Q180 / 180 · COMEDK 2018 · Physics

How many times more intense is a $60 \mathrm{~dB}$ sound than a $30 \mathrm{~dB}$ sound?

  • Correct answer
    A
    1000
  • B
    2
  • C
    100
  • D
    4

Solution

Loudness of sound is given as $$ L=10 \log _{10} \frac{I}{I_{0}} $$ So, we get $L_{2}-L_{1}=10 \log _{10} \frac{I_{2}}{I_{1}}$ here $L_{1}=30 \mathrm{~dB}, L_{2}=60 \mathrm{~dB}$ $\begin{array}{ll}\therefore & 60-30=10 \log _{10} \frac{I_{2}}{I_{1}} \\ \Rightarrow & 3=\log _{10} \frac{I_{2}}{I_{1}} \\ \Rightarrow & \frac{I_{2}}{I_{1}}=10^{3} \Rightarrow I_{2}=1000 I_{1}\end{array}$

Hence, the correct answer is option: 'A'.

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