Formatting page ...
Question Bank Sound 1. What is an echo? Ans. The phenomenon due to which the repetition of sound is heard, after reflection from a distant object (such as high building or hillock), after the original sound from a given source dies off, is called an echo. 2. State three conditions for the formation of echo. Ans. 1. The minimum distance between the source of sound and reflecting body should be 17 m. 2. The wavelength of sound should be less than the height of reflecting body. 3. The intensity of sound should be sufficient, so that it could be heard after reflection. 3. Derive a relation between speed of sound, distance of the source of sound from a reflecting body and the time for hearing an echo. Ans. Let v be the velocity of sound, t the time after which an echo is heard and d the distance between source of sound and reflecting body. Total distance travelled by sound before an echo is heard = (d + d) = 2d. In time t, distance travelled by sound = 2d. Class-X 1 Question Bank In time 1s, distance travelled by sound = 2d t But distance travelled by sound in one second is wave velocity, v. V= 2d t 4. Briefly describe how will you find the speed of sound by the method of echoes. or Explain how the speed of sound can be determined by the method of echoes. Ans. An observer selects a high rise building at a distance of at least half a kilometre or more and stands at A which is at a distance x from building. He fires the gun and records the time in which echo is heard, let the time be t1. Speed of sound V = 2x t1 or 2x = vt1 Class-X ...(i) 2 Question Bank The observer then moves a known distance d (not less than 150 m) backward and again fires the gun and records time in which echo is heard. Let the time be t2. Speed of sound v = or 2( x + d ) t2 2x + 2d = vt2 ...(ii) Substracting (i) from (ii) 2x + 2d 2x = vt2 vt1 2d = v (t2 t1) v= 2d t2 t1 As d , t2 and t1 are known, the velocity of sound can be calculated. 5. State two practical uses of echoes. Ans.(1) Echoes are used for locating fish in the sea. (2) Echoes are used by army to locate the enemy gun positions. 6. The echo of a sound made by us is not heard in a small room, but is heard distinctly in a big hall. Explain, why? Ans. In small rooms the distance between the source of sound and the walls is less than 17 m. Thus, no echo is heard. However, in case of big halls, the distance between the source of sound and the walls is more than 17 m. Thus, an echo is formed. Class-X 3 Question Bank 7. What do you understand by the term reverberations? Give an example. Ans. Reverberations : When series of reflections fall on ear from various reflectors one after another in a closed space, thereby forming a continuous rolling sound, reverberations are said to take place. For example, if someone speaks in an empty hall, a continuous rolling sound is heard because of multiple reflections taking place from the walls. 8. Why is an echo not heard when distance between the source of sound and reflecting body is less than 10 m? Ans. The impression of sound lasts on the eardrum for second the sound travels a distance of 332 1 1 of a second. In of a 10 10 m 1 s = 32.2 m. Thus, the s 10 minimum distance between the source of sound and the reflecting body should be 33.2 2 = 16.6 m. As the distance of 10 m is far less than 16.6 m, therefore, the ear cannot make out when the original sound has died and echo has been received. Thus, no echo is heard. Class-X 4 Question Bank 9. State the use of an echo by : (i) a bat, (ii) a dolphin, (iii) a fisherman, (iv) and a warship. Ans. (i) Bat uses echoes to locate its prey and to avoid obstacles in its flight path. (ii) Dolphin uses echoes to establish contact with other members of the family and to detect its enemies. (iii) Fisherman uses echoes to locate the shoals of fish in the open sea. (iv) Warship uses echoes to detect the enemy submarines. 10. What kind of waves are used by a bat to locate its prey? Ans. It uses ultrasonic waves. 11. What is sonar? State the principle on which it is based. What kind of waves are used in it. Ans. Sonar is a device fitted in the sailing ships, trawlers, warships, etc., to locate the submarines or the shoals of fish or the depth of ocean bed. It is based on the principle of reflection of sound from the solid distant objects in the form of echoes. Ultrasonic waves are used in the sonar. Class-X 5 Question Bank 12. Explain Briefly the working of radar. Ans. Radar : In this instrument radio waves of very short wavelength are used to locate the enemy aircraft or the ship. A concave transmitter mounted on rotating platform sends radio waves in all directions. The radiowaves on striking the aircraft or the ship are reflected back. The reflected radiowaves are received by the concave receiver, which is mounted at a small angle with the transmitter. On receiving the radiowaves, the position of enemy air-craft can be located on a monitor screen as a bright spot. 13. What do you understand by the following terms : (a) Free vibrations, (b) Natural frequency. (c) Natural time period. Ans.