A source of sound S emitting waves of frequency 100 Hz and an observer O are located at some distance from each other. The source is moving with a speed of 19.4 ms^-1 at an angle of 60° with the source observer line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer is (velocity of sound in air is 330 ms^-1)

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

103 Hz

100 Hz

97 Hz

106 Hz

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Here, original frequency of sound, f₀= 100 Hz

Speed of source V_s = 19.4 cos 60° = 9.7

From Doppler’s formula,

$f = f_{0} (\frac{V - V_{0}}{V - V_{s}}) f = 100 (\frac{V - 0}{V - (+ 9.7)}) f = 100 \frac{V}{V (1 - \frac{9.7}{V})} f = 100 (1 + \frac{9.7}{330}) = 103 H z$

Apparent frequency f = 103 Hz

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th