This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 551. |
How does a bat make use of ultrasonics waves to find its way? |
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Answer» Use of ultrasonics by bats : Bats have no eyes. But they easily move about without colliding with any object (or obstacle). The reason is that they produce ultrasonic sound as they fly. When this ultrasonic sound comes back after reflection from any object (or obstacle) in their way, they hear it and thus they detect the presence of the object (or obstacle). |
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| 552. |
Match the following:1. Ultrasonicsa. Frequency below 20Hz2. Speed of sound in airb. Needs material medium3. Infrasonicsc. Needs material medium4. Sound propagationd. Frequency more than 20000 Hz |
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Answer» 1. Ultrasonics – Frequency more than 20000 Hz 2. Speed of sound in air – 330 m 3. Infrasonics – Frequency below 20Hz 4. Sound propagation – Needs material medium |
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| 553. |
Match the following.Column - IColumn - II(a) Tuning fork(i) The point where density of air is maximum(b) sound(ii) Maximum displacement from the equilibrium position(c) Compressions(iii) The sound whose frequency is greater than 20, 000 Hz(d) Amplitude(iv) Longitudinal wave(e) Ultrasonics(v) Production of sound |
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Answer» (a) (v) (b) (i) (c) (iv) (d) (ii) (e) (iii) |
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| 554. |
What is mean by Pitch. |
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Answer» Pitch : The pitch of sound depends on the frequency of sound (vibration). It is directly proportional to its frequency. Greater the frequency, higher is the pitch and lesser the frequency, lower is the pitch. |
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| 555. |
Write the Applications of Ultrasound. |
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Answer» • It is used for medical diagnosis and therapy and also as a surgical tool. • Bats and porpoises use ultrasound for navigation and to locate food in darkness. • It is used to detect defective foetus. • It is used as a tool in the treatment of muscular pain. • Ultra sonography (is a technique of 3-dimensional photographs with the help of ultrasonic waves) is used to locate the exact position of an eye tumour. • Ultrasound is generally used to clean spiral tubes,electronic components etc. • Ultrasound are used to detect cracks and flaws in metal blocks. |
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| 556. |
Determination of the depth of an ocean |
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Answer» The depth of an ocean is determined with the help of SONAR. Sonar uses ultrasonic waves to detect and locate objects under water. Ultrasonic waves produced from the transmitter kept in a ship are directed towards the ocean floor. The ocean floor reflects these waves. By measuring the time interval t between the generation of the wave and reception of the echo, we calculate the depth of the ocean by using the relation S = (v x t)/2 Where v is the velocity of ultrasonic waves. |
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| 557. |
Explain SONAR. |
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Answer» One of the most important applications of the reflection of sound is oceanographic studies. For this purpose, we use a system called the SONAR. The SONAR is abbreviated form of Sound Navigation and Ranging. The SONAR system is used for detecting the presence of unseen under water object, such as submerged submarine, a sunken ship, iceberg and locating them. In Sonar ultrasonic waves are sent in all directions from the ship and are then received on their return after reflection. |
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| 558. |
For a sine wave passing through a medium, let y be the displacement of a particle, v be its velocity and a be its acceleration. (a) y, v and a are always in the same phase. (b) y and a are always in opposite phase. (c) Phase difference between y and v is π/2. (d) Phase difference between v and a is π/2. |
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Answer» Correct Answer is: (b, c, & d) |
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| 559. |
The displacement due to a wave moving in the positive x-direction is given by y = 1/(1 +x2) at time t = 0 and by y = 1 / [1 + (x - 1)2] at t = 2 seconds, where x and y are in metres. The velocity of the wave in m/s is(a) 0.5 (b) 1 (c) 2 (d) 4 |
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Answer» Correct Answer is: (a) 0.5 We rewrite the given equation as y = 1/1 +(x - vt)2. For t = 0, this becomes y = 1 / (1 + x2), as given. For t = 2, this becomes y = 1 / [1 + (x - 2v)2] = 1 / [1 + (x - 1)2]. ∴ 2v = 1 or v = 0.5 m/s. |
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| 560. |
A sine wave has an amplitude A and a wavelength λ. Let V be the wave velocity, and v be maximum velocity of a particle in the medium. (a) V cannot be equal to v (b) V = v, if A = λ/2π (c) V = v, if A = 2πλ(d) V = v, if λ = A/π |