(a) Free vibrations : The vibrations produced in a body, on being slightly disturbed from its mean position, are called free vibrations. (b) Natural frequency : The number of vibrations executed by a freely vibrating body in one second is called natural frequency. (c) Natural time period : The time period of a body executing one free vibration is called natural time period. 14. Distinguish between free vibrations and forced vibrations. Ans. (a) When a body vibrates with its natural frequency, it is said to produce free vibrations. However, if a body vibrates with a frequency, other than its natural frequency, it is said to produce forced vibrations. (b) No continuous external force is required to produce free vibrations. However, a continuous external force is required to produce forced vibrations. Class-X 6 Question Bank 15. When does resonance occur? Give one example based on the phenomenon of resonance. Ans.When the natural frequency of a given body corresponds to the frequency of the vibrations impressed upon it, such that it vibrates with a large amplitude, resonance is said to take place. Example : It is a common experience that sometimes a wine glass start rattling when a note of some particular frequency is played from a musical instrument. At this moment the natural frequency of the wine glass corresponds to the frequency of the note impressed on it. Thus, resonance takes place, with the result, the wine glass vibrates with an increased amplitude and hence rattles. 16. The stem of a vibrating tuning fork is pressed against the table top. Answer the following questions : (i) Will the above action produce any audible sound? (ii) Does the above action cause the table to set into vibration? (iii) If the answer in (ii) is yes, what kind of vibrations are they? (iv) Under what conditions, the above action leads to resonance? Ans.(i) Yes. It will produce audible sound. (ii) Yes. The above action sets the table in vibrating mode. (iii) The vibrations produced in table are forced vibrations. (iv) When the frequency of tuning fork is same as that of the natural frequency of table, the resonance takes place. Class-X 7 Question Bank 17. In the given diagram, A, B, C and D are four pendulums suspended from same elastic wire PQ, such that lengths of pendulums B and D are same. The pendulum D is set to motion. What is your observation? State reason for your observation. Ans. (i) The pendulums B and D will vibrate in same phase, i.e., they will describe exactly same kind of motion. (ii) The pendulums A and C will vibrate out of phase, compared to pendulums B and D. Reason : The time period, and hence, the frequency of pendulums B and D is same. Thus, when pendulum D is set in motion, its frequency is impressed on pendulum B. Hence, B will vibrate in the same phase as D, due to resonance. The time period and hence the frequency of pendulums A and C is not the same as compared to pendulum D. Thus, these pendulums will vibrate with forced vibrations. Class-X 8 Question Bank 18. A stretched wire 10 m long is made to vibrate in two different modes as shown in diagrams (A) and (B) given below. (a) If the wavelength of wave produced in (A) is 2 m, what is the wavelength of wave produced in mode (B) in the diagram? (b) In which case is the note produced louder? Give reason for your answer. (c) In which case is the pitch of the note produced higher? Give reason for your answer. Ans.(a) Wavelength of B is 1 m. (b) In case of A, note is louder. It is because loudness (Amplitude)2. (c) In case of B, pitch of note is higher. It is because the frequency of B is double than A, and hence, higher is the pitch. Class-X 9 Question Bank 19. (a) Write down the factors on which frequency of a vibrating string depends. (b) What adjustments will you make for tuning stringed instruments, such as violin, for it to emit a desired pitch? Ans. (a) Frequency of vibrating string is 1 Length of wire Tension in wire 1 Mass per unit length of wire (b) The tension in the wire is increased or decreased by tuning the knobs on violin, till it gives note of desired pitch. 