A submarine (A) travelling at \(18 \mathrm{~km} / \mathrm{hr}\) is being chased along the line of its velocity by another submarine (B) travelling at \(27 \mathrm{~km} / \mathrm{hr}\). B sends a sonar signal of 500 Hz to detect A and receives a reflected sound of frequency \(v\). The value of \(v\) is close to : (Speed of sound in water \(=1500 \mathrm{~ms}^{-1}\) )

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Waves & Sound

A submarine (A) travelling at

18 km / hr

is being chased along the line of its velocity by another submarine (B) travelling at

27 km / hr

. B sends a sonar signal of 500 Hz to detect A and receives a reflected sound of frequency

v

. The value of

v

is close to : (Speed of sound in water

= 1500 ms^{- 1}

)

Answer: (B)

First convert the speeds: submarine A moves at

18 km/h = 5 m/s

and B at

27 km/h = 7.5 m/s

. The sound travels out from B to A and reflects back, so apply the Doppler formula twice. The frequency reaching A is

f_{1} = 500 \frac{v - v _{A}}{v - v _{B}} = 500 \frac{1500 - 5}{1500 - 7.5}

, and the echo received back at B is

f_{2} = f_{1} \frac{v + v _{B}}{v + v _{A}} = 500 \frac{1500 - 5}{1500 - 7.5} \cdot \frac{1500 + 7.5}{1500 + 5} \approx 502 Hz

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option