Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
2.6k views
in Physics by (138k points)
A piece of copper having a rectangular cross section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?

1 Answer

0 votes
by (93.8k points)
selected by
 
Best answer

Length of the piece of copper, l = 19.1 mm = 19.1 × 10–3 m

Area of the copper piece:

A = l × b = 19.1 × 10–3 × 15.2 × 10–3 = 2.9 × 10–4 m2

Tension force applied on the piece of copper, F = 44500 N

Modulus of elasticity of copper, η = 42 × 109 N/m2

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...