Maharashtra Board Class 9 Science Chapter 12 Study of Sound Solutions

Std 9 Science Chapter 12 Study Of Sound Question Answer Maharashtra Board

Class 9 Science Chapter 12 Study Of Sound Question Answer Maharashtra Board

 

Question 1. Fill in the blanks and explain.
a. Sound does not travel through - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
b The velocity of sound in steel is - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - than the velocity of sand in water.
c. The incidence of - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - in daily life shows that the velocity of sound is less than the velocity of light.
d. To discover a sunken ship or objects deep inside the sea, - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - technology is used.
Answer: (Answers not provided in the original content for fill in the blanks)
In simple words: This question asks to fill in the blanks regarding basic properties of sound travel and applications, such as its inability to travel through a vacuum, its varying speed in different mediums, and the technology used for underwater exploration.

🎯 Exam Tip: Understanding the fundamental principles of sound propagation and its practical applications is crucial for scoring well in such questions.

 

Question 2. Explain giving scientific reasons.
a. The roof of a movie theatre and a conference hall is curved.
Answer:
• Sound waves get reflected from the walls and roof of a room multiple times. This causes a single sound to be heard not once but continuously. This is called reverberation.
• Due to reverberation, some auditoriums or some particular seats in an auditorium have inferior sound reception. This can be compensated with curtains.
• Ceilings of these halls are made curved so that sound after reflecting from the ceiling, reaches all parts of the hall and the quality of sound improves.
In simple words: Curved ceilings in halls ensure that sound reflects evenly to all areas, preventing poor sound quality from reverberation and ensuring every listener hears clearly.

🎯 Exam Tip: Focus on explaining the role of reverberation and how architectural design (like curved ceilings) addresses sound distribution for optimal acoustics.

 

b. The intensity of reverberation is higher in a closed and empty house.
Answer:
• Reverberation occurs due to multiple reflections of sound.
• The furniture in the house acts as a sound-absorbing material.
• So if the house is closed and empty, a reflection of sound will be maximum and hence, intensity of reverberation is higher.
In simple words: An empty room lacks furniture or other soft materials that absorb sound, leading to more reflections and thus a higher intensity of reverberation.

🎯 Exam Tip: When explaining reverberation, highlight the role of sound absorption; fewer absorbing surfaces lead to more reflections and increased reverberation.

 

c. We cannot hear the echo produced in a classroom.
Answer:
• For distinct echoes, the minimum distance of the reflecting surface from the source of sound must be 17.2 m.
• Benches in the classroom are sound absorbing materials which prevent echo of sound.
• Because of these two reasons echo is not heard in a classroom.
In simple words: Echoes are not heard in classrooms because the distance to reflecting surfaces is usually less than 17.2 meters, and sound-absorbing materials like benches prevent clear reflections.

🎯 Exam Tip: Remember the two key conditions for hearing a distinct echo-minimum distance and the nature of reflecting surfaces.

 

Question 3. Answer the following questions in your own words.
a. What is an echo? What factors are important to get a distinct echo?
Answer:
• An echo is the repetition of the original sound because of reflection by some surface.
• At 22°C, the velocity of sound in air is 344 m/s.
• Our brain retains a sound for 0.1 seconds Thus, for us to be able to hear a distinct echo, the sound should take more than 0.1 seconds after starting from the source to get reflected and. come back to us.
• We know that,
Distance = speed x time
= \(344 \text{ m/s} \times 0.1 \text{ s}\)
= \(34.4 \text{ m}\)
• Thus, to be able to hear a distinct echo, the reflecting surface should be at a minimum distance of half of the above, i.e. 17.2 m.
• As the velocity of sound depends on the temperature of air, this distance depends on the temperature.
In simple words: An echo is a repeated sound due to reflection. For a distinct echo, the reflecting surface must be at least 17.2 meters away, as sound needs to travel for more than 0.1 seconds (brain's sound retention time) to be perceived as separate from the original.

🎯 Exam Tip: Clearly state the definition of echo and meticulously explain the 0.1 second time delay and the resultant minimum distance calculation (17.2 m), emphasizing temperature dependence.

 

b. Study the construction of the Golghumat at Vijapur and discuss the reasons for the multiple echoes produced there.
Answer:
• Goighumat with a height of 51 metres and diameter of 37 metres with 3 metres thick walls is spread over approximately 1700 square metres.
• This meets the conditions for echo i.e. : 17.2 metres minimum.
• The dome of the golghumat is curved and hence, sound reflects multiple times before reaching the observer.
• This is the reason for multiple echoes being produced.
In simple words: The Golghumat's large size (over 17.2m minimum distance) and its curved dome structure cause sound to reflect multiple times, leading to the distinct multiple echoes heard there.

🎯 Exam Tip: When discussing Golghumat, focus on its dimensions and curved dome as key architectural features that facilitate multiple reflections and echoes.

 

c. What should be the dimensions and the shape of classrooms so that no echo can be produced there?
Answer:
1. Dimensions: The distance between opposite walls in a classroom must be less than 17.2 m so that the reflected sound returns to the observer within 0.1 s.
2. Shape: The classrooms should have curved ceilings and walls so that the reflected sound is directed towards the observer instantly within 0.1 s
In simple words: To prevent echoes in classrooms, dimensions should keep reflecting surfaces less than 17.2m apart, and ceilings/walls should be curved to direct sound quickly to listeners, avoiding delayed reflections.

🎯 Exam Tip: For classroom acoustics, emphasize the critical distance (less than 17.2m) and the use of curved surfaces to control sound reflection and prevent echoes.

 

Question 4. Where and why are sound-absorbing materials used?
Answer:
The sound absorbing materials are used in :
• School, cinema hall, concert hall, houses or places where quality of sound is important.
• In the absence of sound absorbing material the sound will undergo multiple reflection causing reverberation of sound.
In simple words: Sound-absorbing materials are used in places like schools and concert halls to prevent excessive sound reflection (reverberation), which can degrade sound quality.

🎯 Exam Tip: Highlight both the locations where sound-absorbing materials are used and the specific problem (reverberation) they are designed to solve to improve sound quality.

 

Question 5. Solve the following examples.
a. The speed of sound in air at O °C is 332 m/s. If it increases at the rate of 0.6 m/s per degree, what will be the temperature when the velocity has increased to 344 m/s?
Answer:
Given:
Initial speed of sound at 0°C \( = 332 \text{ m/s}\).
Final speed of sound \( = 344 \text{ m/s}\).
Rate of increase per degree rise in temp. \( = 0.6 \text{m/s}\)
To find:
Temperature when speed is \(344 \text{m/s}\)
Formulae:
Increase in temperature
Increase in temperature
\[ = \frac{\text{Increase in speed of sound}}{\text{Rate of increase per degree rise in temperature}} \]
Solution:
Increase in temperature
\[ = \frac{344 - 332}{0.6} \]
\[ = \frac{12}{0.6} \]
\[ = 20^\circ\text{C} \]
Temperature when the speed of sound is \(344 \text{ m/s}\) is \(20^\circ\text{C}\)
In simple words: To find the temperature at which sound speed reaches 344 m/s, we calculate the total increase in speed from 332 m/s, then divide it by the given rate of increase per degree Celsius.

🎯 Exam Tip: Clearly state the given values, formula used, and show all calculation steps. Ensure the units are correct throughout the solution.

 

b. Nita heard the sound of lightning after 4 seconds of seeing it. What was the distance of the lightning from her? (The velocity of sound in air is 340 m/s?)
Answer:
Given : Speed of sound \( (v) = 340 \text{ m/s}\)
Time taken \( (t) = 4 \text{ sec}\)
To find : Distance \( (s) = ?\)
Formulae: Velocity \( = \frac{\text{distance}}{\text{time}}\)
Solution:
Velocity \( = \frac{\text{distance}}{\text{time}}\)

\(\implies\) Distance \( = \text{velocity} \times \text{time}\)
\( = 340 \times 4 = 1360 \text{ m}\)
The lightning has struck at a distance of \(1360 \text{ m}\) from the observer.
In simple words: To find the distance of the lightning, multiply the given speed of sound in air by the time difference Nita observed between seeing the flash and hearing the thunder.

