Join Sarthaks eConnect Today - Largest Online Education Community!
0 votes
asked in Physics by (28.3k points)

A steel wire of diameter 0.8 mm and length 1 m is clamped firmly at two points A and B which are 1 m apart and in the same horizontal plane. A body is hung from the middle point of the wire such that the middle point sags 1 cm lower from the original position. Calculate the mass of the body. Given Young’s modulus of the material of wire = 2 × 1012 dynes/cm2.

1 Answer

+1 vote
answered by (24.7k points)
selected by
Best answer

Let the body be hung from the middle point C so that it sags through 1 cm to the point D as shown in the figure

∴ AD2 = AC2 + CD2 = (50)2 + (1)2

or AD = 50.01 cm

Increase in length = 0.01

Strain = 0.01/50 = 2 x 10-4

Stress = 2 × 1012 × 2 × 10–4

= 4 × 108 dynes/cm2

Tension T = Stress × Area of cross-section

= 4 × 108 × π × (0.08)2

Since the mass m is in equilibrium,

mg = 2T cosθ or m = 2T cosθ/g

Related questions

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.

One Thought Forever

“There is a close connection between getting up in the world and getting up in the morning.“
– Anon