A small speaker delivers 2 W of audio output. At what distance from the speaker will one detect 120 dB intensity sound? [Given reference intensity of sound as $$10^{-12}$$ W/m$$^2$$]

We start with the acoustic power output of the speaker, which is given as $$P = 2 \text{ W}$$.

For a point source that radiates sound uniformly in all directions, the intensity $$I$$ at a distance $$r$$ from the source is obtained from the definition of intensity,

$$I \;=\;\frac{\text{Power passing normally through a surface}}{\text{Area of that surface}}.$$

The surface through which the power spreads is an imaginary sphere of radius $$r$$. The area of this sphere is $$4\pi r^{2}$$. Hence, using the above definition, we can write

$$I \;=\;\frac{P}{4\pi r^{2}}.$$

The loudness of sound is usually expressed in decibels (dB). The relation between the sound‐level $$\beta$$ in decibels and the intensity $$I$$ is

$$\beta \;=\;10\,\log_{10}\!\left(\frac{I}{I_{0}}\right),$$

where $$I_{0}=10^{-12}\,\text{W m}^{-2}$$ is the standard (reference) intensity.

We are told that the required sound‐level is $$\beta = 120 \text{ dB}$$. Substituting this value in the above formula, we have

$$120 \;=\;10\,\log_{10}\!\left(\frac{I}{10^{-12}}\right).$$

Dividing both sides by $$10$$, we obtain

$$12 \;=\;\log_{10}\!\left(\frac{I}{10^{-12}}\right).$$

Now, by the definition of the logarithm, this means

$$\frac{I}{10^{-12}} \;=\;10^{12},$$

so

$$I \;=\;10^{12}\times10^{-12} \;=\;1\;\text{W m}^{-2}.$$

Thus, to perceive a 120 dB sound, the intensity incident on the listener’s ear must be $$1\;\text{W m}^{-2}$$.

Next, we substitute this required intensity into the inverse‐square expression $$I = \dfrac{P}{4\pi r^{2}}$$. Writing the equality explicitly, we get

$$1 \;=\;\frac{2}{4\pi r^{2}}.$$

Multiplying both sides by $$4\pi r^{2}$$, we have

$$4\pi r^{2} \;=\;2.$$

Dividing both sides by $$4\pi$$ yields

$$r^{2} \;=\;\frac{2}{4\pi} \;=\;\frac{1}{2\pi}.$$

Hence

$$r \;=\;\sqrt{\frac{1}{2\pi}} \;=\;\sqrt{\frac{1}{6.283}} \;\approx\;\sqrt{0.159}\;\text{ m} \;\approx\;0.399\;\text{ m}.$$

Expressing this distance in centimetres,

$$0.399\;\text{ m}=39.9\;\text{ cm}\approx40\;\text{ cm}.$$

Hence, the correct answer is Option A.