20. A 0.6 m long stretched wire is made to vibrate in two different modes as shown in the figures below. Copy and mark the points with N for the points of least displacement. If the frequency of the note produced in diagram (ii) is n, what is the frequency in case of (i)? By stating a reason, explain which of two notes (i) or (ii) is louder? Class-X 10 Question Bank Ans. (i) (ii) The frequency of note in (i) is n . 2 (iii) The note produced in (i) is louder, as it has a larger amplitude. The loudness (Amplitude)2. 21. Why are soldiers asked to walk out of step while crossing bridges? Ans. When the soldiers walk in step, they produce same fixed frequency. If this frequency corresponds to the natural frequency of the bridge, the resonance can take place. This in turn will make the bridge to vibrate violently, and hence, make it collapse. To avoid such a situation, the soldiers are asked not to march in step. 22. Why does a wine glass start rattling, when a note of some particular frequency is struck by a piano? Ans. When the glass rattles, at that moment, its natural frequency corresponds with the frequency of piano note. Thus, resonance takes place which makes the glass to vibrate violently. Class-X 11 Question Bank 23. Why does rear-view mirror of a motor bike start vibrating violently, at some particular speed of motor bike? Ans. When the frequency of the engine of motor bike corresponds with the natural frequency of rear-view mirror, the resonance takes place. Thus, rear-view mirror vibrates with a larger amplitude violently. 24. A tuning fork (vibrating) is held close to the ear. One hears a faint hum. The same (vibrating) tuning fork is placed on table, such that its handle is in contact with table, one hears a loud sound. Explain? Ans. When the tuning fork is held close to the ear, then small amount of air is disturbed, and hence, sound is faint. When the handle of the vibrating tuning fork is held against table, it makes the table top vibrate, with forced vibrations. As the table top has a larger surface area, therefore large volume of air is set into vibrations, thereby producing a loud sound. 25. Why are stringed musical instruments provided with large sound boxes? Ans. The large sound box contains large amount of trapped air. When the vibrations of the vibrating string are impressed upon the air, it starts vibrating with forced vibrations. As large volume of enclosed air vibrates, a loud sound is produced. 26. A person walking past a railway line, at the middle of night hears a ringing sound along with the sound of his footsteps. Why? Ans.When the vibrations produced by the feet of the person are impressed on the rails, they vibrate with forced vibrations, thereby producing a ringing sound. Class-X 12 Question Bank 27. Fill in the blanks. (i) A wire is stretched between two fixed supports. It is plucked exactly in the middle and then released. The string executes ...................................... . (ii) When a body vibrates with its natural frequency the force acting on body is directly ................... to its ............................... . Ans. (i) A wire is stretched between two fixed supports. It is plucked exactly in the middle and then released. The string executes free vibrations. (ii) When a body vibrates with its natural frequency the force acting on body is directly proportional to its displacement. 28. Name one factor on which the frequency of sound emitted due to vibration in air column depends. Ans.The frequency of sound is inversely proportional to the length of air column. 29. What adjustments would you make for tuning a stringed instrument for it to emit a note of desired frequency? Ans.The string of the instrument is tightened or loosened with the help of tuning knob, till a note of desired frequency is produced. 30. Explain, why are the strings of different thickness provided on a stringed instrument. Ans.The frequency of note emitted by a string is inversely proportional to the square root of the mass of string. Thus, thicker the string, more is its mass and hence lesser is its frequency. Thus, to produce notes of different frequencies we use strings of different thickness. Class-X 13 Question Bank 31. Draw a sketch showing displacement of a body executing damped vibrations against time. Ans. 32. How is it possible to detect the filling of bottle under tap by hearing sound at distance. Ans. Initially the frequency of sound is bass. However, as the length of air column in the bottle goes on decreasing the frequency rises and becomes shrill. Thus, when the frequency is shrillest and then suddenly falls, it means that bottle is completely filled with water. 