Question

The ratio of intensities of sounds A
and B is 100: 1, then the intensity
level of A is greater than that of B
(a) 100 times (b) 10 times
(c) 2 times (d) 2 times​

Answer 1

(a) 100 Times

is a correct answer

I hope it's help you

please mark me brainlist

Answer 2

The intensity level of A is greater than that of B by 20 times.

Given: The ratio of intensities of sounds A and B is 100: 1.

To Find: The intensity level of A that is greater than that of B.

Solution:

• We know that the other name for intensity level is loudness.

• Loudness can be calculated using the formula,

         Δ L = 10 × log ( I1 / I2 )                                              ...(1)

Where Δ L = change in loudness, I1 = intensity of A, I2 = intensity of B.

Coming to the numerical,

let the intensity of A be ' I1 '

and let the intensity of B be ' I2 '.

So, we are given that,

      I1 / I2 = 100

So, putting respective values in (1), we get;

      Δ L = 10 × log ( I1 / I2 )      

 ⇒  Δ L = 10 × log 100

 ⇒  Δ L = 10 × log 10²

 ⇒  Δ L = 10 × 2 log 10

 ⇒  Δ L = 20

So, the intensity level of A is greater than that of B by 20 times.

#SPJ2