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NEET Class neet Answered

please answer this (multiple options are correct)
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Asked by Prashant DIGHE | 29 May, 2020, 10:17: PM
answered-by-expert Expert Answer
In this question the plate behaves like moving observer. So we will solve the problem using the concepts of doppler effect.
B y space d o p p l e r apostrophe s space f o r m u l a f equals f subscript 0 open parentheses fraction numerator V plus-or-minus V subscript 0 over denominator V plus-or-minus V subscript s end fraction close parentheses f equals f subscript 0 open parentheses fraction numerator V plus V subscript 0 over denominator c end fraction close parentheses A s space t h e space o b j e c t s space a r e space m o v i n g space c l o s e space s o space t h e space space f r e q u e n c y space w i l l space i n c r e a s e s f equals f subscript 0 open parentheses fraction numerator c plus v over denominator c end fraction close parentheses N o w space f o r space f e l e c t i o n space f r o m space t h e space p l a t e comma space t h e space p l a t e space b e c o m e s space t h e space s o u r c e T h e space o r i g i n a l space f r e q u e n c y space o f space t h e space s o u r c e space i s space n o w space f subscript 0 open parentheses fraction numerator V plus c over denominator V end fraction close parentheses comma space a l s o space t h e space s o u r c e space i s space m o v i n g. s o space w e space c a n space w i t e space t h e space e q u a t i o n space u sin g space d o p p l e r apostrophe s space e f f e c t f equals f to the power of apostrophe open parentheses fraction numerator V plus-or-minus V subscript 0 over denominator V plus-or-minus V subscript s end fraction close parentheses space w h e r e space f to the power of apostrophe equals f subscript 0 open parentheses fraction numerator V plus c over denominator V end fraction close parentheses comma f equals f subscript 0 open parentheses fraction numerator c plus v over denominator c end fraction close parentheses open parentheses fraction numerator c over denominator c minus v end fraction close parentheses O n space s i m p l i f i c a t i o n f equals f subscript 0 open parentheses fraction numerator c plus v over denominator c minus v end fraction close parentheses W a v e l e n g t h space v equals f lambda c equals f subscript 0 open parentheses fraction numerator c plus v over denominator c minus v end fraction close parentheses lambda c over f subscript 0 open parentheses fraction numerator c minus v over denominator c plus v end fraction close parentheses equals lambda N u m b e r space o f space b e a t s space equals f subscript 2 minus f subscript 1 f subscript 0 open parentheses fraction numerator c plus v over denominator c minus v end fraction close parentheses minus f subscript 0 f subscript 0 open parentheses fraction numerator c plus v over denominator c minus v end fraction minus 1 close parentheses f subscript 0 open parentheses fraction numerator 2 v over denominator c minus v end fraction close parentheses
So a,b and c are correct
Answered by Utkarsh Lokhande | 30 May, 2020, 08:54: AM

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