#1
as per the formula of level of sounds we know that
[tex]L = 10 Log\frac{I}{I_o}[/tex]
now for two different levels we can use that
[tex]L_2 - L_1 = 10 Log\frac{I_2}{I_1}[/tex]
now plug in data in above equation
[tex]50 - 20 = 10 Log \frac{I_2}{I_1}[/tex]
[tex]10^3 = \frac{I_2}{I_1}[/tex]
now we know that
[tex]\frac{I_2}{I_1} = \frac{r_1^2}{r_2^2}[/tex]
[tex]10^3 = \frac{15^2}{r^2}[/tex]
[tex]r = 47 cm[/tex]
#4)
level of sound is given as 110 dB
now we can say
[tex]L = 10 Log\frac{I}{I_0}[/tex]
[tex]110 = 10 Log\frac{I}{10^{-12}}[/tex]
[tex]I = 0.10 W/m^2[/tex]
now the energy received per second is given as
[tex]P = Intensity \times area[/tex]
[tex]P = 0.10 \times \pi r^2[/tex]
[tex]P = 0.10 \times \pi(0.003)^2[/tex]
[tex]P = 2.8 \mu J[/tex]
#5
length of the canal is 2.4 cm
resonating frequency of first harmonic is given as
[tex]f = \frac{v}{4L}[/tex]
[tex]f = \frac{340}{4\times 0.024}[/tex]
[tex]f = 3.6 kHz[/tex]
