QBManager

WAVES

172929 A car $P$ travelling at $20 {ms}^{- 1}$ sounds its horn at a frequency of $400 Hz$ . Another car $Q$ is travelling behind the first car in the same direction with a velocity $40 {ms}^{- 1}$ . The frequency heard by the passenger of the car $Q$ is approximately [Take, velocity of sound $= 360$ ${ms}^{- 1}$ ]

1

514 Hz

2

421 Hz

3

485 Hz

4

471 Hz

Explanation:

B Given
$f_{P} = 400 Hz$
$v_{s} = 360 m / s$
$v_{Q} = 40 m / s$
$v_{P} = 20 m / s$
We know that
$f_{Q} = f_{P} (\frac{v_{s} + v_{Q}}{v_{s} + v_{P}})$
$f_{Q} = 400 (\frac{360 + 40}{360 + 20})$
$f_{Q} = 400 (\frac{400}{380}) = 421 Hz$

JEE Main-11.04.2023

WAVES

172930 An observer moves towards a stationary source of sound, with a velocity equal to one-fifth of the velocity of sound. The percentage change in the frequency will be:

1

20 %

2

10 %

3

5 %

4

0 %

Explanation:

A Given, $v_{0} = \frac{v}{5}, v_{s} = 0$
Then, apparent frequency $(f^{'}) = f_{0} (\frac{v + v_{0}}{v - v_{s}})$
$f^{'} = f_{0} (\frac{v + v / 5}{v}) = \frac{6}{5} f_{0}$
Hence, percentage increase in frequency
$% f = \frac{(\frac{6}{5} - 1) f_{0}}{f_{0}} \times 100$
$\frac{1}{5} \times 100 = 20 %$
$% f = 20 %$

JEE Main-27.06.2022

WAVES

172932 A car is approaching a factory siren that has a frequency of $510 Hz$ . The speed of the sound in air is $340 {ms}^{- 1}$ . If the apparent frequency of the sound as heard by the car driver is $600 Hz$ . Then the speed of the car is

1

60 {ms}^{- 1}

2

33.3 {ms}^{- 1}

3

50.5 {ms}^{- 1}

4

80 {ms}^{- 1}

Explanation:

A Given that, frequency of sound hear by driver $(f^{'}) = 600 Hz$ , frequency of siren $(f_{o}) = 510 Hz$ , velocity of sound $(v_{s}) = 340 m / \sec$ and $v_{1} = 0$
Let, $v_{o} =$ velocity of car
We know that, $f^{'} = f (\frac{v_{s} + v_{o}}{v_{s} - v_{1}})$
$600 = 510 (\frac{340 + v_{0}}{340 - 0})$
$\frac{600 \times 2}{3} = 340 + v_{0}$
$400 = 340 + v_{0}$
$v_{0} = 60 m / \sec .$

AP EAMCET-11.07.2022

WAVES

172933 Two trains $A$ and $B$ are moving towards each other with speeds $72 {kmh}^{- 1}$ and $36 {kmh}^{- 1}$ respectively. The train-A whistles at $640 Hz$ frequency. Before the trains meet, frequency of sound heard by a passenger in Train-B is (Speed of sound in air $= 340 {ms}^{- 1}$ )

1

500 Hz

2

600 Hz

3

700 Hz

4

800 Hz

Explanation:

C $V_{A} = 72 km / h = 72 \times \frac{5}{18} = 20 m / \sec$
$V_{B} = 36 km / h = 36 \times \frac{5}{18} = 10 m / \sec$
$f_{B} = f_{A} [\frac{V + V_{B}}{V - V_{A}}]$
$= 640 [\frac{340 + 10}{340 - 20}]$
$= 640 \times \frac{350}{320} = 700 Hz$
$f_{B} = 700 Hz$

AP EAMCET-07.07.2022

WAVES

172934 A police car moving at $5.4 km / hr$ sounds siren emitting frequency of $550 Hz$ which is reflected back from a stationary object some distance ahead of the car. The number of beats heard per second by an observer sitting in the car is (Assume velocity of sound in air $= 330 m / \sec$ )

1 4

2 5

3 6

4 0

Explanation:

B Given that,
$f = 550 Hz$ ,
$v_{0} = v_{s} = 5.4 km / h = 1.5 m / s$
$v = 330 m / s$
$f^{'} = f (\frac{v + v_{0}}{v - v_{s}})$
$= 550 (\frac{330 + 1.5}{330 - 1.5})$
$= 555 Hz$
$Beat = 555 - 550$
$= 5$