🎯 Exam Tip: For simple distance-speed-time problems, correctly identify the given values and formula, and ensure units are consistent for calculation.

 

c. Sunil is standing between two walls. The wall closest to him is at a distance of 360 m. If he shouts, he hears the first echo after 4 s and another after another 2 seconds.
1. What is the velocity of sound in air?
2. What is the distance between the two walls? (Ans: 330 m/s; 1650 m)
Answer:
Given:
Distance of the closer wall \((\text{S}_1) = 360 \text{ m}\)
Time of echo from closer wall \( = 4 \text{ sec}\)

\(\therefore\) Time taken \((\text{t}_1) = 4/2 \text{ sec} = 2 \text{ sec}\)
Time of echo from distant wall \( = 6 \text{ sec}\)

\(\therefore\) Time taken \((\text{t}_2) = 6/2 \text{ sec} = 3 \text{ sec}\)
To find :
Velocity of sound in air \( (v) = ?\)
Distance between two walls \( (\text{S}_1 + \text{S}_2) = ?\)
Formulae:
Velocity \( = \frac{\text{distance}}{\text{time}}\)
Solution:
(a) Velocity \( = \frac{\text{distance}}{\text{time}}\)

\(\therefore v_1 = \frac{\text{S}_1}{\text{t}_1}\)
\[ = \frac{660}{2} = 330 \text{ m/s} \]
(b) Also, \(v_1 = v_2\)

\(\therefore v_2 = 330 \text{ m/s}\)

\(\therefore \frac{v_1}{v_2} = \frac{\text{S}_2}{t_2}\)
\[ \frac{330}{v_2} = \frac{\text{S}_2}{3} \]
\[ 330 \times 3 = \text{S}_2 \]
\[ \text{S}_2 = 990\text{m} \]

\(\therefore\) Distance between two wall \( = \text{S}_1 + \text{S}_2\)
\( = 660 + 990\)
\( = 1650 \text{ m}\)
The velocity of sound in air is 330 m/s and the distance between two walls is 1650 m.
In simple words: Sunil hears two echoes, allowing us to calculate the velocity of sound using the first echo's time and then determine the second wall's distance using that velocity and the second echo's time, finally summing both distances for total wall separation.

🎯 Exam Tip: Pay close attention to the time taken for *each* echo (round trip) versus the time for sound to reach *one* wall (half the echo time). Clearly distinguish between S1 and S2 distances.

 

d. Hydrogen gas is filled in two identical bottles, A and B, at the same temperature. The mass of hydrogen in the two bottles is 12 gm and 48 gm respectively. In which bottle will sound travel faster? How may times as fast as the other? (Ans: In A; Twice)
Answer:
In A; Thrice
In simple words: Since sound travels faster in less dense mediums, and density is directly related to mass when volume is constant, sound will travel faster in bottle A (12 gm) than in bottle B (48 gm).

🎯 Exam Tip: Remember that sound velocity is inversely proportional to the square root of density. A lower mass for the same volume means lower density and faster sound travel.

 

e. Helium gas is filled in two identical bottles A and B. The mass of the gas in the two bottles is 10 gm and 40 gm respectively. If the speed of sound is the same in both bottles, what conclusions will you draw? (Ans: Temperature of B is 4 times the temperature of A.)
Answer:
Given:
Mass of Helium in bottle A \( (\text{m}_\text{A}) = 10\text{gm}\)
Mass of Helium in bottle B \( (\text{m}_\text{B}) = 40\text{gm}\)
Formulae:
\[ v \propto \frac{1}{\sqrt{M}}, v \propto \sqrt{T}, \rho = \frac{m}{V} \]
Solution:
\[ v_A \propto \frac{\sqrt{T_A}}{\sqrt{m_A}} \text{ ...(i)} \]
\[ v_B \propto \frac{\sqrt{T_B}}{\sqrt{m_B}} \text{ ...(ii)} \]
From (i) and (ii),

\(\therefore\) \(\frac{v_A}{v_B} = \frac{\sqrt{T_A}}{\sqrt{m_A}} \times \frac{\sqrt{m_B}}{\sqrt{T_B}}\)
\[ \frac{\sqrt{T_A}}{\sqrt{m_A}} = \frac{\sqrt{T_B}}{\sqrt{m_B}} \quad \text{...}[v_A = v_B] \]

\(\therefore\) \(\frac{T_A}{m_A} = \frac{T_B}{m_B}\)
\[ \frac{T_A}{10} = \frac{T_B}{40} \]
\[ T_B = \frac{40}{10} T_A \]
\[ T_B = 4T_A \]
The temperature of B is 4 times the temperature of A
In simple words: If the speed of sound is the same in both helium bottles despite different masses, it implies a relationship between their temperatures. Since velocity is affected by mass and temperature, for speeds to be equal with mass B being four times mass A, temperature B must be four times temperature A to balance the equation.

🎯 Exam Tip: When dealing with gas properties and sound velocity, remember the relationship: sound velocity is proportional to the square root of temperature and inversely proportional to the square root of molecular mass/density. Apply these proportionalities carefully in derivations.

 

Class 9 Science Chapter 12 Study Of Sound Intext Questions And Answers

 

Question 1. How does the velocity of sound depend on its frequency?
Answer:
The velocity of sound is directly proportional to its frequency
\( v = \upsilon\lambda \)
when \(v = \text{velocity}\)
\(u = \text{frequency}\)
\( \lambda = \text{wavelength}\)
In simple words: The velocity of sound depends on its frequency and wavelength, as expressed by the formula \(v = \upsilon\lambda \), where velocity equals frequency multiplied by wavelength.

🎯 Exam Tip: State the wave equation \(v = \upsilon\lambda \) clearly and define each variable. Mention that while velocity depends on the medium, for a given medium, higher frequency means shorter wavelength, maintaining the constant velocity.

 

Question 2. The molecular weight of oxygen gas (O2) is 32 while that of hydrogen gas (H2) is 2. Prove that under the same physical conditions, the velocity of sound in hydrogen is four times that in oxygen.
Answer:
Given:
Molecular wt of Oxygen \((\text{M}_\text{O}) = 32\)
Molecular wt of hydrogen \((\text{M}_\text{H}) = 2\)
To Find:
\( \text{V}_\text{H} = 4 \text{ V}_\text{O}\)
Formulae:
\[ v \propto \frac{1}{\sqrt{M}} \]

\(\therefore \text{V}_\text{O} = \frac{1}{\sqrt{\text{M}_\text{O}}}\) ...(i)

\(\therefore \text{V}_\text{O} \propto \frac{1}{\sqrt{32}}\)
\[ \text{V}_\text{H} \propto \frac{1}{\sqrt{\text{M}_\text{H}}} \]

\(\therefore \text{V}_\text{H} \propto \frac{1}{\sqrt{2}}\) ...(ii)
Dividing equation (i) by (ii) we get,
\[ \frac{\text{V}_\text{O}}{\text{V}_\text{H}} = \frac{1/\sqrt{32}}{1/\sqrt{2}} \]
\[ \frac{\text{V}_\text{O}}{\text{V}_\text{H}} = \frac{\sqrt{2}}{\sqrt{32}} \]
\[ \frac{\text{V}_\text{O}}{\text{V}_\text{H}} = \frac{1}{\sqrt{16}} \]
\[ \frac{\text{V}_\text{O}}{\text{V}_\text{H}} = \frac{1}{4} \]

\(\therefore \text{V}_\text{H} = 4 \text{ V}_\text{O}\)
Hence, proved that velocity of sound in hydrogen is four times that in oxygen.
In simple words: The velocity of sound is inversely proportional to the square root of molecular weight. Since hydrogen's molecular weight is 16 times less than oxygen's, sound travels four times faster in hydrogen than in oxygen.

