Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

A SONAR system fixed in a submarine operates at a frequency of 60 KHz. An enemy submarine moves towards the SONAR with a speed of 720 Km/h. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1600 m/s

Option: 1
7224.3 KHZ

Option: 2
7714.2 KHZ

Option: 3
7625.2 KHZ

Option: 4
7494.2 KHZ

Answers (1)

best_answer

Here, frequency of SONAR (source)

$V=60 \mathrm\,{KHZ}=60 \times 10^3$

speed of observers, $v_0=720 \mathrm{~km} / \mathrm{h}$
$=720 \times \frac{5}{18}=40 \times 5=200 \mathrm{~m} / \mathrm{s}$
Since the source is at rest and observer moves towards the source ( SONAR )

$\therefore \quad V^{\prime} =\frac{V+V_0}{V} \cdot v \\$

$V^{\prime} =\frac{1600+200}{1600} \times 60 \times 10^3 \\$

$V^{\prime} =\frac{1800}{1600} \times 60 \times 10^3 \\$

$V^{\prime} =\frac{18}{16} \times 60 \times 10^3 \\$

$V^{\prime} =\frac{1080}{16} \times 10^3 \\$

$V^{\prime} =67.5 \times 10^3 \mathrm{~Hz}$

This frequency (v') is reflected by the enemy ship and is observed by the SONAR (which now acts as observer)

Therefore, in this case $V_s=720 \mathrm{~km} / \mathrm{h}$ $=200 \mathrm{~m} / \mathrm{s}$

$\therefore \quad$ Apparent frequency $V^{\prime \prime}=\frac{V}{V-VS} \cdot V^{\prime}$

$=\frac{1600}{1600-200} \times 67.5 \times 10^3 \\$

$=\frac{1600}{1400} \times 67.5 \times 10^3 \\$

$=\frac{16}{14} \times 67.5 \times 10^3 \\$

$=\frac{8 \times 675 \times 10^3}{7 \times 10}=\frac{8 \times 675}{7} \times 10^2 \\$

$=\frac{5400}{7} \times 10^2 \\$

$=771.42 \times 10^2 \mathrm{~Hz}=7714.2 \mathrm\,{KHZ}$

Posted by

Divya Prakash Singh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

BTech Admission 2025 Live: JEE Main correction window closes at 11:50 pm; VITEEE, SRMJEE, AEEE dates out

anam-khan

JEE Main 2025 application form correction begins at jeemain.nta.nic.in; editable fields

anam-khan

JEE Main 2025 application form correction window opens today; apply at jeemain.nta.nic.in

Latest Question