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

Q: 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 .... $dB$

a (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 ${L_2} - {L_1} = 10{\log _{10}}\left( {\frac{{{I_2}}}{{{I_1}}}} \right)$ = $10{\log _{10}}\left( {\frac{{{P_2}}}{{{P_1}}}} \right)$= $10{\log _{10}}\left( {\frac{{400}}{{20}}} \right)$= $10{\log _{10}}20$ =$10\log (2 \times 10)$ = $10(0.301 + 1) = 13dB$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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 .... $dB$

A$13$
B$10$
C$20$
D$800$

Diffcult

STD 11 Science
NEET
STD 11 - 14. waves and sound
Physics

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