The power of sound from the speaker of radio is 10 W . Now, the power of sound from the speaker of the radio is increased to 400 W by increasing the volume of the radio. The power increased in dB as compared to original power is nearly

Question

The power of sound from the speaker of radio is 10 W . Now, the power of sound from the speaker of the radio is increased to 400 W by increasing the volume of the radio. The power increased in dB as compared to original power is nearly (A) 8 (B) 12 (C) 13 (D) 16

Tardigrade Admin · Accepted Answer

P1=10log|(10/10- 12)| P1=130 dB P2=10log|(400/10- 12)| =10log|4 × 1014| =140+10log4 P2=146 dB Δ P=(.146-130.)=16 dB

Q. The power of sound from the speaker of radio is $10 W$ . Now, the power of sound from the speaker of the radio is increased to $400 W$ by increasing the volume of the radio. The power increased in $d B$ as compared to original power is nearly

$8$

$12$

$13$

$16$

$P_{1} = 10 l o g ∣ ∣ \frac{10}{1 0 ^{- 12}} ∣ ∣$
$P_{1} = 130 d B$
$P_{2} = 10 l o g ∣ ∣ \frac{400}{1 0 ^{- 12}} ∣ ∣$
$= 10 l o g ∣ ∣ 4 \times 1 0^{14} ∣ ∣$
$= 140 + 10 l o g 4$
$P_{2} = 146 d B$
$Δ P = (146 - 130) = 16 d B$

Q. The power of sound from the speaker of radio is 10W . Now, the power of sound from the speaker of the radio is increased to 400W by increasing the volume of the radio. The power increased in dB as compared to original power is nearly

8

12

13

16

P1​=10log∣∣​10−1210​∣∣​ P1​=130dB P2​=10log∣∣​10−12400​∣∣​ =10log∣∣​4×1014∣∣​ =140+10log4 P2​=146dB ΔP=(146−130)=16dB

Q. The power of sound from the speaker of radio is $10 W$ . Now, the power of sound from the speaker of the radio is increased to $400 W$ by increasing the volume of the radio. The power increased in $d B$ as compared to original power is nearly

$8$

$12$

$13$

$16$

$P_{1} = 10 l o g ∣ ∣ \frac{10}{1 0 ^{- 12}} ∣ ∣$
$P_{1} = 130 d B$
$P_{2} = 10 l o g ∣ ∣ \frac{400}{1 0 ^{- 12}} ∣ ∣$
$= 10 l o g ∣ ∣ 4 \times 1 0^{14} ∣ ∣$
$= 140 + 10 l o g 4$
$P_{2} = 146 d B$
$Δ P = (146 - 130) = 16 d B$