33. (a) What do you understand by the terms : (i) Musical sound, (ii) Noise? (b) State three differences between musical sound and noise. (c) State three characteristics of musical sound and define each of them. Ans. (a) Musical sound : A sound which has a pleasant effect on the ears and is generally acceptable is called musical sound. Class-X 14 Question Bank Noise : A sound which has a jarring effect on the ears and is generally unacceptable is called noise. (b) Musical Sound Noise 1. It has a pleasant effect on the ears. It has an unpleasant effect on the ears. 2. The wave form proceeds at regular The wave form proceeds at irregular intervals of time in quick succession. intervals of time. 3. The wavelength, the frequency and the amplitude do not change suddenly. The wavelength, the frequency and the amplitude change suddenly. (c) Three characteristics of musical sound are : (i) Pitch, (ii) Loudness, (iii) Quality. Pitch : The characteristic which enables us to differentiate between two sounds of equal loudness coming from different sources and having different frequencies. Loudness of sound : The rate of flow of sound energy per unit area and its effect on the ears is collectively called loudness. Quality or timbre : The property by which two notes of same frequency and same loudness can be distinguished from each other because of the difference in the wave forms is called quality or timbre. Class-X 15 Question Bank 34. State three factors which determine the loudness of sound. Ans. Loudness of sound is : (i) Directly proportional to the square of the amplitude. (ii) Inversely proportional to the square of distance between the source of sound and observer. (iii) Directly proportional to the density of medium, in which sound energy travels. 35. (a) What determines pitch of sound? (b) Why do the qualities of sound of same pitch differ when emitted by different musical instruments? Ans. (a) The frequency of a vibrating body and its effect in the ear determines pitch of sound. (b) It is because, in addition to fundamental note, other notes are produced, which depend upon the shape and size of instrument. It is the presence of these notes which gives a special wave form to the note. Thus, the notes can be distinguished. Class-X 16 Question Bank 36. In what respect does frequency of noise pattern differ from musical sound? Show by drawing diagrams. Ans. 37. What do you understand by the term quality of musical note? Illustrate your answer with diagram. Ans.Quality of note is the property by which two notes of same frequency and same loudness, can be distinguished from each other, because of the difference in wave forms. Class-X 17 Question Bank 38. How do you account for the fact that two strings can be used to give notes of same pitch and loudness, but of different quality? Ans.The final quality of note produced depends upon the shape and size of instrument on which the string is mounted. It is because, harmonics are produced, which impart final quality and hence the notes can be differentiated. 39. (a) State two differences between light wave and sound wave. (b) Explain, why lightning flash is seen before the crack of thunder. Ans.(a) Light wave 1. They travel with the speed of 3 108 ms 1. 2. They do not need a material medium for propagation. Sound wave They travel with the speed of 332 ms 1 in air at 20 C. They need a material medium for propagation. (b) Light travels at a speed of 3 108 ms 1, and hence lightning is seen at once. However, sound travels at a speed of 332 ms 1, and hence, takes longer time to reach observer. 40. Two persons are playing on identical stringed instruments, whose strings are adjusted to give notes of same pitch. Will the quality of two notes be same? Give a reason for your answer. Class-X 18 Question Bank Ans. Quality of notes will not be the same. It is because, each instrument produces other harmonics in addition to fundamental note, which are determined by the shape and size of instrument. It is the production of different harmonics which produces notes of different quality. 41. (i) Name the unit in which loudness of sound is measured. (ii) What does the unit named by you signify? (iii) What is the range of loudness of sound which is picked by the human ears? (iv) What is the normal range of loud sound for human ears? Ans.