Question

‌‌The‌ ‌power‌ ‌of‌ ‌a‌ ‌loudspeaker‌ ‌is‌ ‌increased‌ ‌from‌ ‌20W‌ ‌to‌ ‌400W.‌ ‌What‌ ‌is‌ ‌the‌ ‌intensity‌ ‌increase‌ ‌
as‌ ‌compared‌ ‌to‌ ‌the‌ ‌original‌ ‌value?‌ ‌
A.‌ ‌13‌ ‌dB‌ ‌
B.‌ ‌7‌ ‌dB‌ ‌
C.‌ ‌4‌ ‌dB‌ ‌
D.‌ ‌2‌ ‌dB‌ ‌

Answer 1

Sound is a wave. For any wave, Intensity is defined as the power delivered per unit area. The intensity level or loudness of sound is related to intensity as given below. And since intensity is directly proportional to the power of the sound, in the given difference of intensities formula, intensity can be substituted with Power. This formula, with power in it, can be used to find the difference in intensities.
Formula used:

& L=10lo{{g}_{10}}\dfrac{I}{{{I}_{o}}} \\\ & {{L}_{2}}-{{L}_{1}}=10lo{{g}_{10}}\dfrac{{{I}_{2}}}{{{I}_{1}}}~ \\\ & {{L}_{2}}-{{L}_{1}}=10lo{{g}_{10}}\dfrac{{{P}_{2}}}{{{P}_{1}}} \\\ \end{aligned}$$ **Complete answer:** The loudness of a speaker is nothing but the intensity of the sound waves. The unit of intensity is dB or decibel. The Intensity of a wave, such as sound, is the power delivered by the wave per unit area. And as such Intensity is directly proportional to the Power. $$I\propto P$$ Now the relationship between sound intensity levels or loudness and the intensity of the sound waves is given as $$L=10lo{{g}_{10}}\dfrac{I}{{{I}_{o}}}$$ Here I am the intensity of the sound waves and $${{I}_{o}}$$ is the intensity of reference. The change in Intensity levels or loudness can be written as the difference between two intensity levels or $${{L}_{2}}-{{L}_{1}}=10lo{{g}_{10}}\dfrac{{{I}_{2}}}{{{I}_{1}}}$$ Since Intensity is nothing but Power per unit area, the relation can be written as $${{L}_{2}}-{{L}_{1}}=10lo{{g}_{10}}\dfrac{{{P}_{2}}}{{{P}_{1}}}$$ Where$${{P}_{1}}$$ is the initial power and $${{P}_{2}}$$ is the final power. In our question, the initial and final powers are given as 20W and 400W respectively Plugging in the value $$\begin{aligned} & {{L}_{2}}-{{L}_{1}}=10lo{{g}_{10}}\dfrac{{{P}_{2}}}{{{P}_{1}}} \\\ & \Rightarrow {{L}_{2}}-{{L}_{1}}=10lo{{g}_{10}}\dfrac{400}{20} \\\ & \Rightarrow {{L}_{2}}-{{L}_{1}}=10lo{{g}_{10}}20 \\\ \end{aligned}$$ And we know that $${{\log }_{10}}20=1.3$$ So, $$\begin{aligned} & {{L}_{2}}-{{L}_{1}}=10\times 1.3 \\\ & \Rightarrow {{L}_{2}}-{{L}_{1}}=13dB \\\ \end{aligned}$$ The intensity increases as compared to the original value of 13dB. **So, the correct answer is “Option A”.** **Note:** The term $${{I}_{o}}$$ in the intensity level or loudness formula is the intensity of reference. It tells us how we perceive the sound to be loud in a relative manner. In busy traffic, the talking sound seems very low, but the same talking sound in a quiet classroom seems very loud. The intensity of reference also differs with the medium and is generally taken to be in the air.