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Answer» Correct Answer is: (b) V = v, if A = λ/2π y = Asin(ωt - kx) ω = 2πn, k = 2π/λ, where n = frequency and λ = wavelength. Wave velocity = V = nλ = ω/k. Particle velocity = dy/dt = Aωcos(ωt - kx). ∴ maximum particle velocity = v = Aω. For V = v1, ω/k = Aω or A = 1/k = λ/2π. |
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| 561. |
What is echo ? How is echo formed ? |
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Answer» The repetition of sound caused by the reflection of sound waves is called an echo. • We can hear echo when there is a time gap of 0.1 second in original sound and echo (reflected sound). • Echo is produced when sound reflected from a hard surface (i.e., brick wall, mountain etc.) as soft surface tends to absorb sound. |
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| 562. |
Write any three applications of ultrasound. |
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Answer» Applications of Ultrasound: (i) It is used to detect cracks in metal blocks in industries without damaging them. (ii) It is used in industries to clean ‘hard to reach’ parts of objects such as spiral tubes, odd shaped machines etc. (iii) It is used to investigate the internal organs of human body such as liver, gall bladder, kidneys, uterus and heart. |
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| 563. |
A student notes down the observations in the two spring balances and the measuring cylinder. .A. 64 ccB. 36 ccC. 30 ccD. 100 cc |
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Answer» ( c) Loss in weight of the solid = difference in two readings of spring balance `= 65 g - 35 g - 30 g` Volume of water displaced `= cc (as 1 cc of water weights 1 g)` Volume of the solid `= 30 cc`. |
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| 564. |
While determining the density of a copper piece using a spring balance and a measuring cylinder, Seema carried out the following procedure : (i) noted the water level in the measuring cylinder without the copper piece. (ii) immersed the copper piece in the water. (iii) noted the water level in the measuring cylinder with the copper piece inside it. (iv) removed the copper piece from the water and immediately weighed it using a spring balance. The wrong step in the proceedure is :A. step (i)B. step (ii)C. step (iii)D. step (iv) |
| Answer» (d) It is wrong to weigh the wet copper piece in air. | |
| 565. |
Which of the following statements is correct ?A. All vibrations produce sound.B. Vibrations of frequency less than 20 Hz only produce sound.C. Vibrations of frequency more than 20 Hz only produce sound.D. Vibrations of frequency between 20 Hz and 20,000 Hz only produce sound. |
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Answer» Correct Answer - A All vibrations produce sound but we can hear sound only when the frequency is between 20 Hz and 20,000 Hz |
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| 566. |
The velocity of sound in vacuum is _______ `ms^(-1)`.A. `332`B. `330`C. `3 xx 10^(8)`D. zero |
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Answer» Correct Answer - D The velocity of sound is zero in vacuum. Sound cannot be propagated through vacuum. |
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| 567. |
What is rarefaction? |
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Answer» As a sound wave travels through air, the region in which air is at low pressure and low density is called rarefaction. |
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| 568. |
What is the relation between the frequency of vibration and the vibrating length (or height) of the air column? |
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Answer» The greater the vibrating length (or height) of the air column, the less is the frequency of vibration of the column. |
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| 569. |
State the relation between time period and frequency. |
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Answer» The time period (T) and the frequency (f) are inversely proportional to each other T ∝ \(\frac{1}{f}\) |
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| 570. |
Name the characteristic property of sound through which we can differentiate between the sound of male and female. |
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Answer» The shrillness of sound is the characteristic property of sound which differentiates the sound between male and female. Usually, the voice of a woman has higher frequency and shriller than the voice of a man. |
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| 571. |
As age increases, the audible range of frequency of sound …………… A) increases B) decreases C) both A & B D) none of these |
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Answer» Correct option is B) decreases |
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| 572. |
Mention the range of audible sound of us. |