(i) Loudness of sound is measured in decibel (dB). (ii) The decibel signifies the pressure level of the sound on human ears. (iii) The human ears can pick sound between 10 dB-180 dB. (iv) The loudness of sound is considered normal, if it is between the range of 5 dB-60 dB. 42. (i) What do you understand by the term noise pollution? (ii) Name two factors which contribute to noise pollution at a particular place. (iii) Name four causes of noise each in : (a) homes, (b) surroundings. Ans.(i) The disturbance produced in the environment by undesirable, loud and harsh sound from various sources is called noise pollution. (ii) At any place the factors responsible for noise are (a) loudness of sound, (b) duration (time) of noise at that particular place. Class-X 19 Question Bank (iii) Causes of noise in homes are : (i) television, (ii) mixer grinder, (iii) vacuum cleaner, (iv) telephone. (iv) Causes of noise in sourroundings are : (i) loudspeakers, (ii) hawkers in street, (iii) noise produced by the construction of houses (iv) exploding crackers on various occasions. 43. (a) State four harmful effects of noise pollution. (b) State four ways of minimising noise with tolerable limits. Ans. (a) (i)Noise in the surrounding interferes in conversation with another person. (ii) A long exposure of noise may result in loss of hearing or deafness. (iii) Noise pollution reduces concentration at work and hence results in loss of work effeciency. (iv) Noise pollution causes anger, tension and interferes with sleep patterns of human beings. (b) (i) Machines should be redesigned, so that they produce minimum noise. (ii) Automobiles, portable home electric generator should be provided with improved silencers. (iii) Heavy vehicles should not be allowed to have pressure horn and should not be allowed to enter in the residential areas. (iv) At homes, T.V., radios, music systems should be played at low volume. Class-X 20 Question Bank Numerical Problems. 1. What should be the minimum distance between a source of sound and a reflector in water, so that echo is heard distinctly. [Speed of sound in water is 1400 ms 1] Ans. Speed of sound in water = 2 minimum distance Time Now the minimum time for hearing an echo is 0.1 s. 2 minimum distance 1400 ms 1 = 0.1 s 1400 ms 1 0.1 s Minimum distance = 2 = 70 m 2. A man standing 48 m away from a wall fires a gun. Calculate the time after which an echo is heard. [Speed of sound in air = 320 ms 1] Ans. Speed of sound (v) = 320 ms 1; Distance (d) = 48 m; Time (t) = ? Now, t = = 2d 2 48 m 320 ms 1 = 0.30 s Class-X 21 Question Bank 3. A ship on the surface of water sends a signal and receives it back from a submarine inside water after 4 s. Calculate the distance of submarine from the ship. [Speed of sound in water = 1450 ms 1] Ans. Speed of sound (v) = 1450 ms 1 Distance Time (d ) = ? (t) = 4 s 2d v= Now t t d= 2 1450 ms 1 4s = 2 = 2900 m = 2.9 km 4. A pendulum has a frequency of 5 vibrations per second. An observer starts the pendulum and fires a gun simultaneously. He hears an echo from a cliff after 8 vibrations of the pendulum. If the velocity of sound in air is 340 ms 1, what is the distance between the cliff and the observer? Ans. Time for 5 vibrations = 1 s 8 Time for 8 vibrations = s. 5 Distance between the observer and cliff t d= 2 340 ms 1 8 s = 2 5 = 272 m Class-X 22 Question Bank 5. A person fires a gun while standing at a distance of 167 m from a building. If the speed of sound is 334 ms 1, calculate the time in which the person hears an echo of the gunfire. Ans. Distance of person from the building (d) = 167 m Speed of sound in air = 334 ms 1 Time in which echo is heard, t= = 2d 2 167 m 334 ms 1 = 1s. 6. An echo is heard by a person in 0.8 s, when he fires a cracker at a distance 132.8 m from a high building. Calculate the speed of sound in air. Ans. Time after which echo is heard (t) = 0.8s Distance of source of sound from building = 132.8 m Velocity of sound in air, v = = 2d t 2 132.8 m 0.8 s = 332 ms 1. Class-X 23 Question Bank 7. The speed of sound in air is 310 ms 1. A person fires a gun and hears its echo after 1.5 s. Calculate the distance of person from cliff from which the echo takes place. Speed of sound in air (v) = 310 ms 1 Ans. Time after which echo is heard (t) = 1.5 s Distance of person from cliff (d) = = t 2 310 1.5 2 = 232.5 m. 8. A man stands between two high rise buildings and blows a whistle. He hears two successive echoes after 0.4 s and 1.6 s. Calculate the distance between the two buildings [Speed of sound is 336 ms 1]. Ans. (i) Time for hearing the first echo (t1) = 0.4 s Let the