🎯 Exam Tip: Clearly state the relationship between sound velocity and molecular weight. Show the ratio calculation step-by-step for full credit, ensuring correct square root manipulation.

 

Answer The Following:

 

Question 1. How will you reduce reverberation in public halls or buildings?
Answer:
(i) Reverberation in public halls or buildings will be reduced by using sound absorbing materials like curtains on wall, carpets on the floor.
(ii) By keeping the windows open, as sound will not get reflected.
In simple words: Reverberation can be reduced by using soft, sound-absorbing materials like carpets and curtains on walls and floors, or by opening windows to allow sound to escape instead of reflecting.

🎯 Exam Tip: Focus on practical solutions for sound absorption. Mentioning both material use and architectural adjustments (like open windows) demonstrates a comprehensive understanding.

 

Question 2. How is ultrasound used in medical science?
Answer:
• Sonography: Sonography technology uses ultrasonic sound waves to generate images of internal organs of the human body.
• Echocardiography: Echocardiography is a test that uses ultrasonic sound waves to produce live images of your heart.
In simple words: Ultrasound is used in medicine through sonography to create images of internal organs and in echocardiography to visualize the heart, both employing high-frequency sound waves.

🎯 Exam Tip: Provide specific examples like sonography and echocardiography, briefly explaining what each procedure does with ultrasound for medical diagnosis.

 

Question 3. To hear the echo distinctly, will the distance from the source of sound to the reflecting surface be same at all temperatures? Explain your answer.
Answer:
• No,the distance from the source of sound to the reflecting surface will not be the same at all temperatures.
• Velocity of sound is directly proportional to the square root of temperature.
• One of the conditions of echo is that the time interval between the original and reflected sound should be more than 0.1 sec.
• So if the temperature increases, the velocity of sound increases and the reflected sound reaches in less than 0.1 sec.
• So for echo to be heard the distance between the observer and the reflecting surface has to increase.
In simple words: No, the distance required for a distinct echo changes with temperature because the velocity of sound varies with temperature; higher temperatures mean faster sound and thus a greater distance needed to maintain the 0.1 second delay for echo perception.

🎯 Exam Tip: Crucially link the dependence of sound velocity on temperature to the minimum distance required for an echo, explaining how changes in one affect the other in relation to the brain's sound retention time.

 

Question 4. When is the reflection of sound harmful?
Answer:
• Reflected sound of high intensity called as noise is disturbing and harmful to the ears.
• When sound reverberates i.e it undergoes multiple reflections, poor quality of sound is produced.
In simple words: Reflection of sound can be harmful when it creates loud, disturbing noise or when excessive reverberation degrades sound quality, making it unclear or unpleasant to hear.

🎯 Exam Tip: Distinguish between harmful reflections causing noise and those causing poor sound quality (reverberation). Both are negative impacts of sound reflection.

 

Question 5. What kind of waves are created when a stone is dropped in water?
Answer:
• When a stone is dropped in water, the particles of water oscillate up and down.
• These oscillations are perpendicular to the direction of propagation of the wave, such waves are called transverse waves.
In simple words: When a stone drops into water, it creates transverse waves where water particles move up and down, perpendicular to the wave's outward propagation.

🎯 Exam Tip: Define transverse waves and explicitly state that the particle oscillation is perpendicular to the wave propagation direction in this context.

 

Answer The Following Question:

 

Question 1. Observe the graph/ diagram and discuss your observation.
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र ध्वनि तरंगों को दर्शाता है। इसमें ध्वनि की घनत्व और दबाव में परिवर्तन, साथ ही संपीड़न (compression) और विरलन (rarefaction) क्षेत्रों को स्पीकर से उत्पन्न होते हुए दिखाया गया है। चित्र A घनत्व में बदलाव, चित्र B दबाव में बदलाव और चित्र C घनत्व/दबाव में तरंगीय बदलाव को दर्शाता है जिसमें क्रेस्ट और ट्रफ शामिल हैं।
1. Fig. A shows changes in density. The region where particles are crowded is called compression and where they are far apart are rarefaction.
2. Fig. B show change in pressure. The lines represent layers of air. The regions when lines are crowded are high pressure regions while when they are far apart are of low pressure.
3. Fig. C shows changes in density or pressure. The crest represents high pressure region while trough represents low pressure region.
In simple words: The diagram illustrates how sound waves cause changes in air density and pressure: crowded regions are compressions (high density/pressure), sparse regions are rarefactions (low density/pressure), and these propagate as waves with crests (high pressure) and troughs (low pressure).

🎯 Exam Tip: Clearly differentiate between compression and rarefaction zones in terms of density and pressure. Explain how these relate to the crests and troughs of a wave representation.

 

Answer The Following Question:

 

Question 1. How are the frequencies of notes sa, re, ga, ma, pa, dha, ni related to each other?
Answer:
The frequencies of notes sa, re, ga, ma, pa, dha, ni are related in the ratio.

Symbols	Sa	Re	Ga	Ma	Pa	Dha	Ni	Sa
Ratio with Sa	1	9/8	5/4	4/3	3/2	5/3	15/8	2

e.g. Sa \( = 240 \text{ Hz}\)
Re \( = 240 \times \frac{9}{8} = 270 \text{ Hz}\)
Ga \( = 240 \times \frac{5}{4} = 300 \text{ Hz}\)
Ma \( = 240 \times \frac{4}{3} = 320 \text{ Hz}\)
Pa \( = 240 \times \frac{3}{2} = 360 \text{ Hz}\)
Dha \( = 240 \times \frac{5}{3} = 400 \text{ Hz}\)
Ni \( = 240 \times \frac{15}{8} = 450 \text{ Hz}\)
Sa \( = 240 \times 2 = 480 \text{ Hz}\)
i.e if first Sa is 240Hz then the next Sa will be 240 x 2 = 480Hz
In simple words: The frequencies of the musical notes Sa, Re, Ga, Ma, Pa, Dha, Ni are related by specific mathematical ratios, meaning each note's frequency is a precise multiple of the base Sa frequency.

🎯 Exam Tip: Present the frequency ratios in a clear table and demonstrate how to calculate the actual frequencies for each note based on a given base frequency for 'Sa'.

 

Question 2. What is the main difference between the frequencies of the voice of a man and that of a woman?
Answer:
• Voice of a woman is high pitch i.e shorter wavelength and higher frequency
• Voice of man is low pitch i.e larger wavelength and smaller frequency.
In simple words: Women's voices generally have a higher pitch due to shorter wavelengths and higher frequencies, while men's voices have a lower pitch with longer wavelengths and lower frequencies.

🎯 Exam Tip: Clearly define pitch in terms of frequency and wavelength, and then apply these concepts to describe the difference between male and female voices.

 

Question 3. Try This;

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र ध्वनि के परावर्तन को दर्शाता है। इसमें एक ध्वनि स्रोत (घड़ी) से निकलने वाली ध्वनि तरंगों को दो कार्डबोर्ड ट्यूबों के माध्यम से दीवार से टकराकर परावर्तित होते हुए दिखाया गया है, जहाँ आपतन कोण (\( \theta_1 \)) और परावर्तन कोण (\( \theta_2 \)) को दर्शाया गया है।
(a) In the above activity, what will happen if you lift one of the tubes to some height?
Answer:
If one of the tubes is lifted, angle of incidence will not be equal to angle of reflection, hence, the sound will not be clearly audible.
In simple words: Lifting one tube changes the angle of incidence, causing it to no longer equal the angle of reflection, which means the sound will not be clearly heard.

