Question

17. A microphone of surface area 2.0 cm’ absorbs 1.1 mW of sound. What is the intensity
of sound hitting the microphone?
A) 2.2 x 10-5 W/m2
B) 0.55 W/m²
C) 2.2 W/m2
D 2.8 W/m2
E) 5.5 W/m2

Answer 1

Answer:

$I=0.055\ W/m^2$

Explanation:

Given that,

A microphone of surface area 2.0 cm’ absorbs 1.1 mW of sound.

Surfrace area, A = 2 cm = 0.02 m

Power absorbed, P = 1.1 mW = 0.0011 W

We need to find the intensity of sound hitting the microphone. Power per unit area is called the intensity of sound. It can be given by :

$I=\dfrac{P}{A}\\\\=\dfrac{0.0011\ W}{0.02\ m^2}\\\\=0.055\ W/m^2$

So, the intensity of sound is $0.055\ W/m^2$. Hence, the correct option is (b).

Answer 2

The intensity  of sound hitting the microphone is equal to: E. 5.5 $W/m^2$

Given the following data:

Surface area = 2.0 $cm^2$

Power = 1.1 milliwatt to watt = 0.0011 watt.

Conversion

1 $cm^2$ = 0.0001 $m^2$

2.0 $cm^2$ = 0.0002 $m^2$

To determine the intensity  of sound hitting the microphone:

Mathematically, the intensity of a sound wave is given by the formula:

$I=\frac{P}{A}$

Where:

• I is the intensity of a wave.

• P is the power.

• A is the surface area.

Substituting the given parameters into the formula, we have;

$I=\frac{0.0011}{0.0002}$

Intensity, I = 5.5 $W/m^2$

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