How many time are taken intense is 90dB sound than 40dB sound?

Options

A. 5
B. 50
C. 500
D. `10^(5)`

Answer

To solve the problem of how many times more intense a 90 dB sound is compared to a 40 dB sound, we can follow these steps:

### Step 1: Understand the relationship between sound intensity and decibels
The sound level in decibels (dB) is related to the intensity of the sound (I) by the formula:
\[ L = 10 \log_{10} \left( \frac{I}{I_0} \right) \]
where \( L \) is the sound level in decibels, \( I \) is the intensity of the sound, and \( I_0 \) is a reference intensity (usually taken as \( 10^{-12} \, \text{W/m}^2 \)).

### Step 2: Set up the equation for the two sound levels
Let \( L_1 = 40 \, \text{dB} \) and \( L_2 = 90 \, \text{dB} \). We can use the difference in sound levels to find the ratio of their intensities:
\[ L_2 - L_1 = 10 \log_{10} \left( \frac{I_2}{I_1} \right) \]

### Step 3: Substitute the values into the equation
Substituting the values of \( L_1 \) and \( L_2 \):
\[ 90 - 40 = 10 \log_{10} \left( \frac{I_2}{I_1} \right) \]
This simplifies to:
\[ 50 = 10 \log_{10} \left( \frac{I_2}{I_1} \right) \]

### Step 4: Divide both sides by 10
Dividing both sides by 10 gives:
\[ 5 = \log_{10} \left( \frac{I_2}{I_1} \right) \]

### Step 5: Convert from logarithmic form to exponential form
To find the ratio \( \frac{I_2}{I_1} \), we convert from logarithmic form:
\[ \frac{I_2}{I_1} = 10^5 \]

### Conclusion
Thus, the intensity of the 90 dB sound is \( 10^5 \) times greater than that of the 40 dB sound.

### Final Answer
The answer is \( 10^5 \).

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