🎯 Exam Tip: Link the change in tube height directly to the alteration of the angles of incidence and reflection, and then to the consequence of unclear sound perception.

 

(b) Measure the angle of incidence 01 and the angle of reflection 02. Try to see if they are related in any way.
Answer:
Angle of incidence is same as the angle of reflection.
In simple words: According to the law of reflection, the angle at which sound (or light) hits a surface (angle of incidence) is always equal to the angle at which it bounces off (angle of reflection).

🎯 Exam Tip: State the law of reflection clearly: the angle of incidence is equal to the angle of reflection.

 

Class 9 Science Chapter 12 Study Of Sound Additional Important Questions And Answers

 

Can You Recall?

 

Question 1. How is the direction of the oscillation of the particles of the medium related to the direction of propagation if the sound wave?
Answer:
• Sound travels as a longitudinal wave.
• In a longitudinal wave, the particle of the medium oscillate parallel to the direction of propagation of the wave.
In simple words: Sound waves are longitudinal, meaning the particles of the medium oscillate back and forth in the same direction that the wave energy travels.

🎯 Exam Tip: Emphasize that sound is a longitudinal wave and define this by stating that particle oscillation is parallel to wave propagation.

 

Choose And Write The Correct Option:

 

Question 1. The unit of frequency is ............
(a) Hertz
(b) m/s²
(c) Decibels
(d) m/s
Answer: (a) Hertz
In simple words: The unit for frequency, which measures the number of cycles or oscillations per second, is Hertz.

🎯 Exam Tip: Know the SI units for fundamental physical quantities. Hertz (Hz) is the standard unit for frequency.

 

Question 2. The normal hearing range for humans is .............
(a) 0 Hz to 20 Hz
(b) greater than 20,000 Hz
(c) 20 Hz to 20,000 Hz
(d) none of these
Answer: (c) 20 Hz to 20,000 Hz
In simple words: Humans typically hear sounds with frequencies between 20 Hertz (very low pitch) and 20,000 Hertz (very high pitch).

🎯 Exam Tip: Memorize the standard audible frequency range for humans, as it's a common knowledge question.

 

Question 3. Sound will not travel through ............
(a) Vacuum
(b) Liquid
(c) Solid
(d) Gases
Answer: (a) Vacuum
In simple words: Sound requires a medium (like solid, liquid, or gas) to travel because it propagates through the vibration of particles, so it cannot travel through a vacuum where no particles exist.

🎯 Exam Tip: Understand that sound is a mechanical wave, meaning it requires a material medium for its propagation; it cannot travel in the absence of particles, like in a vacuum.

 

Question 4. Sl unit of ............ is Hertz (Hz).
(a) Wavelength
(b) Frequency
(c) Speed of wave
(d) Velocity
Answer: (b) frequency
In simple words: Hertz (Hz) is the standard international unit used to measure frequency, which indicates how many cycles of a wave occur per second.

🎯 Exam Tip: Associate Hertz directly with frequency. Be clear on the units for other wave properties like wavelength (meters) and speed (m/s).

 

Question 5. The velocity of sound is inversely proportional to the ...........
(a) Pressure
(b) Square root of temperature
(c) Square root of density
(d) Humidity
Answer: (c) square root of density
In simple words: The speed at which sound travels is slower in denser mediums; specifically, it is inversely proportional to the square root of the medium's density.

🎯 Exam Tip: Remember the primary factors affecting sound velocity, particularly its inverse relationship with the square root of the medium's density. This is a key formula in acoustics.

 

Question 6. Sound waves with frequency greater than 20 kHz are called .............
(a) Infrasound
(b) Ultrasound
(c) Sonic
(d) Damped sound
Answer: (b) ultrasound
In simple words: Sound waves with frequencies above 20,000 Hertz, which are beyond the range of human hearing, are known as ultrasound.

🎯 Exam Tip: Distinguish between audible sound (sonic), infrasound (below 20 Hz), and ultrasound (above 20 kHz) by their frequency ranges.

 

Question 7. The loudness of a sound depends upon ............
(a) Amplitude
(b) Speed
(c) Density
(d) Wavelength
Answer: (a) Amplitude
In simple words: The loudness or intensity of a sound is directly determined by its amplitude, which is the maximum displacement or intensity of the wave.

🎯 Exam Tip: Associate loudness directly with amplitude. A larger amplitude means a louder sound.

 

Question 8. ............ are used in sonography.
(a) High frequency ultrasound
(b) Stationary waves
(c) High frequency infrasound
(d) High frequency micro waves
Answer: (a) High frequency ultrasound
In simple words: Sonography utilizes high-frequency ultrasound waves to create images of internal body structures, as these waves can penetrate tissues and reflect back to form detailed visuals.

🎯 Exam Tip: Understand that sonography specifically uses high-frequency ultrasound for medical imaging due to its ability to create detailed images without using ionizing radiation.

 

Question 9. The - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - receives the vibrations coming from the membrane and
Answer: (Options and answer are on the next page, outside the specified processing range.)
In simple words: This question asks to identify the part of the ear that receives vibrations from the membrane (eardrum) to further process sound.

🎯 Exam Tip: Familiarize yourself with the parts of the human ear and their functions, especially the sequence of sound transmission from the eardrum inwards.

Find The Odd One Out:

Question 1. Bats, rats, cats, dolphins
Answer: Cats: cannot produce ultrasonic sound.
In simple words: Cats cannot produce ultrasonic sounds, unlike bats, rats, and dolphins.

🎯 Exam Tip: Identifying animals that use specific sound ranges (ultrasound/infrasound) is a common knowledge-based question.

 

Question 2. Clothes, paper, curtains, mirror
Answer: Mirror: is a good reflector of sound, while others are poor reflectors.
In simple words: A mirror reflects sound well, whereas clothes, paper, and curtains absorb sound and are poor reflectors.

🎯 Exam Tip: Understanding the properties of materials regarding sound reflection and absorption is key for these types of questions.

 

Question 3. Submarines, icebergs, internal organ, sunken ships.
Answer: Internal organ: sonography is used, while for others sonar system is used.
In simple words: Sonography is used for internal organs, while a SONAR system is used for detecting submarines, icebergs, and sunken ships.

🎯 Exam Tip: Differentiate between medical imaging techniques (sonography) and underwater detection technologies (SONAR).

 

Question 4. Temperature, density, molecular weight, pressure
Answer: Pressure: for a fixed temperature, the speed of sound does not depend on the pressure of the gas, all other factors affect speed of sound.
In simple words: For a constant temperature, the speed of sound in a gas is independent of its pressure, while temperature, density, and molecular weight do influence it.

🎯 Exam Tip: Remember the primary factors affecting the speed of sound in gases, and note that pressure is not one of them at a fixed temperature.

Answer In One Sentence:

Question 1. How can one produce sound?
Answer: Vibration set up in an object produces sound (or) sound is produced when an object is disturbed and starts vibrating.
In simple words: Sound is produced when an object vibrates or is set into motion, causing oscillations that create sound waves.

🎯 Exam Tip: The fundamental principle of sound production is vibration; ensure you state this clearly.

 

Question 2. What is velocity of sound wave ?
Answer: The distance covered by a point on the wave in unit time is the velocity of the sound wave.
In simple words: The velocity of a sound wave is how much distance a point on the wave travels in a given amount of time.

🎯 Exam Tip: A precise definition of velocity in the context of waves is crucial, focusing on distance per unit time.

 

Question 3. What is the minimum distance of the reflecting surface to hear an echo ?
Answer: To be able to hear a distinct echo, the reflecting surface should be at a minimum distance of 17.2 m.
In simple words: For a clear echo, the reflecting surface must be at least 17.2 meters away from the sound source.