UPSEE 2020

WAVES

172929 A car $P$ travelling at $20 {ms}^{- 1}$ sounds its horn at a frequency of $400 Hz$ . Another car $Q$ is travelling behind the first car in the same direction with a velocity $40 {ms}^{- 1}$ . The frequency heard by the passenger of the car $Q$ is approximately [Take, velocity of sound $= 360$ ${ms}^{- 1}$ ]

1

514 Hz

2

421 Hz

3

485 Hz

4

471 Hz

Explanation:

B Given
$f_{P} = 400 Hz$
$v_{s} = 360 m / s$
$v_{Q} = 40 m / s$
$v_{P} = 20 m / s$
We know that
$f_{Q} = f_{P} (\frac{v_{s} + v_{Q}}{v_{s} + v_{P}})$
$f_{Q} = 400 (\frac{360 + 40}{360 + 20})$
$f_{Q} = 400 (\frac{400}{380}) = 421 Hz$

JEE Main-11.04.2023

WAVES

172930 An observer moves towards a stationary source of sound, with a velocity equal to one-fifth of the velocity of sound. The percentage change in the frequency will be:

1

20 %

2

10 %

3

5 %

4

0 %

Explanation:

A Given, $v_{0} = \frac{v}{5}, v_{s} = 0$
Then, apparent frequency $(f^{'}) = f_{0} (\frac{v + v_{0}}{v - v_{s}})$
$f^{'} = f_{0} (\frac{v + v / 5}{v}) = \frac{6}{5} f_{0}$
Hence, percentage increase in frequency
$% f = \frac{(\frac{6}{5} - 1) f_{0}}{f_{0}} \times 100$
$\frac{1}{5} \times 100 = 20 %$
$% f = 20 %$

JEE Main-27.06.2022

WAVES

172932 A car is approaching a factory siren that has a frequency of $510 Hz$ . The speed of the sound in air is $340 {ms}^{- 1}$ . If the apparent frequency of the sound as heard by the car driver is $600 Hz$ . Then the speed of the car is

1

60 {ms}^{- 1}

2

33.3 {ms}^{- 1}

3

50.5 {ms}^{- 1}

4

80 {ms}^{- 1}

Explanation:

A Given that, frequency of sound hear by driver $(f^{'}) = 600 Hz$ , frequency of siren $(f_{o}) = 510 Hz$ , velocity of sound $(v_{s}) = 340 m / \sec$ and $v_{1} = 0$
Let, $v_{o} =$ velocity of car
We know that, $f^{'} = f (\frac{v_{s} + v_{o}}{v_{s} - v_{1}})$
$600 = 510 (\frac{340 + v_{0}}{340 - 0})$
$\frac{600 \times 2}{3} = 340 + v_{0}$
$400 = 340 + v_{0}$
$v_{0} = 60 m / \sec .$

AP EAMCET-11.07.2022

WAVES

172933 Two trains $A$ and $B$ are moving towards each other with speeds $72 {kmh}^{- 1}$ and $36 {kmh}^{- 1}$ respectively. The train-A whistles at $640 Hz$ frequency. Before the trains meet, frequency of sound heard by a passenger in Train-B is (Speed of sound in air $= 340 {ms}^{- 1}$ )

1

500 Hz

2

600 Hz

3

700 Hz

4

800 Hz

Explanation:

C $V_{A} = 72 km / h = 72 \times \frac{5}{18} = 20 m / \sec$
$V_{B} = 36 km / h = 36 \times \frac{5}{18} = 10 m / \sec$
$f_{B} = f_{A} [\frac{V + V_{B}}{V - V_{A}}]$
$= 640 [\frac{340 + 10}{340 - 20}]$
$= 640 \times \frac{350}{320} = 700 Hz$
$f_{B} = 700 Hz$

AP EAMCET-07.07.2022

WAVES

172934 A police car moving at $5.4 km / hr$ sounds siren emitting frequency of $550 Hz$ which is reflected back from a stationary object some distance ahead of the car. The number of beats heard per second by an observer sitting in the car is (Assume velocity of sound in air $= 330 m / \sec$ )

1 4

2 5

3 6

4 0

Explanation:

B Given that,
$f = 550 Hz$ ,
$v_{0} = v_{s} = 5.4 km / h = 1.5 m / s$
$v = 330 m / s$
$f^{'} = f (\frac{v + v_{0}}{v - v_{s}})$
$= 550 (\frac{330 + 1.5}{330 - 1.5})$
$= 555 Hz$
$Beat = 555 - 550$
$= 5$

