How Many times more intense is a 60 dB sound than a 30 dB sound ?

Options

A. 1000
B. 2
C. 100
D. 4

Answer

To solve the problem of how many times more intense a 60 dB sound is compared to a 30 dB sound, we can use the formula that relates sound intensity levels in decibels (dB) to intensity.

### Step-by-Step Solution:

1. **Understand the Decibel Formula**:
The loudness of sound in decibels (dB) is given by the formula:
\[ L = 10 \log_{10} \left(\frac{I}{I_0}\right) \]
where \( L \) is the sound level in dB, \( I \) is the intensity of the sound, and \( I_0 \) is a reference intensity.

2. **Set Up the Problem**:
Let \( L_1 = 30 \, \text{dB} \) and \( L_2 = 60 \, \text{dB} \). We want to find the ratio of the intensities \( \frac{I_2}{I_1} \).

3. **Write the Equations for Both Levels**:
For \( L_1 \):
\[ 30 = 10 \log_{10} \left(\frac{I_1}{I_0}\right) \]
For \( L_2 \):
\[ 60 = 10 \log_{10} \left(\frac{I_2}{I_0}\right) \]

4. **Rearranging the Equations**:
Dividing both equations by 10 gives:
\[ 3 = \log_{10} \left(\frac{I_1}{I_0}\right) \]
\[ 6 = \log_{10} \left(\frac{I_2}{I_0}\right) \]

5. **Convert Logarithmic Equations to Exponential Form**:
From the first equation:
\[ \frac{I_1}{I_0} = 10^3 = 1000 \quad \Rightarrow \quad I_1 = 1000 I_0 \]
From the second equation:
\[ \frac{I_2}{I_0} = 10^6 = 1000000 \quad \Rightarrow \quad I_2 = 1000000 I_0 \]

6. **Finding the Ratio of Intensities**:
Now, we can find the ratio of the two intensities:
\[ \frac{I_2}{I_1} = \frac{1000000 I_0}{1000 I_0} = \frac{1000000}{1000} = 1000 \]

7. **Conclusion**:
Therefore, a 60 dB sound is **1000 times more intense** than a 30 dB sound.

### Final Answer:
A 60 dB sound is **1000 times** more intense than a 30 dB sound.

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