🎯 Exam Tip: This specific distance (17.2 m) is a key numerical value related to human perception of echoes and should be memorized.

Match The Columns:

Question 1.

Column 'A'	Column B'	Column C
(1) Transverse wave	(a) Particles oscillate parallel to direction of propagation	(i) Wave produced in a slinky
(2) Longitudinal wave	(b) Particles oscillate perpendicular to direction of propagation	(ii) Frequency less than 20 Hz
(3) Ultrasound	(c) Echo formation is heard under particular conditions	(iii) Wave produced in string
(4) Infrasound	(d) High frequency waves	(iv) Frequency between 20 Hz to 20000 Hz
(5) Audible frequency	(e) Low frequency waves	(v) Frequency greater than 20000 Hz

Answer:
(1-b- iii),
(2- a- i),
(3 – d – v),
(4 – e – ii),
(5 -c- iv)
In simple words: Matching the wave types with their characteristics and examples helps understand sound wave classifications.

🎯 Exam Tip: Pay close attention to the definitions of transverse and longitudinal waves and the frequency ranges for infrasound, audible sound, and ultrasound.

 

Question 2.

Column A'	Column 'B'	Column C
(1) Amplitude	(a) T	(i) Pitch of sound
(2) Frequency	(b) A	(ii) Loudness of sound
(3) Wavelength	(c) υ	(iii) Reciprocal of frequency
(4) Time period	(d) λ	(iv) ν/υ

Answer:
(1 -b - ii),
(2 -c - i),
(3-d – iv),
(4 - a - iii)
In simple words: This match-the-column exercise connects fundamental sound wave properties (amplitude, frequency, wavelength, time period) with their respective symbols, related concepts, or mathematical relationships.

🎯 Exam Tip: Understand the definitions and units of these basic wave characteristics, as they are foundational to the study of sound.

Name The Following:

Question 1. A form of energy which produces sensation of hearing in our ears.
Answer: Sound energy
In simple words: The energy that allows us to hear is called sound energy.

🎯 Exam Tip: Define energy types by their effects or primary functions.

 

Question 2. Repetitions of sound due to reflection .
Answer: Echo
In simple words: When a sound repeats because it has bounced off a surface, it's called an echo.

🎯 Exam Tip: Clearly distinguish between reverberation (continuous reflection) and echo (distinct reflection).

 

Question 3. The audible range of sound for human being.
Answer: 20 Hz to 20,000 Hz
In simple words: Humans can typically hear sounds with frequencies between 20 Hertz and 20,000 Hertz.

🎯 Exam Tip: Memorize the human audible frequency range (20 Hz - 20 kHz) as it's a fundamental concept.

 

Question 4. A method to obtain images of internal organs of the human body.
Answer: Sonography
In simple words: Sonography is a medical technique that uses sound waves to create images of organs inside the body.

🎯 Exam Tip: Understand the application of ultrasonic sound waves in medical imaging.

 

Question 5. The matter or substance through which sound gets transmitted.
Answer: Solid, liquid, gases
In simple words: Sound needs a medium to travel and can pass through solids, liquids, and gases.

🎯 Exam Tip: Remember that sound is a mechanical wave and requires a material medium for propagation.

 

Question 6. Three major parts of the ear.
Answer: External ear, the middle ear and the inner ear.
In simple words: The human ear is divided into three main sections: the outer, middle, and inner ear.

🎯 Exam Tip: Knowing the basic anatomical divisions of the ear is important for understanding hearing.

 

Question 7. Any two examples in which infrasound is produced.
Answer: Pendulum, earthquake.
In simple words: Infrasound, which is sound below human hearing, can be produced by a pendulum or during an earthquake.

🎯 Exam Tip: Be able to give examples of phenomena that produce sound outside the audible range.

 

Question 8. Name the living beings that can produce ultrasound.
Answer: Bats, dolphins, mice.
In simple words: Bats, dolphins, and mice are examples of animals that produce and use ultrasound for navigation or communication.

🎯 Exam Tip: Recognize common animals that utilize ultrasound for biological purposes.

Give Scientific Reasons:

Question 1. Bats can navigate in dark.
Answer:
• The ultrasonic sound produced by bats, gets reflected on hitting an obstacle.
• This reflected sound is received by their ears and they can locate the obstacle and estimate its distance even in the dark.
• Hence, bats can navigate in dark.
In simple words: Bats navigate in the dark by emitting high-frequency ultrasonic sounds that bounce off objects, allowing them to map their surroundings through echolocation.

🎯 Exam Tip: Explain the process of echolocation and how bats utilize ultrasonic waves for navigation, including sound emission, reflection, and reception.

 

Question 2. A SONAR system is installed in a ship.
Answer:
• A SONAR system determines the depth of the sea.
• It locates underwater hills, valleys, icebergs, submarines and sunken ships. It also locates the positions of other ships or submarines.
• Hence a SONAR system is installed in a ship.
In simple words: SONAR systems are installed in ships to use ultrasonic waves to detect underwater objects, measure sea depth, and locate other vessels or submerged structures.

🎯 Exam Tip: Describe the primary functions of SONAR (Sound Navigation and Ranging) related to underwater detection and mapping.

 

Question 3. Sound travels faster in iron than in air.
Answer:
• Sound requires a material medium for its propagation and travels in the form of a longitudinal wave.
• The denser the medium, faster is the propagation of sound.
• Hence, sound travels faster in iron than in air.
In simple words: Sound travels faster in iron than in air because iron is a much denser and more elastic medium, allowing sound vibrations to propagate more quickly.

🎯 Exam Tip: Remember that sound speed is generally higher in denser and more elastic media (solids) compared to less dense media (gases).

Solve The Following:

Formula:
(i) Velocity = \( \frac{\text{distance}}{\text{time}} \)
Type - A

Question 1. Ultrasonic waves are transmitted downwards into the sea with the help of a SONAR. The reflected sound is received after 4 s. What is the depth of the sea at that place? (Velocity of sound in seawater = 1550 m/s)
Answer:
Given:
Time to hear echo = 4 sec
Time taken by sound waves to reach the bottom of sea (t) = 4/2 sec = 2 sec
Velocity of sound in sea water (v) = 1550 m/s
To find:
Depth of sea(s) = ?
Formulae :
Velocity = \( \frac{\text{distance}}{\text{time}} \)
Solution:
Velocity = \( \frac{\text{distance}}{\text{time}} \)
\( \implies \) Distance = velocity \( \times \) time
= 1550 \( \times \) 2 = 3100 m
The depth of the sea at that place is 3100 m.
In simple words: The SONAR sends a pulse, which travels to the seabed and back. Since the total travel time is 4 seconds and the speed of sound in seawater is 1550 m/s, the one-way distance (depth) is calculated as speed multiplied by half the total time, resulting in a depth of 3100 meters.

🎯 Exam Tip: For echo-related problems, always remember to halve the total time for a one-way distance calculation, as the sound travels to and from the object.

 

Question 2. A person standing near a hill fires a gun and hears the echo after 1 second. If speed of sound in air is 340 m/s. Find the distance between the hill and the person.
Answer:
Given:
Time to hear echo = 1 sec
Time taken (t) = 1/2 sec
Velocity of sound (v) = 340 m/s
To find:
Distance (s) = ?
Formulae:
Velocity = \( \frac{\text{distance}}{\text{time}} \)
Solution:
Velocity = \( \frac{\text{distance}}{\text{time}} \)
\( \implies \) Distance = velocity \( \times \) time
= 340 \( \times \frac{1}{2} \)
= 170 m
Distance between the person and hill is 170 m.
In simple words: An echo heard after 1 second means the sound traveled to the hill and back in that time. Since the speed of sound is 340 m/s, the one-way distance to the hill is half the total distance covered, which is 170 meters.

🎯 Exam Tip: Similar to SONAR problems, remember to use half the echo time to calculate the distance to the reflecting surface.

