Use app×
Ask a Question
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

The equation of a transverse travelling on a rope is given by `y=10sinpi (0.01x-2.00t)` where y and x are in cm and t in seconds. The maximum transver

← Prev Question Next Question →
0 votes
143 views
in Physics by (79.4k points)
closed by
The equation of a transverse travelling on a rope is given by `y=10sinpi (0.01x-2.00t)` where y and x are in cm and t in seconds. The maximum transverse speed of a particle in the rope is about
A. `63 cm//s`
B. `75 cm//s`
C. `100 cm//s`
D. ` 121 cm//s`
Play Quiz Game >

1 Answer

0 votes
by (86.6k points)
selected by
 
Best answer
Correct Answer - a
Standard equation of travelling wave `y=A sin(kx-omegat)` by comparing with the given equation `y=10 sin (0.01pix-2pit)`
`A=10 cm,omega=2pi`
maximum particle velocity `=Aomega=2pixx10=63 cm//s`
← Prev Question Next Question →

Find MCQs & Mock Test

  1. JEE Main 2025 Test Series
  2. NEET Test Series
  3. Class 12 Chapterwise MCQ Test
  4. Class 11 Chapterwise Practice Test
  5. Class 10 Chapterwise MCQ Test
  6. Class 9 Chapterwise MCQ Test
  7. Class 8 Chapterwise MCQ Test
  8. Class 7 Chapterwise MCQ Test

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.

Categories

User Contribution to Sarthaks eConnect
Managed by Sarthaks Digitizing India
...