---
title: "Mean free path and reflections in room"
subject: "Engineering Physics"
chapter: 3
key: "semester_1/engineering_physics:room-mean-free-path-reflections"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/room-mean-free-path-reflections
site: BCT Engineering
---

# Mean free path and reflections in room

Engineering Physics · Chapter 3: Acoustics

- Asked in: 2073 Chaitra, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/room-mean-free-path-reflections
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-3

## Question

A room has dimensions 6×4×5 m. Find (a) the mean free path of the sound wave in the room (b) the number of reflections per second made by the sound wave with the walls of the room. (Given, velocity of sound in air is 330 ms⁻¹)

## Short answer

(a) Mean free path d = 4V/S = 4×120/148 ≈ 3.24 m. (b) Reflections per second n = v/d = 330/3.24 ≈ 102 s⁻¹.

## Answer

The **mean free path** of sound in a room is the average distance travelled between two successive reflections from the walls: *d* = 4V/S, where V is the volume and S the total surface area.

**Given:** l = 6 m, b = 4 m, h = 5 m, v = 330 m/s.

#### (a) Mean free path
1. Volume V = 6 × 4 × 5 = 120 m³
2. Surface area S = 2(lb + bh + lh) = 2(24 + 20 + 30) = 148 m²
3. Mean free path d = 4V/S = 4 × 120 / 148 = **3.24 m**

#### (b) Reflections per second
Number of reflections per second n = v/d = 330 / 3.243 = 101.75

**Answer: (a) mean free path ≈ 3.24 m; (b) ≈ 102 reflections per second.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The mean free path of sound in a room is
   A. d = V/S
   B. d = 4V/S
   C. d = 4S/V
   D. d = S/4V
   Answer: B. d = 4V/S, where V is volume and S total surface area. (Key point: Mean free path formula)

2. For a 6 × 4 × 5 m room, the total surface area is
   A. 120 m²
   B. 74 m²
   C. 148 m²
   D. 296 m²
   Answer: C. S = 2(24 + 20 + 30) = 148 m². (Key point: Surface area)

3. The mean free path is about
   A. 3.24 m
   B. 0.81 m
   C. 1.62 m
   D. 6.48 m
   Answer: A. d = 4 × 120/148 = 3.24 m. (Key point: Mean free path value)

4. The number of reflections per second (v = 330 m/s) is about
   A. 3.24
   B. 10.2
   C. 1070
   D. 102
   Answer: D. n = v/d = 330/3.243 ≈ 102. (Key point: Reflections per second)

## Close variants set in other exams

- [Mean free path in room (350 m/s)](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aroom-mean-free-path-reflections-alt) — Engineering Physics, chapter 3, asked 2×

## Related in Engineering Physics

- [Sound energy flow per square cm](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Asound-energy-flow-frequency-amplitude) — Engineering Physics, chapter 3

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