LIVE Course for free

Rated by 1 million+ students
Get app now
JEE MAIN 2023
JEE MAIN 2023 TEST SERIES
NEET 2023 TEST SERIES
NEET 2023
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
7.5k views
in Physics by (25.8k points)

A load of 981 N is suspended from a steel wire of radius 1 mm. What is the maximum angle through which the wire with the load can be deflected so that it does not break when the load passes through the equilibrium position. Breaking stress is 7.85 x 108 Nm-2

1 Answer

+1 vote
by (26.1k points)
selected by
 
Best answer

Let θ be the maximum angular deflection them

1/2mv2 = mg (l - lcosθ)

and T - mg = mv2/l

T = mg + mv2/l

T/πr2 = mg + 2 mg (l - lcosθ)

= 3mg - 2mg cosθ

For maximum value of θ, stress (=T/πr2) must be equal to breaking stress,

i.e., T/πr2 = 7.85 x 108

or, (3mg - 2mg cosθ)/(π x (10-3)2) = 7.85 x 108

or, cosθ = 0.2436 = cos 75.9°

or, θ = 75.9°

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

...