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NEET PHYSICSWAVE OPTICSMedium

Question

The driver of a car travelling at a speed of 30 m/s30 \text{ m/s} towards a hill sounds a horn of frequency 600 Hz600 \text{ Hz}. If the velocity of sound in air is 330 m/s330 \text{ m/s}, the frequency of reflected sound as heard by the driver is:

A

550 Hz550 \text{ Hz}

B

555.5 Hz555.5 \text{ Hz}

C

720 Hz720 \text{ Hz}

D

500 Hz500 \text{ Hz}

Step-by-Step Solution

Let the velocity of sound be v=330 m/sv = 330 \text{ m/s}, the velocity of the car be vc=30 m/sv_c = 30 \text{ m/s}, and the actual frequency of the horn be f=600 Hzf = 600 \text{ Hz}. When the sound travels towards the hill, the car acts as a moving source approaching a stationary reflector (the hill). The apparent frequency f1f_1 received by the hill is: f1=f(vvvc)f_1 = f \left( \frac{v}{v - v_c} \right) When the sound is reflected from the hill, the hill acts as a stationary source emitting frequency f1f_1, and the car acts as an observer moving towards the source. The frequency heard by the driver is: f=f1(v+vcv)f' = f_1 \left( \frac{v + v_c}{v} \right) Substituting f1f_1 into the second equation gives the formula for the apparent frequency of an echo heard by a moving source: f=f(v+vcvvc)f' = f \left( \frac{v + v_c}{v - v_c} \right) f=600(330+3033030)f' = 600 \left( \frac{330 + 30}{330 - 30} \right) f=600(360300)f' = 600 \left( \frac{360}{300} \right) f=600×1.2=720 Hzf' = 600 \times 1.2 = 720 \text{ Hz} Therefore, the driver hears a frequency of 720 Hz720 \text{ Hz}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WAVE OPTICS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWAVE OPTICSdrivertravellingtowardssoundsfrequency

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