UPSEE 2020

WAVES

172929 A car $P$ travelling at $20 {ms}^{- 1}$ sounds its horn at a frequency of $400 Hz$ . Another car $Q$ is travelling behind the first car in the same direction with a velocity $40 {ms}^{- 1}$ . The frequency heard by the passenger of the car $Q$ is approximately [Take, velocity of sound $= 360$ ${ms}^{- 1}$ ]

1

514 Hz

2

421 Hz

3

485 Hz

4

471 Hz

Explanation:

B Given
$f_{P} = 400 Hz$
$v_{s} = 360 m / s$
$v_{Q} = 40 m / s$
$v_{P} = 20 m / s$
We know that
$f_{Q} = f_{P} (\frac{v_{s} + v_{Q}}{v_{s} + v_{P}})$
$f_{Q} = 400 (\frac{360 + 40}{360 + 20})$
$f_{Q} = 400 (\frac{400}{380}) = 421 Hz$

JEE Main-11.04.2023

WAVES

172930 An observer moves towards a stationary source of sound, with a velocity equal to one-fifth of the velocity of sound. The percentage change in the frequency will be:

1

20 %

2

10 %

3

5 %

4

0 %

Explanation:

A Given, $v_{0} = \frac{v}{5}, v_{s} = 0$
Then, apparent frequency $(f^{'}) = f_{0} (\frac{v + v_{0}}{v - v_{s}})$
$f^{'} = f_{0} (\frac{v + v / 5}{v}) = \frac{6}{5} f_{0}$
Hence, percentage increase in frequency
$% f = \frac{(\frac{6}{5} - 1) f_{0}}{f_{0}} \times 100$
$\frac{1}{5} \times 100 = 20 %$
$% f = 20 %$

JEE Main-27.06.2022

WAVES

172932 A car is approaching a factory siren that has a frequency of $510 Hz$ . The speed of the sound in air is $340 {ms}^{- 1}$ . If the apparent frequency of the sound as heard by the car driver is $600 Hz$ . Then the speed of the car is

1

60 {ms}^{- 1}

2

33.3 {ms}^{- 1}

3

50.5 {ms}^{- 1}

4

80 {ms}^{- 1}

Explanation:

A Given that, frequency of sound hear by driver $(f^{'}) = 600 Hz$ , frequency of siren $(f_{o}) = 510 Hz$ , velocity of sound $(v_{s}) = 340 m / \sec$ and $v_{1} = 0$
Let, $v_{o} =$ velocity of car
We know that, $f^{'} = f (\frac{v_{s} + v_{o}}{v_{s} - v_{1}})$
$600 = 510 (\frac{340 + v_{0}}{340 - 0})$
$\frac{600 \times 2}{3} = 340 + v_{0}$
$400 = 340 + v_{0}$
$v_{0} = 60 m / \sec .$

AP EAMCET-11.07.2022

WAVES

172933 Two trains $A$ and $B$ are moving towards each other with speeds $72 {kmh}^{- 1}$ and $36 {kmh}^{- 1}$ respectively. The train-A whistles at $640 Hz$ frequency. Before the trains meet, frequency of sound heard by a passenger in Train-B is (Speed of sound in air $= 340 {ms}^{- 1}$ )

1

500 Hz

2

600 Hz

3

700 Hz

4

800 Hz

Explanation:

C $V_{A} = 72 km / h = 72 \times \frac{5}{18} = 20 m / \sec$
$V_{B} = 36 km / h = 36 \times \frac{5}{18} = 10 m / \sec$
$f_{B} = f_{A} [\frac{V + V_{B}}{V - V_{A}}]$
$= 640 [\frac{340 + 10}{340 - 20}]$
$= 640 \times \frac{350}{320} = 700 Hz$
$f_{B} = 700 Hz$

AP EAMCET-07.07.2022

WAVES

172934 A police car moving at $5.4 km / hr$ sounds siren emitting frequency of $550 Hz$ which is reflected back from a stationary object some distance ahead of the car. The number of beats heard per second by an observer sitting in the car is (Assume velocity of sound in air $= 330 m / \sec$ )

1 4

2 5

3 6

4 0

Explanation:

B Given that,
$f = 550 Hz$ ,
$v_{0} = v_{s} = 5.4 km / h = 1.5 m / s$
$v = 330 m / s$
$f^{'} = f (\frac{v + v_{0}}{v - v_{s}})$
$= 550 (\frac{330 + 1.5}{330 - 1.5})$
$= 555 Hz$
$Beat = 555 - 550$
$= 5$

