
A man standing unsymmetrically between two parallel cliffs, claps his hand and starts hearing a series of echoes at intervals of $1{\text{ s}}$. If the speed of sound in air is \[340{\text{ m}}{{\text{s}}^{ - 1}}\], then the distance between the two parallel cliffs is
A. $170{\text{ m}}$
B. $340{\text{ m}}$
C. $510{\text{ m}}$
D. $680{\text{ m}}$
Answer
232.8k+ views
Hint Sound wave is a longitudinal wave and hence travel in straight line motion. So, time taken by a sound to travel a distance $d$ can be given by $t = \dfrac{d}{v}$ where $v$ is the velocity of sound waves in a particular medium.
Complete step by step answer
As given in the question that the man is standing unsymmetrically between the cliffs that means he can be anywhere between the cliffs.
Let the man is standing at a distance ${x_1}{\text{ m}}$ from the nearest cliff and ${x_2}{\text{ m}}$ from the farther cliff so that the distance between the cliffs will be $\left( {{x_1} + {x_2}} \right){\text{ m}}$
Now, as we know that the sound wave is a longitudinal wave and hence travels in straight line motion. So, time taken by a sound to travel a distance $d$ can be given by $t = \dfrac{d}{v}$ where $v$ is the velocity of sound waves in a particular medium.
As given in the question that the man hears the echoes at an interval of $1{\text{ s}}$
So, the first sound he will hear after $1{\text{ s}}$ which was get reflected by the nearest cliff after covering a distance of $2{x_1}$
Therefore from the above equation we have
$1 = \dfrac{{2{x_1}}}{{340}}$
On solving we get ${x_1} = 170{\text{ m}}$
Now, the second echo he will hear after ${\text{2 s}}$ which was get reflected by the farther cliff after covering a distance of $2{x_2}$
Therefore from the above equation we have
$2 = \dfrac{{2{x_2}}}{{340}}$
On solving we get ${x_2} = 340{\text{ m}}$
So, the distance between the cliffs is $\left( {{x_1} + {x_2}} \right) = 170 + 340 = 510{\text{ m}}$
Hence, option C is correct.
Note As sound wave is a longitudinal wave it has compressions and rarefactions while moving. The compressions and rarefactions after striking a rigid wall or cliff, reflects back as compressions and rarefaction respectively as there is phase difference of $180^\circ $ . This phenomenon is known as reflection of sound waves.
Complete step by step answer
As given in the question that the man is standing unsymmetrically between the cliffs that means he can be anywhere between the cliffs.
Let the man is standing at a distance ${x_1}{\text{ m}}$ from the nearest cliff and ${x_2}{\text{ m}}$ from the farther cliff so that the distance between the cliffs will be $\left( {{x_1} + {x_2}} \right){\text{ m}}$
Now, as we know that the sound wave is a longitudinal wave and hence travels in straight line motion. So, time taken by a sound to travel a distance $d$ can be given by $t = \dfrac{d}{v}$ where $v$ is the velocity of sound waves in a particular medium.
As given in the question that the man hears the echoes at an interval of $1{\text{ s}}$
So, the first sound he will hear after $1{\text{ s}}$ which was get reflected by the nearest cliff after covering a distance of $2{x_1}$
Therefore from the above equation we have
$1 = \dfrac{{2{x_1}}}{{340}}$
On solving we get ${x_1} = 170{\text{ m}}$
Now, the second echo he will hear after ${\text{2 s}}$ which was get reflected by the farther cliff after covering a distance of $2{x_2}$
Therefore from the above equation we have
$2 = \dfrac{{2{x_2}}}{{340}}$
On solving we get ${x_2} = 340{\text{ m}}$
So, the distance between the cliffs is $\left( {{x_1} + {x_2}} \right) = 170 + 340 = 510{\text{ m}}$
Hence, option C is correct.
Note As sound wave is a longitudinal wave it has compressions and rarefactions while moving. The compressions and rarefactions after striking a rigid wall or cliff, reflects back as compressions and rarefaction respectively as there is phase difference of $180^\circ $ . This phenomenon is known as reflection of sound waves.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

