Home JEE Main - Previous Year Papers JEE MAIN 2020 PAPER-6 A string of mass per unit length is fixed a…
JEE Main - Previous Year Papers JEE MAIN 2020 PAPER-6 Single Correct MCQ
Published on: August 22, 2026

A string of mass per unit length is fixed at both ends under the tension . If the string is in resonance with consecutive frequencies and . Then what would be the length of the string?

A
B

C
D

Share this question

For Instagram sharing, use “Apps” on mobile or copy the link.

Text Solution

Verified by Experts
The correct answer is:
B

Key Idea: The difference of two consecutive resonant frequencies is the fundamental resonant frequency.

Fundamental frequency

──────────────────────────────────────────────────────────────────────────────────────────

Prepare Smarter with CGP Edu

Get practice questions, solutions, and test series in one place.

Write a Review

Share your experience with this question and solution.

Commentary

Send your comment, doubt, correction, or feedback to admin.

Student Reviews

What students say about this solution

No reviews yet. Be the first to write a review.

Similar Questions

Explore conceptually related problems

CG
CGP Question Assistant Question Bank + AI Help
Hi! Type your question or upload one screenshot. First I will search related questions from CGP Edu Question Bank. If none match, type YES and I will solve it with AI.
Upload only one screenshot at a time. Flow: Question Bank first → If not matched, type YES for AI solution.