UPSEE 2020

WAVES

172929 A car $P$ travelling at $20 {ms}^{- 1}$ sounds its horn at a frequency of $400 Hz$ . Another car $Q$ is travelling behind the first car in the same direction with a velocity $40 {ms}^{- 1}$ . The frequency heard by the passenger of the car $Q$ is approximately [Take, velocity of sound $= 360$ ${ms}^{- 1}$ ]

1

514 Hz

2

421 Hz

3

485 Hz

4

471 Hz

Explanation:

B Given
$f_{P} = 400 Hz$
$v_{s} = 360 m / s$
$v_{Q} = 40 m / s$
$v_{P} = 20 m / s$
We know that
$f_{Q} = f_{P} (\frac{v_{s} + v_{Q}}{v_{s} + v_{P}})$
$f_{Q} = 400 (\frac{360 + 40}{360 + 20})$
$f_{Q} = 400 (\frac{400}{380}) = 421 Hz$

JEE Main-11.04.2023

WAVES

172930 An observer moves towards a stationary source of sound, with a velocity equal to one-fifth of the velocity of sound. The percentage change in the frequency will be:

1

20 %

2

10 %

3

5 %

4

0 %

Explanation:

A Given, $v_{0} = \frac{v}{5}, v_{s} = 0$
Then, apparent frequency $(f^{'}) = f_{0} (\frac{v + v_{0}}{v - v_{s}})$
$f^{'} = f_{0} (\frac{v + v / 5}{v}) = \frac{6}{5} f_{0}$
Hence, percentage increase in frequency
$% f = \frac{(\frac{6}{5} - 1) f_{0}}{f_{0}} \times 100$
$\frac{1}{5} \times 100 = 20 %$
$% f = 20 %$

JEE Main-27.06.2022

WAVES

172932 A car is approaching a factory siren that has a frequency of $510 Hz$ . The speed of the sound in air is $340 {ms}^{- 1}$ . If the apparent frequency of the sound as heard by the car driver is $600 Hz$ . Then the speed of the car is

1

60 {ms}^{- 1}

2

33.3 {ms}^{- 1}

3

50.5 {ms}^{- 1}

4

80 {ms}^{- 1}

Explanation:

A Given that, frequency of sound hear by driver $(f^{'}) = 600 Hz$ , frequency of siren $(f_{o}) = 510 Hz$ , velocity of sound $(v_{s}) = 340 m / \sec$ and $v_{1} = 0$
Let, $v_{o} =$ velocity of car
We know that, $f^{'} = f (\frac{v_{s} + v_{o}}{v_{s} - v_{1}})$
$600 = 510 (\frac{340 + v_{0}}{340 - 0})$
$\frac{600 \times 2}{3} = 340 + v_{0}$
$400 = 340 + v_{0}$
$v_{0} = 60 m / \sec .$

AP EAMCET-11.07.2022

WAVES

172933 Two trains $A$ and $B$ are moving towards each other with speeds $72 {kmh}^{- 1}$ and $36 {kmh}^{- 1}$ respectively. The train-A whistles at $640 Hz$ frequency. Before the trains meet, frequency of sound heard by a passenger in Train-B is (Speed of sound in air $= 340 {ms}^{- 1}$ )

1

500 Hz

2

600 Hz

3

700 Hz

4

800 Hz

Explanation:

C $V_{A} = 72 km / h = 72 \times \frac{5}{18} = 20 m / \sec$
$V_{B} = 36 km / h = 36 \times \frac{5}{18} = 10 m / \sec$
$f_{B} = f_{A} [\frac{V + V_{B}}{V - V_{A}}]$
$= 640 [\frac{340 + 10}{340 - 20}]$
$= 640 \times \frac{350}{320} = 700 Hz$
$f_{B} = 700 Hz$

AP EAMCET-07.07.2022

WAVES

172934 A police car moving at $5.4 km / hr$ sounds siren emitting frequency of $550 Hz$ which is reflected back from a stationary object some distance ahead of the car. The number of beats heard per second by an observer sitting in the car is (Assume velocity of sound in air $= 330 m / \sec$ )

1 4

2 5

3 6

4 0

Explanation:

B Given that,
$f = 550 Hz$ ,
$v_{0} = v_{s} = 5.4 km / h = 1.5 m / s$
$v = 330 m / s$
$f^{'} = f (\frac{v + v_{0}}{v - v_{s}})$
$= 550 (\frac{330 + 1.5}{330 - 1.5})$
$= 555 Hz$
$Beat = 555 - 550$
$= 5$

