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JEE Main 201912 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
12 April
Shift
Morning Shift
Subject
Physics
Chapter
Waves and Sound

JEE Main 2019 · 12 April · Morning Shift · Physics · Waves and Sound

A submarine A travelling at 18km hr ^ - 1 is being chased along the...

A submarine AA travelling at 18kmhr−118km{hr}^{- 1} is being chased along the line of its velocity by another submarine BB travelling at 27kmhr−127km{hr}^{- 1} . BB sends a sonar signal of 500Hz500Hz to detect AA and receives a reflected sound of frequency vv. The value of vv is closed to (Speed of sound in water 1500ms−11500m{s}^{- 1} )

Options

  1. A504Hz504Hz
  2. B499Hz499Hz
  3. C502Hz502Hz
  4. D507Hz507Hz

Correct answer

C. 502Hz502Hz

Step-by-step solution

VA=18kmh−1=5ms−1{V}_{A} = 18km{h}^{- 1} = 5m{s}^{- 1} ; VB=27kmh−1=7.5ms−1{V}_{B} = 27km{h}^{- 1} = 7 .5 m{s}^{- 1}

Frequency received by AA, fA=f0(1500−51500−7.5){f}_{A} = {f}_{0}\left(\dfrac{1500 - 5}{1500 - 7 .5}\right)

Frequency of reflected wave as heard by BB.

f′B=fA(1500+7.51500+5){f}^{'}B = {f}_{A}\left(\dfrac{1500 + 7 .5}{1500 + 5}\right)

=f0(1500−51500−7.5)(1500+7.51500+5)= {f}_{0}\left(\dfrac{1500 - 5}{1500 - 7.5}\right)\left(\dfrac{1500 + 7.5}{1500 + 5}\right)

=500(14951492.5)(1507.51505)= 500\left(\dfrac{1495}{1492.5}\right)\left(\dfrac{1507.5}{1505}\right)

=501.67Hz≈502Hz= 501.67Hz \approx 502Hz
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