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(B) The sound level (in decibels) is defined as $\beta = 10 \log_{10} \left( \frac{I}{I_0} \right)$,where $I$ is the intensity of sound and $I_0$ is t

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$100$ times that at $1 \, kHz$

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(B) The sound level (in decibels) is defined as $\beta = 10 \log_{10} \left( \frac{I}{I_0} ight)$,where $I$ is the intensity of sound and $I_0$ is the reference intensity.Taking the antilog,we have $\frac{I}{I_0} = 10^{\beta/10}$,which implies $I = I_0 	imes 10^{\beta/10}$.Given the linear relationship $\beta = kf + c$,where $f$ is the frequency in $kHz$:At $f = 1 \, kHz$,$\beta = 20 \implies 20 = k(1) + c \quad (i)$.At $f = 9 \, kHz$,$\beta = 60 \implies 60 = k(9) + c \quad (ii)$.Subtracting $(i)$ from $(ii)$: $40 = 8k \implies k = 5$.Substituting $k = 5$ into $(i)$: $20 = 5 + c \implies c = 15$.Thus,at $f = 5 \, kHz$,$\beta = 5(5) + 15 = 40 \, dB$.Now,calculating the intensities:$I_{1 \, kHz} = I_0 	imes 10^{20/10} = I_0 	imes 10^2$.$I_{5 \, kHz} = I_0 	imes 10^{40/10} = I_0 	imes 10^4$.Therefore,the ratio is $\frac{I_{5 \, kHz}}{I_{1 \, kHz}} = \frac{10^4}{10^2} = 100$. The intensity at $5 \, kHz$ is $100$ times that at $1 \, kHz$.

$A$ hearing test is conducted on an aged person. It is found that her threshold of hearing is $20 \, dB$ at $1 \, kHz$ and it rises linearly with frequency to $60 \, dB$ at $9 \, kHz$. The minimum intensity of sound that the person can hear at $5 \, kHz$ is

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