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
54 views
in Physics by (79.2k points)
closed by
A small source of sound vibrating frequency 500 Hz is rotated in a circle of radius `(100)/(pi)` cm at a constant angular speed of `5.0` revolutions per second. The speed of sound in air is `330(m)/(s)`.
Q. If the observer moves towards the source with a constant speed of `20(m)/(s)`, along the radial line to the centre, the fractional change in the apparent frequency over the frequency that the source will have if considered at reat at the centre will be
A. `6%`
B. `3%`
C. `2%`
D. `9%`

1 Answer

0 votes
by (86.4k points)
selected by
 
Best answer
Correct Answer - A
The frequency of the source considered stationary at the centre is `n_0=500Hz` when the observer moves towards the centre with constant speed u, the apparent frequency is
`n_a=((V+u)/(V))500=9330+(20)/(330)xx500=(35)/(33)xx500`
Change in frequency is given by
`n_a-n_0=((35)/(33)xx500-500)=(2xx500)/(33)Hz`
Fractional change of frequency is
`((n_s-n_0)/(n_0))=((2xx500)/(33))xx(1)/(500)`
`=(2)/(33)=0.06`
Hence, percentage change is `6%`

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

...