Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

Multiple choice physics acoustics conditions for hearing echo echo reflection of sound and echo ultrasound and its applications

A man is sitting in Gandhi Park of dimension 6m $\times$ 6m. On the left side of it, there is a building adjoining the park, and on the right side of the park, there is a road adjoining by park. A sound is produced on the road by a diwali bomb :

  1. the man will hear the echo of this sound

  2. no sound will be heard by the man

  3. no echo will be heard

  4. none
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The minimum distance between source of sound and obstacle , to hear an echo is 17m , as the dimensions of the park are $l=6m  , b=6m$ , therefore  distance between road and building is 6m , therefore echo of the bomb sound will not be produced as distance between source of sound (bomb) and building is less than 17m i.e. echo will not be heard .

Multiple choice physics study of sound conditions for hearing echo echo reflection of sound and echo ultrasound and its applications

A girl blew a whistle while standing in front of a cliff. She heard an echo after $4s$. Find the distance of the cliff from the boy if velocity of sound in air is $332m{s}^{-1}$

  1. $332m$
  2. $664m$
  3. $166m$
  4. $1328m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Time taken by the sound to go from the boy to the cliff, $t=\dfrac{4}{2}s=2s$

Speed of sound in air, $v=332m{s}^{-1}$ 

So, distance of the cliff from the boy, $s=v\times{t}=332\times2m=664m$.

Multiple choice physics study of sound conditions for hearing echo echo reflection of sound and echo ultrasound and its applications

A boy shouts while standing in front of a hill. He hears an echo after $6s$. If the speed of sound  in air is $340 m/s$, what is the distance of the hill from the position of the body?

  1. $519 m$
  2. $1020 m$
  3. $2040 m$

  4. None
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let distance of hill from the position of the body be $d$.

Echo is heard after $6s$

Speed of sound, $v=340m/s$

So, $d=v \times {\dfrac{t}{2}}$
or, $d=340 \times {\dfrac{6}{2}}=1020m$

Multiple choice physics study of sound conditions for hearing echo echo reflection of sound and echo ultrasound and its applications

An echo returns in $3 s$. What is the distance of the reflecting surface from the source?

  1. $1020 m$
  2. $22.66 m$
  3. $510 m$
  4. $2.28m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Time taken by sound to travel from the source to the deflecting surface $t=\dfrac{3}{2}=1.5s$

Speed of sound $v=340m/s$

So, distance between the surface and the source $=v \times t=340 \times 1.5=510m$

Multiple choice physics study of sound conditions for hearing echo echo reflection of sound and echo ultrasound and its applications

An echo returned in $3 s$. What is the distance of the reflecting surface from the source, given that the speed of sound is ${ 342ms }^{ -1 }$?

  1. $520 m$
  2. $515 m$
  3. $530 m$
  4. $513 m$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If $d$ be the distance between reflecting surface and the source. 

For the echo, the sound travels distance $2d$ in $3s$. 
Thus, $2d=vt$ or $d=vt/2=\dfrac{342\times 3}{2}=513 m$ 

Multiple choice physics study of sound conditions for hearing echo echo reflection of sound and echo ultrasound and its applications

Sound produced by a thunderstorm is heard 10 s after the lightning is seen. The approximate distance of the thunder cloud :
(Given speed of sound $= 340 m s^{1}$)

  1. 3.4 Km

  2. 340 m

  3. 3.4 m

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Lightning and thunder are produced simultaneously, but the thunder is heard a few seconds after the lightning is seen. This is because the speed of light in air is more than the speed of sound in air.
The speed of light is given as $3\times { 10 }^{ 8 }$ m/s. The speed of sound is 340 m/s. Thus, the entire time of 10 seconds (as mentioned in the question) between seeing the light and hearing the thunderstorm is taken by the sound to travel to the observer. Hence, the distance traveled by the thunder is given as follows.
$s=vt$; where,
s is the distance traveled, v is the velocity of sound and t is the time taken.
That is,  $330m/s\times 10s = 3400 m = 3.4 km.$
Thus, the distance traveled by the thunder is $3.4 km.$

