Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Can someone help me with this, - Properties of Solids and Liquids - JEE Main-2

A steel wire having a radius of 2.0 mm, carrying a load of 4 kg, is hanging from a ceiling. Given that g=3.1\; \pi ms^{-2}, what will be the tensile stress that would be developed in the wire ?

  • Option 1)

    4.8\times10^{6}Nm^{-2}

  • Option 2)

    5.2\times10^{6}Nm^{-2}
     

  • Option 3)

    3.1\times10^{6}Nm^{-2}

     

  • Option 4)

    6.2\times10^{6}Nm^{-2}

 

Answers (1)

best_answer

r=2mm

m=4kg, g=3.1\pi \; ms^{-2}

So, F = mg

Stress = \frac{F}{A}=\frac{mg}{\pi r^{2}}=\frac{m\pi \times3.1}{\pi (4)\times10^{-6}}

Stress = \frac{4\times3.1\times\pi }{\pi \times10^{-6}\times4}=3.1\times10^{6}N/m^{2}

 


Option 1)

4.8\times10^{6}Nm^{-2}

Option 2)

5.2\times10^{6}Nm^{-2}
 

Option 3)

3.1\times10^{6}Nm^{-2}

 

Option 4)

6.2\times10^{6}Nm^{-2}

Posted by

solutionqc

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 Paper 2 Toppers: 4 score 100 percentile; most from Kerala, one female topper
anam-khan

JEE Mains 2026 paper 2 result out on jeemain.nta.nic.in
anam-khan

JEE Mains 2026 session 2 B.Arch, B.Planning final answer key out, no question dropped

Latest Question