distance of building from source of sound = d1 Velocity of sound = 336 ms 1 d1 = = Class-X t1 2 336 0.4 = 67.2 m. 2 24 Question Bank (ii) Time for hearing the second echo (t2) = 1.6 s Let the distance of second building from source of sound = d2 d2 = = t2 2 336 1.6 m 2 = 268.8 m Total distance between building = d1 + d2 = (67.2 + 268.8) m = 336 m. 9. A man stands between two parallel cliffs and explodes a cracker. He hears first echo after 0.6 s and second echo after 2.4 s. Calculate the distance between cliffs. [Speed of sound 336 ms 1] Ans. (i) Let d1 be the distance of nearer cliff and d2 the distance of farther cliff from the source of sound. Time for hearing echo from nearer cliff (t1) = 0.6 s. Time for hearing echo from farther cliff (t2) = 2.4 s. Velocity of sound = 336 ms 1 Class-X 25 Question Bank d1 = t1 2 336 ms 1 0.6 s = 2 = 100.8 m t2 d2 = 2 336 ms 1 2.4 s = 2 = 403.2 m Distance between two cliffs, d = d1 + d2 = (100.8 + 403.2) m = 504 m. 10. A man stands between two cliffs, such that he is at a distance of 133.6 m from nearer cliff. He fires a gun and hears first echo after 0.8 s and second echo after 1.8 s. Calculate (i) speed of sound (ii) distance between two cliffs. Ans. Distance of man from nearer cliff (d) = 133.6 m Time for hearing echo from nearer cliff (t) = 0.8 s Speed of sound = = 2d t 2 133.6 0.8 = 334 ms 1. Class-X 26 Question Bank Let the distance of second cliff from source of sound = d Time for hearing echo from second cliff = 1.8 s Speed of sound = 334 ms 1 d= t 2 334 ms 1 1.8 s = 2 = 300.6 m Distance between two cliffs = (133.6 + 300.6) m = 434.2 m. 11. A person stands between two parallel cliffs which are 99 m apart. He fires a gun and hears two successive echoes after 0.2 s and 0.4 s. Calculate the : (i)distance of person from nearer cliff, (ii) speed of sound. Ans. Distance between two cliffs = 99 m Let the distance of nearer cliff from person = x Distance of farther cliff from person = (99 x). Time in which echo is heard from nearer cliff (t1) = 0.2 s Time in which echo is heard from farther cliff (t2) = 0.4 s Class-X 27 Question Bank 2d t 2x = 0.2 = = Also ...(i) 2d t = 2(99 x) 0.4 ...(ii) Comparing (i) and (ii) 2 x 2(99 x) = 0.2 0.4 2x = 99 x or 3x = 99 x = 33 m. Substituting the value of x in (i) v= 2 33 0.2 = 330 ms 1. 12. A man stands in front of a vertical cliff and fires a gun. He hears an echo after 2.5s. On moving 80 m closer to cliff and again firing a gun he hears an echo after 2s. Calculate : (i) distance of man from cliff from his initial position, (ii) speed of sound. Class-X 28 Question Bank Ans. Let the initial distance of man from cliff = x v= 2d t = 2x 2.5 ...(i) The final distance of man from cliff d = (x 80) v= 2d 2 ( x 80) = 2 t ...(ii) Comparing (i) and (ii) 2 x 2 ( x 80) = 2.5 2 2x = 2.5 x 200 0.5x = 200 x = 400 m. Substituting the value of x in (i) v= 2 400 m 2.5 s = 320 ms 1. Class-X 29 Question Bank 12. A boy stands in front of a cliff on the other side of a river. He fires a gun and hears an echo after 6 seconds. The boy then moves backward by 170 m and again fires the gun. He hears an echo after 7 seconds. Calculate : (i) width of river, (ii) speed of sound. Ans. Let the width of the river = x (a) When the boy stands on the bank of river 2d t 2x v= ...(i) 6 (b) When the boy moves 170 m away from bank. Distance of boy from cliff = x + 170 2 ( x + 170) v= ...(ii) Now, t Comparing (i) and (ii) 2 x 2 x + 340 = 6 7 14x = 12x + 2040 2x = 2040. x = 1020 m. Width of river = 1020 m. Now, v= Substituting the value of x in (i) 2 1020 v= 6 v = 340 ms 1. Class-X 30 Question Bank 14. A person standing between two vertical cliffs produces a sound. Two successive echoes are heard at 4 s and 6 s. Calculate the distance between the cliffs. [Speed of sound in air = 320 ms 1] Ans. Distance of near cliff from the source of sound (d1) = t 2 320 ms 1 4 s = 2 = 640 m Distance of far off cliff from the source of sound (d2) = t 2 320 ms 1 6 s = 2 = 960 m Distance between the cliffs (d) = d1 + d2 = (640 + 960) m = 1600 m Class-X 31 Question Bank 15. 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 an echo after 3 s. Calculate the speed of the sound. Ans. Case I : Let the initial distance of man from cliff =x v= 2d t v= Applying, 2x 5s (i) Case II : New distance of man from cliff = x 310 m v= 2d t v= Applying, 2 ( x 310 m) 3s (ii) Comparing (i) and (ii) 2 x 2 ( x 310 m) = 5s 3s 6x = 10x 3100 m 4x = 3100 m, x = 775 m Class-X 32 Question Bank Substituting the value of x in (i) v= = 2 775 m 5s 1550 m 5s = 310 ms 1. Class-X 33 Question Bank
|
