Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
111 views
in Physics by (97.1k points)
closed by
The fundamental frequency of a closed organ pipe of length `20 cm` is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is
A. 80 cm
B. 100 ccm
C. 120 cm
D. 140 cm

1 Answer

0 votes
by (97.1k points)
selected by
 
Best answer
Correct Answer - c
The fundamental frequencies of closed organ pipe
are given as, `v_(c)=v/(4l)`
`because` Second overtone of an open organ pipe `v_(2)=(3v)/(2l_(1))`
Where `l_(1)=` length of the open pipe
`because v_(c)=v_(2)`
`rArr v/(4l)=(3v)/(2l_(1))rArr l_(1)=6l=6xx20`
`therefore l_(1) = 120` cm

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

...