Numerical For Practice

Question 3. If you hear the thunder 20 seconds after you see the flash of lightning, how far from you has the lightning occurred? (Speed of sound in air = 340 m/s)
Answer: 6800m
In simple words: Since thunder is heard 20 seconds after seeing lightning and sound travels at 340 m/s, the lightning occurred 6800 meters away (distance = speed x time).

🎯 Exam Tip: Light travels much faster than sound, so the time delay between seeing lightning and hearing thunder can be used to calculate distance. No need to divide time by two here as it's a one-way travel.

 

Question 4. Aboy observes smoke from a cannon 3 seconds before he hears the bang. If the cannon is 1020 m from the observer, find the velocity of sound.
Answer: 340 rn/s
In simple words: The sound from the cannon traveled 1020 meters in 3 seconds. To find the velocity, divide the distance by the time, which is 340 m/s.

🎯 Exam Tip: This is a direct application of the distance-speed-time formula; ensure correct units and calculation.

 

Question 5. A soldier standing between the two buildings fires a gun. He heard the echo of the sounds from the first building after 2 seconds and echo from the second building after 3 seconds. Find the distance between two buildings. (Speed of sound in air = 340 m/s)
Answer: 850m
In simple words: The soldier hears echoes from two buildings at different times. Calculating the distance to each building using half the echo time and the speed of sound (340 m/s) will give the individual distances. The sum of these two distances will be the distance between the buildings.

🎯 Exam Tip: This problem involves two separate echo calculations. Calculate the distance to each reflecting surface and then sum them for the total distance between the buildings.

Type - B

Formula : (i)Velocity = Frequency \( \times \) Wavelength (ii)Velocity = \( \frac{\text{distance}}{\text{time}} \)

Question 1. Sound waves of wavelength 1 cm have a velocity of 340 mls in air. What is their frequency? Can this sound be heard by the human ear?
Answer:
Given:
wave length (λ) = 1cm = 1/100 m
Velocity of sound (v) = 340 m/s
To find :
frequency (υ) = ?
Formulae:
v = υλ
Solution:
v = υλ
\( \implies \) υ = \( \frac{\text{V}}{\lambda} \)
\( \implies \) υ = \( \frac{340}{\frac{1}{100}} \)
\( \implies \) υ = 340 \( \times \) 100
\( \implies \) υ = 34000 Hz
The frequency of the sound waves is 34000 Hz. The frequency is higher than 20000 Hz and therefore, this sound cannot be heard by the human ear.
In simple words: Given the sound wave's velocity and wavelength, its frequency is calculated as 34000 Hz. This frequency is above the human audible range of 20,000 Hz, so it cannot be heard by humans.

🎯 Exam Tip: Remember the relationship v = fλ (velocity = frequency × wavelength) and be able to convert units consistently (cm to m). Also, recall the human audible range to determine if a sound is perceivable.

 

Question 2. How long will it take for a sound wave of 25 cm wavelength and 1.5 kHz frequency, to travel a distance of 1.5 km?
Answer:
Given:
frequency (υ) = 1.5 kHz = 1500 Hz
Wavelength (λ) = 25 cm = \( \frac{25}{100} \) m
Distance (s) = 1.5km = 1500m
To find:
time (t)=? Velocity (v)=?
Formulae:
(1) v=υλ (2) velocity = \( \frac{\text{distance}}{\text{time}} \)
Solution:
v = υλ
v = 1500 \( \times \frac{25}{100} \)
v = 15 \( \times \) 25
\( \implies \) v = 375 m/s
time = \( \frac{\text{distance}}{\text{velocity}} \)
\( = \frac{1500}{375} \)
= 4sec
The sound wave takes 4 sec to travel the distance of 1.5 km.
In simple words: First, calculate the velocity of the sound wave using its frequency and wavelength (v = fλ). Then, use this velocity and the given travel distance to find the time taken (t = distance / velocity).

🎯 Exam Tip: This problem requires a two-step calculation. Always convert all units to SI (meters, seconds, Hertz) before performing calculations to avoid errors.

 

Question 3. Calculate distance travelled by a sound wave having frequency 1000 Hz and wavelength 0.25 m, if it travels for 5 seconds in a certain medium.
Answer:
Given:
frequency (υ) = 1000 Hz
wavelength (λ) = 0.25 m
time (t) = 5 seconds
To find :
Distance (d) =?
Formulae:
v = υλ
Solution:
V = υλ
\( \implies \) v = 1000 \( \times \) 0.25
\( \implies \) v = 250 m/s
Also, velocity = \( \frac{\text{distance}}{\text{time}} \)
\( \implies \) distance = velocity \( \times \) time
= 250 \( \times \) 5
= 1250 m.
The distance travelled by the sound wave is 1250 m.
In simple words: First, calculate the speed of the sound wave using its frequency and wavelength (v = fλ). Then, multiply this speed by the time traveled to find the total distance covered.

🎯 Exam Tip: This is another two-step problem. Ensure you correctly calculate velocity first before applying it to find the total distance.

 

Question 4. The audible range of sound is 20 Hz to 20000 Hz. At 22°C in air speed of sound is 344 mls. Express the range of sound in terms of wavelength by calculating the respective values.
Answer:
Given:
frequency (υ₁) 20 Hz
frequency (υ₂) = 20,000 Hz
velocity (v) = 344 rn/s
To find :
Wavelengths λ₁ and λ₂ = ?
Formulae:
v = υλ
Solution:
v= υ₁λ₁
\( \implies \) λ₁ = \( \frac{\text{V}}{\text{υ₁}} \)
\( \implies \) λ₁ = \( \frac{344}{20} \)
\( \implies \) λ₁ = 17.2 m
Also, υ = υ₂λ₂
\( \implies \) λ₂ = \( \frac{\text{V}}{\text{υ₂}} \)
\( \implies \) λ₂ = \( \frac{344}{20000} \)
\( \implies \) λ₂ = 17.2 \( \times \) 10^-3 m
Audible range of wavelength of sound is from 17.2 \( \times \) 10^-3 m to 17.2 m.
In simple words: To find the audible wavelength range, we use the formula wavelength = velocity/frequency. We calculate the wavelength for the minimum (20 Hz) and maximum (20,000 Hz) audible frequencies, using the given speed of sound (344 m/s). This yields a wavelength range of 17.2 meters down to \(17.2 \times 10^{-3}\) meters.

🎯 Exam Tip: Be able to apply the wave equation (v=fλ) to find wavelength from frequency and velocity, and remember to convert units appropriately.

Numerical For Practice

Question 5. A sound wave has frequency 320 Hz and wavelength 0.25 m. How much distance will it travel in 10 second?
Answer: The distance travelled is 800 m.
In simple words: First, calculate the speed of the sound wave using its frequency (320 Hz) and wavelength (0.25 m). Then, multiply this speed by the travel time (10 seconds) to find the total distance.

🎯 Exam Tip: This is a standard calculation combining `v = fλ` and `distance = v * t`. Show all steps clearly.

