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`]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`

The minimum intensity of audibility of sound is `10^(-12) W//m^(2) s`

1.	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`]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» Correct Answer - C `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`

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