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Question

The frequency of a radar is $$ 780\,MHz $$ . After getting reflected from an approaching aeroplane, the apparent frequency is more than the actual frequency by $$ 2.6\,kHz $$ . The aeroplane has a speed of


A
2km/s
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B
1km/s
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C
0.5km/s
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D
0.25km/s
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Solution

The correct option is C $$ 0.5\,km/s $$
$$ f' = \left (\dfrac{c + \nu_a}{c - \nu_a}  \right )f $$
where $$ c $$ is the velocity of the radio wave , an electromagnetic wave , i.e.,  $$ c = 3 \times 10^8\,m/s $$ and $$ \nu_s $$ is velocity of aeroplane .
 
      $$ f' - f = \left [ \dfrac{c + \nu_a}{c - \nu_a} - 1 \right ]f $$ 
$$ \Rightarrow $$             $$ \Delta f = \dfrac{2 \nu_a f}{c - \nu_a} $$ 
Since approaching aeroplane cannot have a speed comparable to the speed of electromagnetic wave , so $$ v_s \ll c . $$ 
$$ \therefore $$                      $$ \Delta f = \dfrac{2 \nu_a f}{c} $$ 
$$ \Rightarrow $$           $$ 2.6 \times 10^3 = \dfrac{2\nu_A(780 \times 10^6)}{3 \times 10^8} $$                               
 $$ \Rightarrow $$                    $$ \nu_A = 0.5 \times 10^3 \,m/s $$
                                 $$ = 0.5\,km/s $$ 

Physics

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