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 \[3\times {{10}^{3}}=\left( \frac{2u}{3\times {{10}^{-8}}-u} \right)9\times {{10}^{9}}\] \[\Rightarrow \] \[3\times {{10}^{8}}-u=\frac{2u\times 9\times {{10}^{9}}}{3\times {{10}^{3}}}\] \[=6u\times {{10}^{6}}\] \[\Rightarrow \] \[6\times {{10}^{6}}u+u=3\times {{10}^{8}}\] \[u(6\times {{10}^{6}}+1)=3\times {{10}^{8}}\] \[u(6000001)=3\times {{10}^{8}}\] \[u=\frac{3\times {{10}^{8}}}{6000001}=49.999\] \[\approx 50\,m/s\]You need to login to perform this action.
You will be redirected in
3 sec