
For a sound source of intensity $I = (W/{m^2})$, corresponding sound level is ${B_o}$ decibel. If the intensity is increased to 4I, a new sound level becomes approximately?
1) $2{B_o}dB$
2) $({B_o} + 3)dB$
3) $({B_o} + 6)dB$
4) $4{B_o}dB$
Answer
232.8k+ views
Hint: Generally one might think that the sound intensity and loudness are the same but they are very different from each other. The sound intensity (I) is defined as power (p) upon Area (A) while the loudness is the ratio of intensity of a given sound upon the intensity at the threshold of hearing. Human ear can only hear sounds in a particular range of intensities. The threshold of human hearing has an intensity of about 120dB. Now, Define the formula for sound level and then establish a proper relation between the current sound level and the final sound level while applying logarithmic properties, Use: (log (a.b) = loga+logb).
Formula Used:
Here we have used the formula
${B_o} = 10\log (\dfrac{I}{{{I_o}}})$
Where
${B_o}$= Sound level
I = Intensity of sound
${I_o}$= ${I_o} = {10^{ - 12}}W/{m^2}$
Given:
I=4I
Find:
Sound level (B) = ??
Complete step by step answer:
Find out the intensity and solve for the unknown
Here we know the formula of sound wave which is
${B_o} = 10\log (\dfrac{I}{{{I_o}}})$
Here ${I_o}$ is given by
${I_o} = {10^{ - 12}}W/{m^2}$
Now we have been given
I=4I
So now, it becomes
$B = 10\log (\dfrac{{4I}}{{{I_o}}})$
Apply trigonometric property (log (a.b) =loga+logb)
B =10(log$\dfrac{I}{{{I_o}}}$) +10log4
Since${B_o} = 10\log (\dfrac{I}{{{I_o}}})$, So
B = ${B_o}$+$10 \times 0.60$
B = (${B_o}$+6) dB
$\therefore$The new sound level becomes option (3) approximately B = (${B_o}$+6) dB.
Additional Information:
The sound energy is defined by the energy (sound) flowing per second through an area (unit) which is held at 90° to the direction of sound wave is known as the intensity of sound. Sound level on the other hand is the measurement of sound vibration or noise using logarithmic functions and their properties
Note: Go step by step first define the term sound intensity and then differentiate between sound intensity and loudness. Apply the formula that relates sound level to sound intensity and solve by using the logarithmic properties.
Formula Used:
Here we have used the formula
${B_o} = 10\log (\dfrac{I}{{{I_o}}})$
Where
${B_o}$= Sound level
I = Intensity of sound
${I_o}$= ${I_o} = {10^{ - 12}}W/{m^2}$
Given:
I=4I
Find:
Sound level (B) = ??
Complete step by step answer:
Find out the intensity and solve for the unknown
Here we know the formula of sound wave which is
${B_o} = 10\log (\dfrac{I}{{{I_o}}})$
Here ${I_o}$ is given by
${I_o} = {10^{ - 12}}W/{m^2}$
Now we have been given
I=4I
So now, it becomes
$B = 10\log (\dfrac{{4I}}{{{I_o}}})$
Apply trigonometric property (log (a.b) =loga+logb)
B =10(log$\dfrac{I}{{{I_o}}}$) +10log4
Since${B_o} = 10\log (\dfrac{I}{{{I_o}}})$, So
B = ${B_o}$+$10 \times 0.60$
B = (${B_o}$+6) dB
$\therefore$The new sound level becomes option (3) approximately B = (${B_o}$+6) dB.
Additional Information:
The sound energy is defined by the energy (sound) flowing per second through an area (unit) which is held at 90° to the direction of sound wave is known as the intensity of sound. Sound level on the other hand is the measurement of sound vibration or noise using logarithmic functions and their properties
Note: Go step by step first define the term sound intensity and then differentiate between sound intensity and loudness. Apply the formula that relates sound level to sound intensity and solve by using the logarithmic properties.
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

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

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

