Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

A uniform heavy rod of mass \mathrm{ 20 \mathrm{~kg},} cross sectional area \mathrm{ 0.4 \mathrm{~m}^{2}} and length \mathrm{ 20 \mathrm{~m}} is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is \mathrm{ x \times 10^{-9} \mathrm{~m}. }The value of \mathrm{ x} is ____________.

\mathrm{ \text { (Given, young modulus } \mathrm{Y}=2 \times 10^{11} \mathrm{Nm}^{-2} \text { and } g=10 \mathrm{~ms}^{-2} \text { ) }}

Option: 1

25


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{M=20 \mathrm{~kg} }

\mathrm{A=0.4 \mathrm{~m}^2 }

\mathrm{\ell=20 \mathrm{~m} }

\mathrm{\Delta \ell=x \times 10^{-9} \mathrm{~m}}

Elongation due to its own weight

\mathrm{\Delta \ell =\frac{M g \ell}{2 A Y} }

\mathrm{=\frac{20 \times 10 \times 20}{2 \times 0.4 \times 2 \times 10^{11}} }

\mathrm{=\frac{4 \times 10^3}{16 \times 10^{10}} }

\mathrm{\Delta \ell =25 \times 10^{-9} \mathrm{~m} }

\mathrm{\therefore x =25}






 

Posted by

Sayak

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, NEET, CUET coaching centres: Assam, Rajasthan have framed laws, but will they work?
anam-khan

JEE Main 2025 withheld results declared; NTA issues show cause notices ahead of action
anam-khan

JCECEB 2025 registration for BTech admission through JEE Main rank begins jceceb.jharkhand.gov.in

Latest Question