An echo is heard in 3s what is the distance of the reflecting

SOLUTION CONCEPT: • Echo: If weshout or clapnear a suitablereflecting objectsuch as atall buildingora mountain, we will hear the same sound again a little later. This sound which we hear is called anecho. • Echoes are heard due to the phenomenon of Reflection of sound waves. • To hear the echo clearly,the reflecting object must be more than 17.2 m from the sound source for the echo to be heard by a person standing at the source. Calculation: Given that, An echo is returned after time (t) = 3s Speed of the sound (v) = 342 m/s Now according to the law's of motion distance traveled by sound waves can be given as Distance travelled (d) = v x t = 342 × 3 = 1026m Distance covered by the sound will double, as the sound will have to travel and source to reflecting surface and back from it hence distance will be ‘d’ is ‘2d. • Now replacing ‘d’ with ‘2d’ we get 2d = 1026 d = 513 m • The distance from the reflecting surface to the source is 513 m.

Speed of sound, V = 342 m s − 1 Echo returns in time, t = 3s Distance travelled by sound = V × t = 342 × 3 = 1026 m In the given time interval, sound has to travel a distance that is twice the distance of the reflecting surface and the source. Hence, the distance of the reflecting surface from the source = 1026 2 m = 513 m .

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Given data: • In the question, it is given that the speed of sound is 342 m / s and the echo returned back to the source in 3 s. • So, V = 342 m / s and t = 3 s • Assume that, the distance between the reflecting surface and the source is d. Concept used: • A sound is said to be an echo when the sound is reflected by a surface and reflected back to the source that generated that sound. • So, according to the question, the echo travels 2 d distance in 3 s with the speed of 342 m / s. Find the required distance: Therefore, 2 d = v × t = 342 × 3 ⇒ 2 d = 1026 ⇒ d = 523 Thus, the distance of the reflecting surface from the source is 523 m .

CONCEPT: • Echo: If weshout or clapnear a suitablereflecting objectsuch as atall buildingora mountain, we will hear the same sound again a little later. This sound which we hear is called anecho. • Echoes are heard due to the phenomenon of Reflection of sound waves. • To hear the echo clearly,the reflecting object must be more than 17.2 m from the sound source for the echo to be heard by a person standing at the source. Calculation: Given that, An echo is returned after time (t) = 3s Speed of the sound (v) = 342 m/s Now according to the law's of motion distance traveled by sound waves can be given as Distance travelled (d) = v x t = 342 × 3 = 1026m Distance covered by the sound will double, as the sound will have to travel and source to reflecting surface and back from it hence distance will be ‘d’ is ‘2d. • Now replacing ‘d’ with ‘2d’ we get 2d = 1026 d = 513 m • The distance from the reflecting surface to the source is 513 m.

SOLUTION CONCEPT: • Echo: If weshout or clapnear a suitablereflecting objectsuch as atall buildingora mountain, we will hear the same sound again a little later. This sound which we hear is called anecho. • Echoes are heard due to the phenomenon of Reflection of sound waves. • To hear the echo clearly,the reflecting object must be more than 17.2 m from the sound source for the echo to be heard by a person standing at the source. Calculation: Given that, An echo is returned after time (t) = 3s Speed of the sound (v) = 342 m/s Now according to the law's of motion distance traveled by sound waves can be given as Distance travelled (d) = v x t = 342 × 3 = 1026m Distance covered by the sound will double, as the sound will have to travel and source to reflecting surface and back from it hence distance will be ‘d’ is ‘2d. • Now replacing ‘d’ with ‘2d’ we get 2d = 1026 d = 513 m • The distance from the reflecting surface to the source is 513 m.

Given data: • In the question, it is given that the speed of sound is 342 m / s and the echo returned back to the source in 3 s. • So, V = 342 m / s and t = 3 s • Assume that, the distance between the reflecting surface and the source is d. Concept used: • A sound is said to be an echo when the sound is reflected by a surface and reflected back to the source that generated that sound. • So, according to the question, the echo travels 2 d distance in 3 s with the speed of 342 m / s. Find the required distance: Therefore, 2 d = v × t = 342 × 3 ⇒ 2 d = 1026 ⇒ d = 523 Thus, the distance of the reflecting surface from the source is 523 m .

CONCEPT: • Echo: If weshout or clapnear a suitablereflecting objectsuch as atall buildingora mountain, we will hear the same sound again a little later. This sound which we hear is called anecho. • Echoes are heard due to the phenomenon of Reflection of sound waves. • To hear the echo clearly,the reflecting object must be more than 17.2 m from the sound source for the echo to be heard by a person standing at the source. Calculation: Given that, An echo is returned after time (t) = 3s Speed of the sound (v) = 342 m/s Now according to the law's of motion distance traveled by sound waves can be given as Distance travelled (d) = v x t = 342 × 3 = 1026m Distance covered by the sound will double, as the sound will have to travel and source to reflecting surface and back from it hence distance will be ‘d’ is ‘2d. • Now replacing ‘d’ with ‘2d’ we get 2d = 1026 d = 513 m • The distance from the reflecting surface to the source is 513 m.

Speed of sound, V = 342 m s − 1 Echo returns in time, t = 3s Distance travelled by sound = V × t = 342 × 3 = 1026 m In the given time interval, sound has to travel a distance that is twice the distance of the reflecting surface and the source. Hence, the distance of the reflecting surface from the source = 1026 2 m = 513 m .

• Course • NCERT • Class 12 • Class 11 • Class 10 • Class 9 • Class 8 • Class 7 • Class 6 • IIT JEE • Exam • JEE MAINS • JEE ADVANCED • X BOARDS • XII BOARDS • NEET • Neet Previous Year (Year Wise) • Physics Previous Year • Chemistry Previous Year • Biology Previous Year • Neet All Sample Papers • Sample Papers Biology • Sample Papers Physics • Sample Papers Chemistry • Download PDF's • Class 12 • Class 11 • Class 10 • Class 9 • Class 8 • Class 7 • Class 6 • Exam Corner • Online Class • Quiz • Ask Doubt on Whatsapp • Search Doubtnut • English Dictionary • Toppers Talk • Blog • Download • Get App Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year papers, NEET previous year papers, NCERT books for classes 6 to 12, CBSE, Pathfinder Publications, RD Sharma, RS Aggarwal, Manohar Ray, Cengage books for boards and competitive exams. Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. G...