Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
$90 \mathrm{~dB}$ sound is $x$ times more intense than $40 \mathrm{~dB}$ sound, then $x$ is
PhysicsWaves and SoundCOMEDKCOMEDK 2016
Options:
  • A 5
  • B 50
  • C $10^{5}$
  • D 500
Solution:
2503 Upvotes Verified Answer
The correct answer is: $10^{5}$
The level of sound is given by
$L=10 \log \left(\frac{I}{I_{0}}\right) \mathrm{dB}$
Let for $40 \mathrm{~dB}$ sound intensity be $I$ and for $90 \mathrm{~dB}$ sound intensity be $I_{x}$, then
$40=10 \log \left(\frac{I}{I_{0}}\right)$
$\Rightarrow \quad \log \left(\frac{I}{I_{0}}\right)=4 ...(i)$
and $\quad 90=10 \log \left(\frac{I_{x}}{I_{0}}\right)$
$\Rightarrow \quad \log \left(\frac{I_{x}}{I_{0}}\right)=9$
Given, $\quad I_{x}=x I$
$\Rightarrow \quad \log \left(\frac{x I}{I_{0}}\right)=9$
or $\quad \log x+\log \left(\frac{I}{I_{0}}\right)=9$
$\Rightarrow \quad \log x+4=9 \quad$ [from Eq. (i)]
$\Rightarrow \quad \log x=5 \quad \Rightarrow \quad x=10^{5}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.