The power of sound from the speaker of radio is 10 W. Now, the power of sound from the speaker of the radio is increased to 400 W by increasing the volume of the radio. The power increased in dB as compared to original power is nearly

Options

A. 8
B. 12
C. 13
D. 16

Answer

To solve the problem of determining the increase in sound power in decibels when the power of sound from a radio speaker is increased from 10 W to 400 W, we can follow these steps:

### Step 1: Understand the formula for sound intensity level in decibels
The sound intensity level in decibels (dB) is given by the formula:
\[ \beta = 10 \log_{10} \left( \frac{I}{I_0} \right) \]
where \( I \) is the intensity of the sound and \( I_0 \) is the reference intensity, which is typically \( 10^{-12} \, \text{W/m}^2 \).

### Step 2: Calculate the initial sound intensity level (\( \beta_1 \))
For the initial power of 10 W, we first need to convert power to intensity. The intensity \( I \) can be calculated as:
\[ I_1 = \frac{P_1}{A} \]
Assuming the area \( A \) remains constant, we can directly use the power values in the logarithmic calculation. Therefore, we can express \( \beta_1 \) as:
\[ \beta_1 = 10 \log_{10} \left( \frac{10}{10^{-12}} \right) \]
This simplifies to:
\[ \beta_1 = 10 \log_{10} (10^{13}) = 10 \times 13 = 130 \, \text{dB} \]

### Step 3: Calculate the new sound intensity level (\( \beta_2 \))
Now, for the increased power of 400 W, we calculate \( \beta_2 \):
\[ \beta_2 = 10 \log_{10} \left( \frac{400}{10^{-12}} \right) \]
This simplifies to:
\[ \beta_2 = 10 \log_{10} (400 \times 10^{12}) = 10 \log_{10} (4 \times 10^{14}) = 10 \left( \log_{10} 4 + 14 \right) \]
Using \( \log_{10} 4 \approx 0.602 \):
\[ \beta_2 = 10 \left( 0.602 + 14 \right) = 10 \times 14.602 = 146.02 \, \text{dB} \]

### Step 4: Calculate the increase in sound intensity level
The increase in sound intensity level in decibels is given by:
\[ \Delta \beta = \beta_2 - \beta_1 \]
Substituting the values we calculated:
\[ \Delta \beta = 146.02 - 130 = 16.02 \, \text{dB} \]

### Conclusion
The increase in power in decibels as compared to the original power is approximately **16 dB**.

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