228 Views

4. A steel rod having circular cross section with diameter 5 cm and length 3 m is utilized to hang a block of 100 kg at its free end. Its upper end is fixed with the ceiling. Steel density is 8000kg/m3 . Find the total elongation of the rod, taking into account its own weight. Take E = 210 GPa.


rahulattri0190 21st Dec, 2020
Answer (1)
Ayush 30th Dec, 2020

Hello Candidate,

The formula for Young's Modulus of Elasticity for steel is given by: Y= Mass Attached* Length/ Elongation* Area of Cross Section.

Here, the value of Young's Modulus=21000 Pascal, Mass Attached=100 kg, Length=3 metres, Area of Cross Section= 22/7*5*5 cm2= .0019 metres2. So, elongation= 1000*3/ .0019*21000= 75.18 metres.

Hope that this answer helps you.

Thank You.

Related Questions

Amity University, Noida Law A...
Apply
700+ Campus placements at top national and global law firms, corporates and judiciaries
Amity University, Noida BBA A...
Apply
Ranked amongst top 3% universities globally (QS Rankings)
VIT Bhopal University | M.Tec...
Apply
M.Tech admissions open @ VIT Bhopal University | Highest CTC 52 LPA | Apply now
Amity University | M.Tech Adm...
Apply
Ranked amongst top 3% universities globally (QS Rankings).
Graphic Era (Deemed to be Uni...
Apply
NAAC A+ Grade | Among top 100 universities of India (NIRF 2024) | 40 crore+ scholarships distributed
Amity University Noida B.Tech...
Apply
Among Top 30 National Universities for Engineering (NIRF 2024) | 30+ Specializations | AI Powered Learning & State-of-the-Art Facilities
View All Application Forms

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books