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An indian submarine is moving in the Arabian sea with constant velocity. To detect enemy it sends out sonar waves which travel with velocity 1050(m)/(s) in water. Initially the waves are getting reflected from a fixed island and the reflected waves are coming back to submarine. The frequency of reflected waves are detected by the submarine and found to be 10% greater than the sent waves. <br> Now an enemy ship comes in front, due to which the frequency of reflected waves detected by submarine becomes 21% greater than the sent waves. <br> Q. The velocity of enemy ship should be <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_VO6_C06_E01_138_Q01.png" width="80%">

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Answer Text

`50(m)/(s)` towards indian submarine`50(m)/(s)` away from indian submarine`100(m)/(2)` towards indian submarine`100(m)/(s)` away from indian submarine

Answer :
A
Solution :
<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_VO6_C06_E01_138_S01.png" width="80%"> <br> `f'=f_0((v+v_2)/(v-50))`,`f''=f'((v+50)/(v-v_2))` <br> `f''=f_0(((v+v_2)(v+50))/((v-v_2)(v-50)))=1.21f_0` <br> [`21%` greater than sent waves] <br> we get `v_2=50(m)/(s)` towards indian submarine.

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