Multiple choice physics acoustics conditions for hearing echo echo reflection of sound and echo ultrasound and its applications

The ceiling and wall behind the stage of good conference halls or concert halls are made:

  1. Straight

  2. Curved

  3. Circular

  4. Rectangular

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In concert or conference halls, music or other sounds that are produced on the stage must be carried through the air to people in the crowd. Some of these sound waves go directly to the people in the crowd, without bouncing off on anything first. However, other sound waves from the stage first go to areas like walls and ceilings, and the sound either bounces off or is absorbed. When sound waves bounce off a surface, the new direction they travel is related to the angle they strike the surface. 
Thus, acoustical engineers and architects sometimes need to place panels on the ceilings and walls to reflect sound in a specific direction, that is, back to the audience. Using panels that are curved have a similar effect. Hence, ceiling and wall behind the stage of good conference halls or concert halls made curved so that sound after reflection reaches all the corners of the hall and the whole audience.

Multiple choice physics study of sound conditions for hearing echo echo reflection of sound and echo ultrasound and its applications

What should be the minimum distance between the listener and the reflector to hear an echo of sound propagating with a speed $v\   m s^{-1}$?

  1. $v/10$ m
  2. $10 v$ m
  3. $v/20$ m
  4. $20v$ m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
if there is a sound reflector at a distance d from the source, then the time interval between original sound and its echo at the site of the source is given as $T=\dfrac { 2d }{ v } $
 $d=\dfrac { v\times t }{ 2 } = \dfrac { v\times 1 }{ 2\times 10 } $ as Time = 0.1 s.
Hence, $d=\dfrac { v }{ 20 }$.

Multiple choice physics study of sound conditions for hearing echo echo reflection of sound and echo ultrasound and its applications

A man starts from rest and moves with constant acceleration towards a hill which is $648 m$ away. He fires a gun at the instant of starting and hears its echo after $4$ seconds. Velocity of sound in air is $320 m/s$

  1. His acceleration is $1m/s^{2}$
  2. His acceleration is $2m/s^{2}$
  3. His velocity at the instant of hearing echo is $4 m/s$
  4. His velocity at the instant of hearing echo is $8 m/s$
Reveal answer Fill a bubble to check yourself
B,D Correct answer
Explanation

Using s = 0.5 * a * t^2, the distance the man travels is 0.5 * a * 16 = 8a. The sound travels to the hill (648 m) and back to the man (648 - 8a). Total distance = 648 + (648 - 8a) = 1296 - 8a. Also, distance = v_sound * t = 320 * 4 = 1280. 1296 - 8a = 1280 implies 8a = 16, so a = 2 m/s^2. Velocity v = a * t = 2 * 4 = 8 m/s.

Multiple choice physics acoustics conditions for hearing echo echo reflection of sound and echo ultrasound and its applications

A man fires a gun and hears its echo after 5 s. The man then moves 310 m towards the hill and fires his gun again. This time he hears  the echo after 3 s. Calculate the speed of sound. 

  1. $330\,m\, s^{-1}$
  2. $350\,m\, s^{-1}$
  3. $210\,m\, s^{-1}$
  4. $310\,m\, s^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The speed of the sound is calculated as follows.
Let d be the distance between the man and the hill in the beginning.
$v=\dfrac { 2\times d }{ t } $ = $\dfrac { 2\times d }{ 5 } \rightarrow eqn\quad 1$
He moves 310 m towards the hill. Therefore the distance will be (d - 310) m. Therefore,
$v=\dfrac { 2\left( d-310 \right)  }{ 3 } \rightarrow eqn\quad 2$
Since velocity of sound is same, equating (1) and (2), we get
$\dfrac { 2d }{ 5 } =\dfrac { 2\left( d-310 \right)  }{ 3 } $
3d = 5d - 1550
2d = 1550
d = 775 m
Hence, the velocity of sound v=$\dfrac { 2\times 775 }{ 5 } $ (substituting in equation 1)
$v=310 m/s$