The loudness, [tex]\( L \)[/tex], measured in decibels (dB), of a sound intensity, [tex]\( I \)[/tex], measured in watts per square meter, is defined as

[tex]\[ L = 10 \log \left( \frac{I}{I_0} \right) \][/tex]

where [tex]\( I_0 = 10^{-12} \)[/tex] is the least intense sound a human ear can hear. What is the approximate loudness of a rock concert with a sound intensity of [tex]\( 10^{-1} \)[/tex] watts per square meter?

A. [tex]\( 2 \)[/tex] dB
B. [tex]\( 22 \)[/tex] dB
C. [tex]\( 60 \)[/tex] dB
D. [tex]\( 110 \)[/tex] dB



Answer :

Sure, let's solve this step by step.

First, let's recall the formula which defines loudness [tex]\( L \)[/tex] in decibels (dB):

[tex]\[ L = 10 \log_{10} \left( \frac{I}{I_0} \right) \][/tex]

where:
- [tex]\( I \)[/tex] is the sound intensity we want to measure,
- [tex]\( I_0 \)[/tex] is a reference intensity, specifically the least intense sound a human ear can hear, usually taken as [tex]\( 10^{-12} \)[/tex] watts per square meter.

In this problem, we are given:
- The sound intensity of the rock concert, [tex]\( I = 10^{-1} \)[/tex] watts per square meter.
- The reference intensity, [tex]\( I_0 = 10^{-12} \)[/tex] watts per square meter.

Let's plug these values into the formula:

[tex]\[ L = 10 \log_{10} \left( \frac{10^{-1}}{10^{-12}} \right) \][/tex]

First, calculate the ratio inside the logarithm:

[tex]\[ \frac{10^{-1}}{10^{-12}} \][/tex]

When dividing powers of 10, we subtract the exponents:

[tex]\[ 10^{-1} / 10^{-12} = 10^{-1 - (-12)} = 10^{-1 + 12} = 10^{11} \][/tex]

So, now we have:

[tex]\[ L = 10 \log_{10} (10^{11}) \][/tex]

Recall that [tex]\( \log_{10} (10^x) = x \)[/tex]. Thus:

[tex]\[ \log_{10} (10^{11}) = 11 \][/tex]

So the equation for loudness simplifies to:

[tex]\[ L = 10 \times 11 \][/tex]

[tex]\[ L = 110 \, \text{dB} \][/tex]

Hence, the approximate loudness of the rock concert is:

[tex]\[ \boxed{110 \, \text{dB}} \][/tex]

Therefore, the correct answer choice is:
[tex]\[ 110 \text{dB} \][/tex]