A SONAR system fixed in submarine operates at a frequency of 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of $360 k m h^{- 1}$ Taking the speed of sound in water as $1450 m s^{- 1}$ , then frequency of the reflected sounds is

46 kHz

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51 kHz

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55 kHz

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60 kHz

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Solution

The correct option is A 46 kHz
freq $= 40 \times 10^{3} H z$
$V = 360 \times \frac{5}{18} = 100 m / s$
$E n e m y \to V$
$V_{s o u n d} = 1450$
$f^{1} = (\frac{V}{V_{S} - V}) f$
$f_{r e f} = (\frac{V}{V_{S} - V}) f (\frac{V_{S} + V}{V}) = (\frac{1550}{1350}) 40$
$f_{r e f} = 46 K H z$

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A SONAR system fixed in submarine operates at a frequency of 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360kmh−1 Taking the speed of sound in water as 1450ms−1, then frequency of the reflected sounds is

The correct option is A 46 kHzfreq =40×103HzV=360×518=100m/sEnemy→VVsound=1450f1=(VVS−V)ffref=(VVS−V)f(VS+VV)=(15501350)40fref=46KHz

A SONAR system fixed in submarine operates at a frequency of 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of $360 k m h^{- 1}$ Taking the speed of sound in water as $1450 m s^{- 1}$ , then frequency of the reflected sounds is

The correct option is A 46 kHz
freq $= 40 \times 10^{3} H z$
$V = 360 \times \frac{5}{18} = 100 m / s$
$E n e m y \to V$
$V_{s o u n d} = 1450$
$f^{1} = (\frac{V}{V_{S} - V}) f$
$f_{r e f} = (\frac{V}{V_{S} - V}) f (\frac{V_{S} + V}{V}) = (\frac{1550}{1350}) 40$
$f_{r e f} = 46 K H z$