Type - C

Formulae:
(i) \( \nu \alpha \frac{1}{\sqrt{M}} \), \( \nu \alpha \sqrt{T} \), \( \rho = \frac{m}{V} \)

Question 1. Hydrogen gas is filled in two identical bottles, A and B, at the same temperature. The mass of hydrogen in the two bottles is 12 gm and 48 gin respectively. In which bottle will sound travel faster? How many times as fast as the other?
Answer:
Given:
Mass of hydrogen in bottle A (m_A) = 12gm
Mass of hydrogen in bottle B(m_B) = 48gm
To find:
In which bottle sound travels faster.
Formulae:
\( \rho = \frac{m}{V} \)
\( \implies \nu \alpha \frac{1}{\sqrt{\rho}} \)
\( \implies \nu \alpha \frac{1}{\sqrt{\frac{m}{V}}} \)
\( \implies \nu \alpha \frac{\sqrt{V}}{\sqrt{m}} \)
Solution:
\( \text{V}_\text{A} \alpha \frac{\sqrt{\text{V}}}{\sqrt{\text{m}_\text{A}}} \) ...(i)
\( \text{V}_\text{B} \alpha \frac{\sqrt{\text{V}}}{\sqrt{\text{m}_\text{B}}} \) ...(ii)
Since both bottles are identical hence, the volume is the same, i.e. V
Dividing (i) and (ii),
\( \frac{\text{V}_\text{A}}{\text{V}_\text{B}} = \frac{\sqrt{\text{m}_\text{B}}}{\sqrt{\text{m}_\text{A}}} \)
\( = \sqrt{\frac{\text{m}_\text{B}}{\text{m}_\text{A}}} \)
\( = \sqrt{\frac{48}{12}} = \sqrt{4} = 2 \)
\( \implies \text{V}_\text{A} = 2\text{V}_\text{B} \)
(i) Vivacity of sound will be more in bottle A.
(ii) Velocity of sound in bottle A (V_A) is twice of that in bottle B (V_B)
In simple words: Since the velocity of sound is inversely proportional to the square root of the gas's mass density (and thus mass for identical volumes), sound will travel faster in bottle A, which has less hydrogen gas (12gm) compared to bottle B (48gm). Specifically, sound will be twice as fast in bottle A as in bottle B.

🎯 Exam Tip: Remember the inverse square root relationship between sound velocity and density (or mass for constant volume) of the medium. Calculations should clearly show this proportionality.

Numerical For Practice

Question 2. Argon gas is filled in two identical bottles X and Y. The mass of the gas in the two bottles is 5 gm and 25gm respectively. If the speed of sound is the same in both bottles, what conclusions will you draw?
Answer: (Temperature of Y is 5 times the temperature of X.)
In simple words: If the speed of sound is the same despite different gas masses, and assuming identical bottles, it implies the temperatures must be different. Since the velocity of sound is proportional to the square root of temperature, to compensate for the higher mass in bottle Y, its temperature must be 5 times that of bottle X.

🎯 Exam Tip: This problem tests the relationship between sound velocity, temperature, and mass. If velocity is constant and mass changes, temperature must compensate. Remember the proportionality `v ∝ √T` and `v ∝ 1/√m`.

Type - D

Numerical For Practice

Question 1. Velocity of sound in air at 0°C is 332nVs. It increases by 0.6ni/s for each °Celsius rise in temperature. At what temperature of ait the velocity will be 359m1s?
Answer: 45°C
In simple words: The velocity of sound in air increases by 0.6 m/s for every degree Celsius rise above 0°C. To reach a velocity of 359 m/s from 332 m/s, the increase needed is 27 m/s, which corresponds to a temperature of 45°C (27 / 0.6).

🎯 Exam Tip: This is a linear relationship problem. Calculate the total increase in velocity, then divide by the rate of increase per degree to find the temperature.

 

Question 2. Velocity of sound In air at 0°C is 332m/s It increases by 0.6mls for each degree Celsius rise In temperature. What will be the velocity of sound at 60°C?
Answer: 368 rn/s
In simple words: Starting from 332 m/s at 0°C, the velocity of sound increases by 0.6 m/s for each degree. At 60°C, the increase will be 60 * 0.6 = 36 m/s, making the total velocity 332 + 36 = 368 m/s.

🎯 Exam Tip: Apply the given rate of increase to the base velocity at 0°C to find the velocity at the specified higher temperature.

Define The Following:

Question 1. Wave length (λ)
Answer: The distance between two consecutive compressions (or crests) or two consecutive rarefactions (or troughs) is called the wavelength.
In simple words: Wavelength is the distance between two identical points on consecutive waves, such as two crests or two troughs.

🎯 Exam Tip: Provide a precise definition of wavelength, specifying it as the distance between corresponding points on successive waves.

 

Question 2. Amplitude (A)
Answer: The maximum value of pressure or density is called amplitude.
In simple words: Amplitude is the maximum displacement or intensity of a wave from its equilibrium position, representing the loudness of a sound.

🎯 Exam Tip: Relate amplitude to the intensity or loudness of sound, emphasizing it as the maximum deviation from equilibrium.

 

Question 3. Frequency (υ)
Answer: The frequency of a sound wave is defined as the number of complete oscillations of density (or pressure of the medium) per second.
In simple words: Frequency is the number of wave cycles or oscillations that occur in one second, determining the pitch of a sound.

🎯 Exam Tip: Define frequency clearly as oscillations per unit time and link it to the pitch of sound.

 

Question 4. Time Period (T)
Answer: The time taken for one complete oscillation of pressure or density at a point in the medium is called the time period.
In simple words: The time period is the duration required for one complete wave cycle or oscillation to pass a given point.

🎯 Exam Tip: Explain time period as the reciprocal of frequency and the time for one complete cycle.

 

Question 5. Echo
Answer: An echo is the repetition of the original sound because of reflection by some surface.
In simple words: An echo is a distinct repeat of a sound, caused by its reflection off a distant surface.

🎯 Exam Tip: Differentiate echo from reverberation; an echo is a clearly distinct reflected sound.

 

Question 6. Transverse waves
Answer: Oscillations of the particles of the medium vibrate at right angles to the direction of propagation of the wave are called transverse waves.
In simple words: In transverse waves, particles of the medium move perpendicularly to the direction the wave is traveling.

🎯 Exam Tip: Focus on the perpendicular relationship between particle oscillation and wave propagation for transverse waves.

 

Question 7. longitudinal waves
Answer: The particles of the medium oscillate about their central or mean position in a direction parallel to the propagation of wave is called as longitudinal waves.
In simple words: In longitudinal waves, particles of the medium move back and forth in the same direction that the wave is traveling.

🎯 Exam Tip: Focus on the parallel relationship between particle oscillation and wave propagation for longitudinal waves.

 

Question 8. Velocity of wave
Answer: The distance covered by a point on the wave (for example the point of highest density or lowest density) in unit time is the velocity of the sound wave.'
In simple words: Wave velocity is the speed at which a specific point on the wave, like a crest or trough, travels through the medium.

🎯 Exam Tip: Define wave velocity as the rate of propagation of a disturbance or energy through a medium.

Distinguish Between:

Question 1. Infrasound and Ultrasound
Answer:

Infrasound	Ultrasound
(i) Longitudinal waves whose are below 20 Hz are called Infrasound waves. frequencies Infrasonic or	(i) Longitudinal waves whose frequencies lie- above 20,000 Hz are called Ultrasonic or ultrasound waves.
(ii) Whales, elephants produce sound in the infrasound range.	(ii) Bats produce (30 kHz to 50 kHz) frequency and dolphins produce ultrasound (100 kHz).

In simple words: Infrasound refers to sound waves with frequencies below 20 Hz, which humans cannot hear, and is produced by large animals like whales. Ultrasound refers to sound waves with frequencies above 20,000 Hz, also inaudible to humans, and used by animals like bats and dolphins.

🎯 Exam Tip: Clearly state the frequency ranges and provide examples of natural phenomena or animals associated with each type of sound.

 

Question 2. Transverse waves and Longitudinal waves
Answer:

Transverse waves	Longitudinal waves
(i) Particles of the medium vibrate at right angles to the direction of propagation of the wave.	(i) Particles of the medium vibrate parallel to the direction of propagation of the wave.
(ii) They produce crests and troughs.	(ii) They produce compression and rarefaction.
(iii) For transverse waves, a wavelength is made up of one crest and one trough.	(iii) For longitudinal waves, a wavelength is made up of one compression and one rarefaction.