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Answer» The range of audible sound is from 20 Hz to 20000 Hz or 20 kHz. |
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| 573. |
1. What is an echo 2. State two conditions for an echo to take place. |
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Answer» 1. The sound heard after reflection from a distant obstacle after the original sound has ceased is called an echo, 2. The conditions for an echo to take place are a. The minimum distance between the source of sound and the reflector in air must be 17m. b. The size of the reflector must be large enough as compared to the wavelength of sound wave. |
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| 574. |
Describe any two applications of reflection of sound. |
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Answer» (i) Megaphone : People use horn shaped metal tubes commonly called megaphones while addressing a group of people in fairs or tourist spots. Sound energy is prevented from spreading out by successive reflections from the horn shaped tubes. (ii) Hearing aid : Its shape is like a trumpet the narrow end is kept in the ear tube of the person who is hard of hearing. Where as the wider end towards the speaker collects the waves and reflects into the narrow end. This increases the intensity of sound energy hence the person who is hard of hearing can hear clearly. |
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| 575. |
What is an echo? |
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Answer» ECHO : “The repeated sound heard after reflection from a distant rigid obstacle (such as cliff, a hill side, wall of a building, edge of forest etc.) after the original sound has ceased is called an ECHO.” |
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| 576. |
What are reverberation? Give two examples. |
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Answer» Reverberation : “Due to repeated reflections at the reflecting surface (reflector is less than 17 metres from original sound) the sound gets prolonged, This effect is known as reverberation.” Example : 1. Speaking in a large empty room. 2. Clapping in tombs like TajMahal. |
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| 577. |
State two conditions necessary for the formation of an echo. |
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Answer» Two conditions for forming an echo : 1. The minimum distance between the source of sound and the reflecting body should be 17 metres. 2. The intensity of sound should be sufficient so that it can be heard after reflection. |
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| 578. |
1. Name the phenomenon involved in tuning a radio set to a particular station. 2. Define the phenomenon named by you in part (i) above. 3. What do you understand by loudness of sound? 4. In which units is the loudness of sound measured? |
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Answer» 1. The phenomenon involved in tuning a radio set to a particular station is called resonance. 2. Resonance : When the frequency of an externally applied periodic force on a body is equal to its natural frequency, the body readily begins to vibrate with an increased amplitude. This phenomenon is known as resonance. 3. Loudness is the property by virtue of which a loud sound can be distinguished from a faint one, both having the same pitch and quality. 4. The loudness of sound is measured in decibel (dB). |
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| 579. |
Define progressive wave. State any four properties. |
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Answer» Waves in which a disturbance created at one place travels to distant points and keeps travelling unless stopped by an external force are known as travelling or progressive waves. Properties of progressive waves are : Amplitude, wavelength, period, double periodicity, frequency and velocity. |
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| 580. |
The intensity of sound of a fire cracker is …….. dB. A) 140 B) 120 C) 60 D) 110 |
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Answer» Correct option is A) 140 |
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| 581. |
It takes `2.4 s` to record the echo of a sonar. If the speed in water is `1450 m//s`, find the depth of the ocean floor. |
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Answer» Here, `t = 2.4 s, v = 1450 m//s` Depth of the ocean floor, `d = (vt)/(2)` or `d = (1450 xx 2.4)/(2) = 1740 m`. |
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| 582. |
When a sound wave goes from one medium to another, the quantity that remains unchanged is :A. speedB. amplitudeC. frequencyD. wavelength |
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Answer» Correct Answer - C Frequency is a characteristic of sound wave which does not change with the change of medium. |
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| 583. |
How is ultrasound used in cleaning? |
