Questions: 0 dB = 20 μPa 70 dB = 200 Pa 0 dB = 20 x 10^-6 pa 80 dB = 2000 pa 0 dB = 2 x 10^-5 Pa 90 dB = 20,000 Pa 10 dB = 2 x 10^-4 Pa 100 dB = 2,00,000 Pa 20 dB = 2 x 10^-3 Pa 110 dB = 2 x 10^6 Pa 30 dB = 2 x 10^-2 Pa 120 dB = 2 x 10^7 pa 40 dB = 2 x 10^-1 Pa 50 dB = 2 pa 60 dB = 20 Pa

Solution Steps

The decibel (dB) scale is a logarithmic scale used to measure sound pressure levels. The formula to convert sound pressure level (SPL) in pascals (Pa) to decibels (dB) is given by:

\[ L_p = 20 \log_{10} \left( \frac{p}{p_0} \right) \]

where:

• \(L_p\) is the sound pressure level in decibels (dB),
• \(p\) is the sound pressure in pascals (Pa),
• \(p_0\) is the reference sound pressure, typically \(20 \times 10^{-6} \, \text{Pa}\) (20 µPa).

We need to verify the given data points using the formula. For example, for 70 dB:

\[ 70 = 20 \log_{10} \left( \frac{200}{20 \times 10^{-6}} \right) \]

Simplifying inside the logarithm:

\[ 70 = 20 \log_{10} \left( \frac{200}{0.00002} \right) \]

\[ 70 = 20 \log_{10} (10^7) \]

\[ 70 = 20 \times 7 \]

\[ 70 = 140 \]

This indicates a discrepancy, so we need to re-evaluate the given data points.

Given the discrepancies, we should correct the data points using the formula. For example, for 70 dB:

\[ 70 = 20 \log_{10} \left( \frac{p}{20 \times 10^{-6}} \right) \]

Solving for \(p\):

\[ 3.5 = \log_{10} \left( \frac{p}{20 \times 10^{-6}} \right) \]

\[ 10^{3.5} = \frac{p}{20 \times 10^{-6}} \]

\[ p = 10^{3.5} \times 20 \times 10^{-6} \]

\[ p = 3162.2777 \times 20 \times 10^{-6} \]

\[ p = 0.0632 \, \text{Pa} \]

This corrected value does not match the given 200 Pa, indicating a need to re-evaluate the entire table.

Final Answer