Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

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

Option: 1

3.22 \, cm^{2}


Option: 2

2.38 \, cm^{2}


Option: 3

0.32 \, cm^{2}


Option: 4

8.23 \, cm^{2}


Answers (1)

best_answer

\begin{aligned}T_{\text {max }}=m(g+a) \\ T_{\text {max }}=(2000)(9.8+1.5) \\ T_{\text {max }}=22600 \mathrm{~N} \end{aligned}\\ \begin{aligned} \text { Maximum stress } & =\frac{T_{\text {max }}}{\text { Area }} \\ \therefore \text { Area } & =\frac{{T_{\text {max }}}}{\text { Maximum Stress }} \\ \text { Area } & =\frac{22600}{7 \times 10^7} \\ & =3.22 \times 10^{-4} \mathrm{~m}^2 \\ & =3.22 \mathrm{~cm}^2 \end{aligned}

Posted by

Nehul

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

How to crack JEE Mains 2025 in 1 month? Follow study plan, revise high-score topics, experts advice
anam-khan

IIT JEE aspirant from Bihar found hanging in Kota hostel; 17th suicide this year
anam-khan

JEE Main 2025 just a month away; section-wise preparation tips, important topics, strategies

Latest Question