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Answer» Ultrasonic cleaning is the rapid and complete removal of contaminants from objects by immersing them in a tank of liquid flooded with high frequency sounds waves. These non-audible sound waves create a scrubbing brush action within the fluid. |
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| 584. |
An observer standing at the sea coast observes `54 waves` reaching the coast per minute. If the wavelength of a wave is `10 m`, its speed is :A. `90 m//s`B. `90 cm//s`C. `9 m//s`D. `900 m//s` |
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Answer» Correct Answer - C As frequency, `v = (54 "waves")/(60 s) = (9)/(10) Hz, lamda = 10 m, v = v lamda = ((9)/(10) xx 10) m//s = 9 m//s`. |
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| 585. |
(a) What type of phenomenon creates sound ? (b) What is the wavelength of a sound wave ? |
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Answer» (a) Sound is a wave phenomenon. (b) Distance travelled by sound wave in its periodic time. |
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| 586. |
The frequencies of four sound waves are given below. Which of these sound waves can be used to measure the depth of sea by the echo method?A. 15,000 HzB. 10 kHzC. 50 kHzD. 10,000 Hz |
| Answer» Correct Answer - C | |
| 587. |
The ultrasound waves can penetrate into matter to a large extent because they have:A. very high speedB. very high freqquencyC. very high wavelengthD. very high amplitude |
| Answer» Correct Answer - B | |
| 588. |
What is the range of frequencies associated with (a) infra sound (b) ultrasound ? |
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Answer» (a) Range of frequencies associated with infrasound is less than 20Hz. (b) Range of frequencies associated with ultrasound is higher than 20,000 Hz. |
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| 589. |
Explain how bats use ultrasound to catch a prey. |
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Answer» Bats emit ultrasonic waves and search their food. Ultrasonic waves having higher pitch reflects from prey and reaches ears of the Bat. Nature of reflection indicates to Bat regarding position and type of prey. |
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| 590. |
If the velocity of sound in hydrogen is 1248 m/s, the velocity of sound in oxygen is[Given: MO = 32 and MH = 2](A) 1248 m/s (B) 624 m/s (C) 312 m/s (D) 300 m/s |
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Answer» Correct option is: (C) 312 m/s |
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| 591. |
If the source is moving away from the observer, then the apparent frequency …………..(A) will increase (B) will remain the same (C) will be zero (D) will decrease |
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Answer» Correct option is: (D) will decrease |
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| 592. |
The working of SONAR is based on ……………(A) resonance (B) speed of a star (C) Doppler effect (D) speed of rotation of sun |
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Answer» Correct option is: (C) Doppler effect |
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| 593. |
For the symbol H, D and T tabulate three subatomic particles found in each of them. |
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| 594. |
Some diwali crackers harmful to ears. Due to they produce a) high pitch sound b) high intensity sound c) high amplitude sound d) high frequency soundA) a, d B) a C) b, c D) d |
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Answer» Correct option is C) b, c |
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| 595. |
Why does a cracker make a loud sound? A) It vibrates air molecule B) It burns with air C) It gives light with heatD) It gives fumes |
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Answer» Correct option is A) It vibrates air molecule |
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| 596. |
The intensity of near total silence is …….. dB.A) 0 B) 60 C)110 D) 120 |
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Answer» Correct option is A) 0 |
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| 597. |
A 30 dB sound is …………. times more powerful than the total silence. A) 10 B) 100 C) 1000 D) 10000 |
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Answer» Correct option is C) 1000 |
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| 598. |
In the above waves, which one has more frequency? A) A B) B C) C D) D |
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Answer» Correct option is A) A |
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| 599. |
BladeVibrationsAmplitudeA15000.005 mB10000.05 inC1000.01 inFrom above table, which blade has more intensity? A) A B) B C)C D) None |
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Answer» Correct option is B) B |
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| 600. |
What experimental evidence is there fro assuming that the speed of sound is the same for all the wavelengths ? |
| Answer» In a given medium, sounds produced by different sources emitting waves of different wavelengths reach us in the same time. | |