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EAMCET Medical EAMCET Medical Solved Paper-2003

  • question_answer
    A radar sends a radio signal of frequency \[9\times {{10}^{9}}Hz,\] towards an aircraft approaching the radar. If the reflected wave shows a frequency shift of \[3\times {{10}^{8}}Hz.\] the speed with which the aircraft is approaching the radar, in m/s: (Velocity of the radio signal is\[3\times {{10}^{8}}m{{s}^{-1}}\])

    A)  150                                       

    B)  100

    C)  50                                         

    D)  20

    Correct Answer: C

    Solution :

                     Given : Frequency of radio signal \[n=9\times {{10}^{9}}Hz\] Frequency shift \[{{n}_{0}}=3\times {{10}^{3}}Hz\] Velocity of radio signal \[v=3\times {{10}^{8}}m/s\] Frequency shift shown by reflected wave is Shift \[=n-n\] \[n-n=\left( \frac{v+u}{v-u} \right)n-n\] \[\text{Shift}={{\left( \frac{2u}{v-u} \right)}^{n}}\] Now, putting the given values in Eq. (i), we get \[3\times {{10}^{3}}=\left( \frac{2u}{3\times {{10}^{-8}}-u} \right)9\times {{10}^{9}}\]                    \[\Rightarrow \]               \[3\times {{10}^{8}}-u=\frac{2u\times 9\times {{10}^{9}}}{3\times {{10}^{3}}}\]                                 \[=6u\times {{10}^{6}}\]                 \[\Rightarrow \]               \[6\times {{10}^{6}}u+u=3\times {{10}^{8}}\] \[u(6\times {{10}^{6}}+1)=3\times {{10}^{8}}\] \[u(6000001)=3\times {{10}^{8}}\] \[u=\frac{3\times {{10}^{8}}}{6000001}=49.999\] \[\approx 50\,m/s\]


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