Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • An elevator cable is to have a maximum stress of <img alt="7 \times 10^7 \mathrm{~N} / \mathrm{m}^2" src="https://entrancecorner.oncodecogs.com/gif.latex?7%20%5Ctimes%2010%5E7%20%5Cmathrm%7B%7

An elevator cable is to have a maximum stress of 7 \times 10^7 \mathrm{~N} / \mathrm{m}^2 to allow for appropriate safety factors. Its maximum upward acceleration is 1.5 \mathrm{~m} / \mathrm{s}^2. If the cable has to support the total weight of 2000kg of a loaded elevator, the area of cross section of the cable should be

Option: 1

3.22 \mathrm{~cm}^2


Option: 2

2.38 \mathrm{~cm}^2


Option: 3

0.32 \mathrm{~cm}^{2}


Option: 4

8.23 \mathrm{~cm}^2


Answers (1)

best_answer

Use, T_{\max }=m(g+a)

\begin{aligned} & T_{\text {max }}=(2000)(9.8+1.5) \\ & T_{\text {max }}=22600 \mathrm{~N} \end{aligned}

Maximum stress = \frac{T_{\text {max }}}{\text { Area }}

\therefore  Area = \frac{T_{\text {max }}}{\text { Maximum stress }}

Area = \frac{22600}{7 \times 10^{7}}

=3.22 \times 10^{-4} \mathrm{~m}^2

=3.22 \mathrm{~cm}^2

Posted by

Divya Prakash Singh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: JCECEB invites fresh applications for Jharkhand state merit list correction

Latest Question