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JEE Main 20206 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
6 September
Shift
Morning Shift
Subject
Physics
Chapter
Waves and Sound

JEE Main 2020 · 6 September · Morning Shift · Physics · Waves and Sound

A sound source S is moving along a straight track with speed v ,...

A sound source S is moving along a straight track with speed vv, and is emitting sound of frequency v0{v}_{0}.An observer is standing at a finite distance, at the point O,O, from the track. The time variation of frequency heard by observer is best represented by : (to represents the instant when the distance between the source and observer is minimum)A sound source S is moving along a straight track with speed v , and is emitting sound of frequency v _ 0 .An observerA sound source S is moving along a straight track with speed v , and is emitting sound of frequency v _ 0 .An observerA sound source S is moving along a straight track with speed v , and is emitting sound of frequency v _ 0 .An observerA sound source S is moving along a straight track with speed v , and is emitting sound of frequency v _ 0 .An observerA sound source S is moving along a straight track with speed v , and is emitting sound of frequency v _ 0 .An observerA sound source S is moving along a straight track with speed v , and is emitting sound of frequency v _ 0 .An observer

Options

  1. AA light ray passing through a prism having a prism angle of 60^ ^ ^ is observed to deviate by 30^ ^ ^ . What limit can
  2. BA light ray passing through a prism having a prism angle of 60^ ^ ^ is observed to deviate by 30^ ^ ^ . What limit can
  3. CA light ray passing through a prism having a prism angle of 60^ ^ ^ is observed to deviate by 30^ ^ ^ . What limit can
  4. DA light ray passing through a prism having a prism angle of 60^ ^ ^ is observed to deviate by 30^ ^ ^ . What limit can

Correct answer

B. A light ray passing through a prism having a prism angle of 60^ ^ ^ is observed to deviate by 30^ ^ ^ . What limit can

Step-by-step solution

A sound source S is moving along a straight track with speed v , and is emitting sound of frequency v _ 0 .An observer

Observed frequency will be v=Vsound+Vcos\thetaVsoundv0v = \dfrac{{V}_{\text{sound}} + Vcos\text{\theta}}{{V}_{\text{sound}}}{v}_{0}

Using this formula we can say that observed frequency will decrease continuously and it becomes equal to original frequency when \theta=90°\text{\theta} = 90 °. So correct graph is

A sound source S is moving along a straight track with speed v , and is emitting sound of frequency v _ 0 .An observer

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