$\beta=10 \log _{10}\left(\frac{I}{I_{0}}\right)$
where, $I=$ intensity of sound and $I_0=$ reference intensity $\left(\sim 10^{-12} \,W / m ^2\right.$ ).
Now, taking antilog, we have
$\frac{I}{I_0}=10^{\left(\frac{\beta}{10}\right)} \Rightarrow I=I_0 \times 10^{\left(\frac{\beta}{10}\right)} \quad \dots(i)$
Here, $\beta=20$ at $1 \,kHz$ and $\beta=60$ at $9 \,kHz$.
'Taking a linear relation, $\beta=k f+c$
where, $k$ and $c$ are constants and $f=$ frequency.
So, we have
$20=k \times 1+c \quad \dots(ii)$
$\text { and }$ $60=k \times 9+c \quad \dots(iii)$
Subtracting, Eqs. $(ii)$ and $(iii)$, we get
$60-20 =9 k-k+c-c$
$40 =8 k+0$
$8 k=40$
$k=5$
Putting the value of $k$ in Eq. $(ii)$, we have
$20=5+c \Rightarrow c=15$
$\Rightarrow \quad c=15$
$\therefore \operatorname{At} f=5 \,kHz ,$
$\beta=k f+c=5(5)+15$
$=40$
Now, width $\beta=20$ and $\beta=40$ from Eq. $(i)$, we have
$\text { Intensity, } \quad I =I_{0} \cdot 10^{\left(\frac{\beta}{10}\right)}$
$\Rightarrow \quad I_{1 \,kHz }=I_0(10)^{\frac{20}{10}}=I_{0} \cdot 10^2$
$I_{5 \,kHz }=I_0(10)^{\frac{40}{10}}=I_0 \cdot 10^4$
$\Rightarrow \quad \frac{I_{5 \,kHz }}{I_{1 \,kHz }}=\frac{10^4}{10^2}=100$
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