Use app×
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
0 votes
287 views
in Physics by (97.1k points)
closed by
A uniform string of length `1.5` m has two successive
harmonics of frequencies 70 Hz and 84 Hz. The
speed of the wave in the string (in `ms^(-1)`) is
A. 84
B. 42
C. 21
D. `10.5`

1 Answer

0 votes
by (97.2k points)
selected by
 
Best answer
Correct Answer - b
`because v_(n)=n/(2l)"v"` ...(i)
and `v_(n+1)=(n+1)/(2l)"v"` ...(ii)
From Equ. (i) and (ii) we get
`v_(n+1)-v_(n)=v/(2l)`
`rArr 84-70=v/(2l`
`therefore v=2 xx1.5xx14=42 " ms"^(-1)`

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

...