The loudspeaker diaphragmm 0.2 m in diameter is vibrating at 1 kHz with an amplitude of `0.01xx10^(-3)m`. Assume that the air molecules in the vicinity have the same amplitude of vibration. Density of air is `1.29kg//m^(3)`. Then match the item given in List 1 to that in list 2. take velocity of sound `=340m//s`

Options

A. A-R,B-P,C-Q
B. A-R,B-S,C-P
C. A-P,B-Q,C-R
D. A-Q,B-P,C-S

Answer

B

Explanation

Pressure amplitude
`P_(0)=rhoomegavA_(0)=1.29xx2pixx10^(3)xx340xx(0.01xx10^(-3))=27.55N//m^(2)`
Intensity `(I)=(1)/(2)rhoomega^(2)A^(2)v=0.865W//m^(2)`
Power=`JA=0.865pi(0.1)^(2)=0.027W`
Intensity at r=10m `=(P_(av))/(A)=2.15xx10^(-5)W//m^(2)`