The intensity of sound from a point source is $1.0 \times 10^{- 6} m^{- 2}$ at a distance of 5.0 m from the source. What will be the intensity at a distance of 25 m from the source?

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Solution

The correct option is D

The power of the sound wave is intensity $\times$ surface area

power $= 1 \times 10^{8} \times 4 π \times 25$

$= 100 π 10^{- 8} = 10^{- 6} π W$

Intensity at 25m $= \frac{p o w e r}{s u r f a c e a r e a}$

$= \frac{10^{- 6} π}{4 π (25)^{2}}$

$= \frac{10^{- 6}}{625 \times 4}$

$= \frac{10^{- 6}}{2500}$

$= 0.04 \times 10^{- 8} W m^{- 2}$

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The intensity of sound from a point source is 1.0 × 10−6m−2 at a distance of 5.0 m from the source. What will be the intensity at a distance of 25 m from the source?

The correct option is D The power of the sound wave is intensity × surface area power =1 × 108 × 4π × 25 =100π 10−8 = 10−6π W Intensity at 25m =powersurfacearea =10−6π4π(25)2 =10−6625 × 4 =10−62500 =0.04 × 10−8 Wm−2

The intensity of sound from a point source is $1.0 \times 10^{- 6} m^{- 2}$ at a distance of 5.0 m from the source. What will be the intensity at a distance of 25 m from the source?