Intensity level of a sound of intensity I is 30 db. Then the ratio `I//I_(0)` is (where `I_(0)` is threshold of hearing )

Options

A. 30
B. 10000
C. 1000
D. 3

Explanation

To solve the problem, we need to find the ratio \( \frac{I}{I_0} \) given that the intensity level of a sound is 30 decibels (dB). Here \( I_0 \) is the threshold of hearing.

### Step-by-Step Solution:

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

2. **Substitute the given intensity level**:
We know that the intensity level \( L \) is 30 dB. Therefore, we can write:
\[ 30 = 10 \log_{10} \left( \frac{I}{I_0} \right) \]

3. **Isolate the logarithmic term**:
To isolate the logarithm, divide both sides by 10:
\[ 3 = \log_{10} \left( \frac{I}{I_0} \right) \]

4. **Convert from logarithmic to exponential form**:
To eliminate the logarithm, we convert it to its exponential form:
\[ \frac{I}{I_0} = 10^3 \]

5. **Calculate \( 10^3 \)**:
Now, calculate \( 10^3 \):
\[ 10^3 = 1000 \]

6. **Conclusion**:
Thus, the ratio of the intensity \( I \) to the threshold intensity \( I_0 \) is:
\[ \frac{I}{I_0} = 1000 \]

### Final Answer:
The ratio \( \frac{I}{I_0} \) is 1000.