Reducing the intensity of noise by a factor of 10 will mean a reduction in noise level by

The correct option is 10 dB.

Understanding the Noise Level Formula
The noise level (or sound intensity level) measured in decibels (dB) is logarithmic and is defined by the following formula:

$$\beta =10{\log }_{10}\left(\frac{I}{I_0}\right)$$

where:
- $\beta$ is the noise level in decibels (dB).
- $I$ is the intensity of the noise.
- $I_0$ is the reference sound intensity (the threshold of human hearing, typically 10^-12 W/m²).

Step-by-Step Derivation
Let the initial noise intensity be $I_1$ and the initial noise level be ${\beta }_1$:

$${\beta }_1=10{\log }_{10}\left(\frac{I_1}{I_0}\right)$$

If the noise intensity is reduced by a factor of 10, the new noise intensity $I_2$ becomes:

$$I_2=\frac{I_1}{10}$$

Now, let us calculate the new noise level ${\beta }_2$ using the new intensity:

$${\beta }_2=10{\log }_{10}\left(\frac{I_2}{I_0}\right)$$

Substitute $I_2$ into the equation:

$${\beta }_2=10{\log }_{10}\left(\frac{I_1}{10I_0}\right)$$

Using the logarithmic property ${\log }_{10}\left(\frac{A}{B}\right)={\log }_{10}\left(A\right)-{\log }_{10}\left(B\right)$, we can rewrite the equation as:

$${\beta }_2=10\left({\log }_{10}\left(\frac{I_1}{I_0}\right)-{\log }_{10}\left(10\right)\right)$$

Since the common logarithm of 10 is 1 (${\log }_{10}\left(10\right)=1$), we simplify this to:

$${\beta }_2=10{\log }_{10}\left(\frac{I_1}{I_0}\right)-10$$

Recognizing that the first term is our initial noise level ${\beta }_1$:

$${\beta }_2={\beta }_1-10$$

Therefore, the change in the noise level is:

$$\Delta \beta ={\beta }_2-{\beta }_1=-10\text{dB}$$

This negative sign indicates a reduction in the noise level by exactly 10 dB.