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 find the increase in sound power in decibels (dB) when the power of sound from a speaker increases from 10 W to 400 W, we can use the formula for sound intensity level in decibels:

\[ L = 10 \log_{10} \left(\frac{P}{P_0}\right) \]

Where:
- \(L\) is the sound level in decibels,
- \(P\) is the power of the sound,
- \(P_0\) is the reference power (which we can consider as the original power).

### Step 1: Calculate the initial sound level \(L_1\)

Using the initial power \(P_1 = 10 \, \text{W}\):

\[ L_1 = 10 \log_{10} \left(\frac{10}{P_0}\right) \]

### Step 2: Calculate the final sound level \(L_2\)

Using the final power \(P_2 = 400 \, \text{W}\):

\[ L_2 = 10 \log_{10} \left(\frac{400}{P_0}\right) \]

### Step 3: Find the difference in sound levels

The increase in sound level \(\Delta L\) is given by:

\[ \Delta L = L_2 - L_1 \]

Substituting the expressions for \(L_1\) and \(L_2\):

\[ \Delta L = 10 \log_{10} \left(\frac{400}{P_0}\right) - 10 \log_{10} \left(\frac{10}{P_0}\right) \]

Using the properties of logarithms:

\[ \Delta L = 10 \left( \log_{10} (400) - \log_{10} (10) \right) \]

\[ \Delta L = 10 \log_{10} \left(\frac{400}{10}\right) \]

\[ \Delta L = 10 \log_{10} (40) \]

### Step 4: Calculate \(\log_{10} (40)\)

Using a calculator or logarithm tables, we find:

\[ \log_{10} (40) \approx 1.602 \]

### Step 5: Calculate the increase in decibels

Now substituting back:

\[ \Delta L = 10 \times 1.602 = 16.02 \, \text{dB} \]

Thus, the power increased in dB as compared to the original power is nearly:

\[ \Delta L \approx 16 \, \text{dB} \]

### Final Answer:
The increase in power is approximately **16 dB**.

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