel wire with cross-section 3 cm23 \mathrm{~cm}^2 has elastic limit 2.4×108 N m−22.4 \times 10^8 \mathrm{~N} \mathrm{~m}^{-2}. The maximum upward acceleration that can be given to a 1200 kg1200 \mathrm{~kg} elevator supported by thiscable wire if the stress is not to exceed one-third of the elastic limit is (Take g=10 m s−2g=10 \mathrm{~m} \mathrm{~s}^{-2} )

Options

  1. A12 m s−212 \mathrm{~m} \mathrm{~s}^{-2}
  2. B10 m s−210 \mathrm{~m} \mathrm{~s}^{-2}
  3. C8 m s−28 \mathrm{~m} \mathrm{~s}^{-2}
  4. D7 m s−27 \mathrm{~m} \mathrm{~s}^{-2}

Correct answer

B. 10 m s−210 \mathrm{~m} \mathrm{~s}^{-2}

Step-by-step solution

Maximum tension an elevator can tolerate is T=13 stress × area of  cross-section =13×(2.4×108)×(3×10−4)=2.4×104N\begin{aligned} T & =\frac{1}{3} \text{ stress } \times \text{ area\text{ of } cross-section } \\ & =\frac{1}{3} \times\left(2.4 \times 10^8\right) \times\left(3 \times 10^{-4}\right)=2.4 \times 10^4 \mathrm{ N} \end{aligned} If aa is the maximum upward acceleration of elevator then T=m(g+a)or2.4×104=1200(10+a)T=m(g+a) or 2.4 \times 10^4=1200(10+a) On solving, a=10ms−2.a=10 \mathrm{ m} \mathrm{ s}^{-2} .
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