In simple words: Transverse waves involve particles oscillating perpendicular to wave direction, forming crests and troughs, while longitudinal waves involve particles oscillating parallel to wave direction, forming compressions and rarefactions.

🎯 Exam Tip: The key distinguishing factor is the direction of particle oscillation relative to wave propagation. Also, remember the specific wave features (crests/troughs vs. compressions/rarefactions).

 

Question 3. Consider two cases
(A) whistle of train (B) roar of a lion
(I) In which case the sound is high pitch?
Answer: Whistle of a train is high pitch as compared to roar of a lion, as the frequency is higher.
In simple words: The train whistle produces a high-pitched sound because it has a higher frequency than the low-pitched roar of a lion.

🎯 Exam Tip: Connect pitch directly to frequency; higher frequency means higher pitch.

 

(II) What is the real cause of sound production? Explain with examples.
Answer:
• Vibrations in the object are responsible to produce a sound.
• Vibration is a rapid to and fro motion of an object.
In simple words: Sound is fundamentally produced by vibrations, which are rapid back-and-forth movements of an object, like a guitar string vibrating to create musical notes.

🎯 Exam Tip: Emphasize that all sound originates from vibrations and provide a relevant example to illustrate this principle.

Answer In Detail:

Question 1. What are the factors on which velocity of sound in gaseous medium depend?
Answer:The velocity of sound in a gaseous medium depends on the physical conditions i.e. the temperature, density of the gas and its molecular weight.
1. Temperature (T): The velocity of sound is directly proportional to the square root of the temperature of the medium. This means that increasing the temperature four times doubles the velocity. \( \nu \propto \sqrt{T} \)
2. Density(\( \rho \)): The velocity of sound is inversely proportional to the square root of density. Thus, increasing the density four times, reduces the velocity to half its value. \( \nu \propto \frac{1}{\sqrt{\rho}} \)
3. Molecular weight (M): The velocity sound is inversely proportional to the square root of molecular weight of the gas. Thus, increasing the molecular weight four times, reduces the velocity to haff its value. \( \nu \propto \frac{1}{\sqrt{M}} \)In simple words: The speed of sound in gas is affected by its temperature (higher temperature, faster sound), density (denser, slower sound), and molecular weight (heavier molecules, slower sound).

🎯 Exam Tip: Remember the direct and inverse proportionalities for each factor (temperature, density, molecular weight) as they are common theoretical questions.

 

Question 2. What are the uses of ultrasonic sound?
Answer:Uses of ultrasonic sound are as follows:

• For communication between ships at sea.
• To join plastic surfaces together.
• To sterilize liquids like milk by killing the bacteria in it so that the milk keeps for a longer duration.
• Echocardiography which studies heartbeats, is based on ultrasonic waves (Sonography technology).
• To obtain images of internal organs in a human body.
• In industry to clean intricate parts of machines where hands cannot reach.
• To locate the cracks and faults in metal blocks.

In simple words: Ultrasonic sound is used for various purposes like underwater communication, cleaning delicate objects, sterilizing liquids, and creating medical images of internal organs and the heart.

🎯 Exam Tip: Focus on understanding the practical applications of ultrasound, especially in medical and industrial fields, as these are frequently asked.

 

Question 3. Explain with the help of a neat labelled diagram the working of human ear.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र मानव कान की संरचना को दर्शाता है, जिसमें बाहरी कान (पिन्ना और ऑडिटरी कैनाल), मध्य कान (इयरड्रम, तीन हड्डियां - मैलियस, इनकस, स्टेपीज़) और आंतरिक कान (कॉक्लिया और अर्धवृत्ताकार नलिकाएं) शामिल हैं। यह विभिन्न भागों के बीच ध्वनि के संचरण को दिखाता है।

• The ear is an important organ of the human body.
• When sound waves fall on the eardrum, it vibrates and these vibrations are converted into electrical signals which travel to the brain through nerves.
• The ear can be divided into three parts:
  (a) Outer ear
  (b) Middle ear
  (c) Inner ear.

(a) Outer ear or Pinna The outer ear collects the sound waves and passes them through a tube to a cavity in the middle ear. Its peculiar funnel-like shape helps to collect and pass sounds into the middle ear.
(b) Middle ear There- is a thin membrane in the cavity of the middle ear called the eardrum. When a compression in a sound wave reaches the eardrum, the pressure outside it increases and it gets pushed inwards. The opposite happens when a rarefaction reaches there. The pressure outside decreases and the membrane gets pulled outwards. Thus, sound waves cause vibrations of the membrane.
(c) Inner ear The auditory nerve connects the inner ear to the brain. The inner ear has a structure resembling the shell of a snail. It is called the cochlea. The cochlea receives the vibrations coming from the membrane and converts them into electrical signals which are sent to the brain through the nerve. The brain analyses these signals.In simple words: The human ear collects sound waves through the outer ear, which then make the eardrum vibrate. These vibrations are amplified by small bones in the middle ear and converted into electrical signals by the cochlea in the inner ear, which are then sent to the brain for interpretation.

🎯 Exam Tip: A well-labelled diagram of the human ear is crucial for this question, along with clear descriptions of how each part processes sound.

 

Question 4. Write a short note on SONAR
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र SONAR तकनीक को दर्शाता है जिसका उपयोग समुद्र के नीचे की वस्तुओं का पता लगाने के लिए किया जाता है। एक जहाज एक ट्रांसमीटर का उपयोग करके अल्ट्रासोनिक पल्स भेजता है, जो पानी के माध्यम से यात्रा करती है, वस्तु से परावर्तित होती है, और जहाज पर एक रिसीवर द्वारा प्राप्त की जाती है।
(i) SONAR is the short form for Sound Navigation and Ranging. It is used to determine the direction, distance and speed of an underwater object with the help of ultrasonic sound waves. SONAR has a transmitter and a receiver, which are fitted on ships or boats.
(ii) The transmitter produces and transmits ultrasonic sound waves. These waves travel through water, strike underwater objects and get reflected by them. The reflected waves are received by the receiver on the ship.
(iii) The receiver converts the ultrasonic sound into electrical signals and these signals are properly interpreted. The time difference between transmission and reception is noted. This time and the velocity of sound in water give the distance from the ship, of the object which reflects the waves.
(iv) SONAR is used to determine the depth of the sea. SONAR is also used to search underwater hills, valleys, submarines, icebergs, sunken ships etc.In simple words: SONAR (Sound Navigation and Ranging) uses ultrasonic sound waves emitted from a ship to detect underwater objects by measuring the time it takes for the sound to reflect back, thus determining their distance and location.

🎯 Exam Tip: Understanding the full form of SONAR and its mechanism (transmission, reflection, reception, and calculation) is key, along with its various applications in marine exploration.

 

Question 5. Write a short note on Sonography. How is it misused?
Answer:

• Sonography technology uses ultrasonic sound waves to generate images of internal organs of the human body.
• This is useful in finding out the cause of swelling, infection, pain, condition of the heart, the state of the heart after a heart attack as well as the growth of foetus inside the womb of a pregnant woman.
• This technique makes use of a probe and a gel.
• The gel is used to make proper contact between the skin and the probe so that the full capacity of the ultrasound can be utilized.
• High-frequency ultrasound is transmitted inside the body with the help of the probe.
• The sound reflected from the internal organ is again collected by the probe and fed to a computer which generates the images of the internal organ.
• As this method is painless, it is increasingly used in medical practice for correct diagnosis.
• This technique is used by many people to find out gender of an unborn baby and this often leads to the incidence of female foeticide.

In simple words: Sonography uses high-frequency ultrasound waves to create images of internal body organs for diagnosis, especially for examining a foetus during pregnancy, but it is misused for illegal sex determination leading to female foeticide.

🎯 Exam Tip: When discussing sonography, ensure you cover both its beneficial medical applications and the ethical concerns surrounding its misuse for sex determination.