Question

The sound level of a dog’s bark is 50 dB. The intensity of a rock concert is 10,000 times that of a dog’s bark. What is the sound level of the rock concert?
A. 50 dB
B. 100 dB
C. 90 dB
D. 60 dB

Answer 1

Firstly, we have to convert the unit of the value of the sound level of a dog’s bark from decibel into Weber per square meter. As the intensity of the rock concert is 10,000 times this value, so, we have to multiply the same. Then, again we have to convert the unit of the obtained result from Weber per square meter into decibel.
Formula used:
${{10}^{n}}{Wb}/{{{m}^{2}}}\;=n\times 10\,dB$

Complete answer:
From the given information, we have the data as follows.
The sound level of a dog’s bark is 50 dB.
The intensity of a rock concert is 10,000 times that of a dog's bark.
Firstly, we have to convert the unit of the value of sound level of a dog’s bark from decibel into Weber per square meter. So, we have the formula as follows,
$n\times 10\,dB={{10}^{n}}{Wb}/{{{m}^{2}}}\;$
Substitute the value in the above formula.

& 50\,dB=5\times 10\,dB \\\ & \Rightarrow 5\times 10\,dB={{10}^{5}}{Wb}/{{{m}^{2}}}\; \\\ \end{aligned}$$ Thus, the sound level of a dog’s bark in terms of the intensity is $${{10}^{5}}{Wb}/{{{m}^{2}}}\;$$. Now we have to multiply this value with the value 10,000. $$\begin{aligned} & I=10000\times {{10}^{5}}{Wb}/{{{m}^{2}}}\; \\\ & \therefore I={{10}^{9}}{Wb}/{{{m}^{2}}}\; \\\ \end{aligned}$$ Now, again we have to convert the unit from Weber per square meter into decibel. So, we have the formula as follows, $${{10}^{n}}{Wb}/{{{m}^{2}}}\;=n\times 10\,dB$$ Substitute the value in the above formula. $$\begin{aligned} & {{10}^{9}}{Wb}/{{{m}^{2}}}\;=9\times 10\,dB \\\ & \therefore {{10}^{9}}{Wb}/{{{m}^{2}}}\;=90\,dB \\\ \end{aligned}$$ **$$\therefore $$ The sound level of the rock concert is 90 dB, thus, option (C) is correct.** **Note:** The unit conversion between the Weber per square meter and the decibel (dB) should be known to solve this problem. Either we can convert the unit of all the values from decibel to Weber per square meter or vice versa. As in the options, the units of the values is the decibel, thus, we have to represent the final answer in terms of decibel.