The intensity level of a sound of intensity I | vedclass

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(B) The formula for the intensity level $L$ in decibels $(dB)$ is given by: $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$ Given that $L = 30 \, dB$,

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$1000$

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(B) The formula for the intensity level $L$ in decibels $(dB)$ is given by: $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$ Given that $L = 30 \, dB$,we substitute this into the equation: $30 = 10 \log_{10} \left( \frac{I}{I_0} \right)$ Divide both sides by $10$: $3 = \log_{10} \left( \frac{I}{I_0} \right)$ To solve for the ratio,we convert the logarithmic equation to its exponential form: $\frac{I}{I_0} = 10^3$ $\frac{I}{I_0} = 1000$

The intensity level of a sound of intensity $I$ is $30 \, dB$. The ratio $\frac{I}{I_0}$ is (where $I_0$ is the threshold of hearing).

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