The minimum intensity of audibility of sound is `10^(-12) W//m^(2) s` and density of air ` = 1.293 kg//m^(3)`. If the frequency of sound in `1000 Hz` , then the corresponding amplitude the vibration of the air particles is
[ Take velocity of sound `= 332 m//s`]

Options

A. `1.1 xx 10^(-7) m`
B. `1.1 xx 10^(-9) m`
C. `1.1 xx 10^(-11) m`
D. `1.1 xx 10^(-14) m`

Answer

C

Explanation

`I = 2 pi ^(2) a^(2) v^(2) rho v`
`a^(2) = (I)/( 2 pi^(2) v^(2) rho v) or a = (1)/( pi v) sqrt((I)/( 2 rho v))`
or ` a = (7)/( 22 xx 1000) sqrt((10^(-12))/( 2 xx 1.293 xx 332)) m`
or ` a = 1.1 xx 10^(-11) m`