The power of a sound from the speaker of a radio is $20 mW$. By turning the knob of the volume control, the power of the sound is increased to $400 mW.$ The power increase in decibels as compared to the original power is .... $dB$
Diffcult
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(a)$P \propto I$
${L_1} = 10{\log _{10}}\left( {\frac{{{I_1}}}{{{I_0}}}} \right)$ and ${L_2} = 10{\log _{10}}\left( {\frac{{{I_2}}}{{{I_0}}}} \right)$
So ${L_2} - {L_1} = 10{\log _{10}}\left( {\frac{{{I_2}}}{{{I_1}}}} \right)$
= $10{\log _{10}}\left( {\frac{{{P_2}}}{{{P_1}}}} \right)$= $10{\log _{10}}\left( {\frac{{400}}{{20}}} \right)$= $10{\log _{10}}20$
=$10\log (2 \times 10)$ = $10(0.301 + 1) = 13dB$
${L_1} = 10{\log _{10}}\left( {\frac{{{I_1}}}{{{I_0}}}} \right)$ and ${L_2} = 10{\log _{10}}\left( {\frac{{{I_2}}}{{{I_0}}}} \right)$
So ${L_2} - {L_1} = 10{\log _{10}}\left( {\frac{{{I_2}}}{{{I_1}}}} \right)$
= $10{\log _{10}}\left( {\frac{{{P_2}}}{{{P_1}}}} \right)$= $10{\log _{10}}\left( {\frac{{400}}{{20}}} \right)$= $10{\log _{10}}20$
=$10\log (2 \times 10)$ = $10(0.301 + 1) = 13dB$
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