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A loudspeaker diaphragm 0.2 m in diameter is vibrating at 1 kHz with an amplitude of $0.01 \times 10^{- 3} m$ . Assume that the air molecules in the vicinity have the same amplitude of vibration. Density of air is $1.29 k g / m^{3}$ . Then match the items given in column I to that in column II. (Take velocity of sound = 340 m/s.)
Column-I Column-II
i) Pressure amplitude immediately in a front of the diaphragm is (in $N / m^{2}$ )
a) $2.7 \times 10^{- 2}$
ii) Sound intensity in front of the diaphragm is ( in $W / m^{2}$ ).
b) $2.15 \times 10^{- 5}$
iii) The acoustic power radiated is (in W) c) 27.55
iv) Intensity at 10 m from the loudspeaker (in $W / m^{2}$ )
d) 0.865

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$i \to b; i i \to a; i i i \to d; i v \to c$

$i \to c; i i \to d; i i i \to a; i v \to b$

$i \to d; i i \to c; i i i \to b; i v \to a$

$i \to a; i i \to b; i i i \to c; i v \to d$

answer is C.

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Detailed Solution

Pressure amplitude is given by, $P_{0} = ρ ω v A_{0}$
= $1.29 \times 2 π \times 10^{3} \times 340 \times (0.01 \times 10^{- 3})$
= $27.55 N / m^{2}$
Intensity is given by, $I = \frac{1}{2} ρ ω^{2} A^{2} v$
= $\frac{1}{2} \times 1.29 \times {(2 π \times 10^{3})}^{2} \times {(10^{- 5})}^{2} \times (340)$
= $0.865 W / m^{2}$
Power,
$P = 1 A = (0.865) π {(0.1)}^{2}$
= $0.027 W = 2.7 \times 10^{- 2} W$
Intensity at distance r m is given by I= $\frac{p_{a v}}{4 π r^{2}}$
I= $\frac{2.7 \times 10^{- 2}}{4 π \times 10^{2}} = 2.15 \times 10^{- 5} W / m^{2}$

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