A string of mass 2.5 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, the disturbance will reach the other end in

(a) one second

(b) 0.5 second

(c) 2 seconds

(d) data given is insufficient.

2 Views|Posted 7 months ago
Asked by Shiksha User
1 Answer
P
7 months ago

This is a multiple choice answer as classified in NCERT Exemplar

(b) m=2.5kg

μ = mass per unit length

=m/l =2.5/20=125/10=0.125kg/m

V= T μ = 200 0.125

L=v * t

20= 125 2 * 10 5 = 20 * 25 * 5 2 * 10 5

= 20 * 25 0.4 * 10 5 = 1 2

Thumbs Up IconUpvote Thumbs Down Icon

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

66K
Colleges
|
1.2K
Exams
|
6.9L
Reviews
|
1.8M
Answers

Learn more about...

Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen 2025

Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen 2025

View Exam Details

Most viewed information

Summary

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

Have a question related to your career & education?

or

See what others like you are asking & answering