UPSEE 2020

WAVES

172929 A car $P$ travelling at $20 {ms}^{- 1}$ sounds its horn at a frequency of $400 Hz$ . Another car $Q$ is travelling behind the first car in the same direction with a velocity $40 {ms}^{- 1}$ . The frequency heard by the passenger of the car $Q$ is approximately [Take, velocity of sound $= 360$ ${ms}^{- 1}$ ]

1

514 Hz

2

421 Hz

3

485 Hz

4

471 Hz

Explanation:

B Given
$f_{P} = 400 Hz$
$v_{s} = 360 m / s$
$v_{Q} = 40 m / s$
$v_{P} = 20 m / s$
We know that
$f_{Q} = f_{P} (\frac{v_{s} + v_{Q}}{v_{s} + v_{P}})$
$f_{Q} = 400 (\frac{360 + 40}{360 + 20})$
$f_{Q} = 400 (\frac{400}{380}) = 421 Hz$

JEE Main-11.04.2023

WAVES

172930 An observer moves towards a stationary source of sound, with a velocity equal to one-fifth of the velocity of sound. The percentage change in the frequency will be:

1

20 %

2

10 %

3

5 %

4

0 %

Explanation:

A Given, $v_{0} = \frac{v}{5}, v_{s} = 0$
Then, apparent frequency $(f^{'}) = f_{0} (\frac{v + v_{0}}{v - v_{s}})$
$f^{'} = f_{0} (\frac{v + v / 5}{v}) = \frac{6}{5} f_{0}$
Hence, percentage increase in frequency
$% f = \frac{(\frac{6}{5} - 1) f_{0}}{f_{0}} \times 100$
$\frac{1}{5} \times 100 = 20 %$
$% f = 20 %$

JEE Main-27.06.2022

WAVES

172932 A car is approaching a factory siren that has a frequency of $510 Hz$ . The speed of the sound in air is $340 {ms}^{- 1}$ . If the apparent frequency of the sound as heard by the car driver is $600 Hz$ . Then the speed of the car is

1

60 {ms}^{- 1}

2

33.3 {ms}^{- 1}

3

50.5 {ms}^{- 1}

4

80 {ms}^{- 1}

Explanation:

A Given that, frequency of sound hear by driver $(f^{'}) = 600 Hz$ , frequency of siren $(f_{o}) = 510 Hz$ , velocity of sound $(v_{s}) = 340 m / \sec$ and $v_{1} = 0$
Let, $v_{o} =$ velocity of car
We know that, $f^{'} = f (\frac{v_{s} + v_{o}}{v_{s} - v_{1}})$
$600 = 510 (\frac{340 + v_{0}}{340 - 0})$
$\frac{600 \times 2}{3} = 340 + v_{0}$
$400 = 340 + v_{0}$
$v_{0} = 60 m / \sec .$

AP EAMCET-11.07.2022

WAVES

172933 Two trains $A$ and $B$ are moving towards each other with speeds $72 {kmh}^{- 1}$ and $36 {kmh}^{- 1}$ respectively. The train-A whistles at $640 Hz$ frequency. Before the trains meet, frequency of sound heard by a passenger in Train-B is (Speed of sound in air $= 340 {ms}^{- 1}$ )

1

500 Hz

2

600 Hz

3

700 Hz

4

800 Hz

Explanation:

C $V_{A} = 72 km / h = 72 \times \frac{5}{18} = 20 m / \sec$
$V_{B} = 36 km / h = 36 \times \frac{5}{18} = 10 m / \sec$
$f_{B} = f_{A} [\frac{V + V_{B}}{V - V_{A}}]$
$= 640 [\frac{340 + 10}{340 - 20}]$
$= 640 \times \frac{350}{320} = 700 Hz$
$f_{B} = 700 Hz$

AP EAMCET-07.07.2022

WAVES

172934 A police car moving at $5.4 km / hr$ sounds siren emitting frequency of $550 Hz$ which is reflected back from a stationary object some distance ahead of the car. The number of beats heard per second by an observer sitting in the car is (Assume velocity of sound in air $= 330 m / \sec$ )

1 4

2 5

3 6

4 0

Explanation:

B Given that,
$f = 550 Hz$ ,
$v_{0} = v_{s} = 5.4 km / h = 1.5 m / s$
$v = 330 m / s$
$f^{'} = f (\frac{v + v_{0}}{v - v_{s}})$
$= 550 (\frac{330 + 1.5}{330 - 1.5})$
$= 555 Hz$
$Beat = 555 - 550$
$= 5$

UPSEE 2020