1. Assume a proportionality relation

   We are told that displacement $s$ depends on:

   • pressure difference $\Delta p$
   • speed of sound $v$
   • density of air $\rho$
   • frequency $f$

   So let $s \propto (\Delta p)^a v^b \rho^c f^d$

2. Use dimensional analysis

   Dimensions of each quantity:

   • $[s] = L$
   • $[\Delta p] = [\text{pressure}] = M L^{-1} T^{-2}$
   • $[v] = L T^{-1}$
   • $[\rho] = M L^{-3}$
   • $[f] = T^{-1}$

   Therefore, $L = (M L^{-1} T^{-2})^a (L T^{-1})^b (M L^{-3})^c (T^{-1})^d$

   Equating powers of $M, L, T$:

   • For mass $M$: $a + c = 0$
   • For length $L$: $-a + b - 3c = 1$
   • For time $T$: $-2a - b - d = 0$

3. Use the condition that displacement is proportional to pressure difference

   Since $s$ is proportional to $\Delta p$, we take $a = 1$

   Then from $a+c=0$, $c = -1$

   From $-a+b-3c=1$, $-1 + b -3(-1) = 1$ $-1 + b + 3 = 1$ $b+2=1$ $b=-1$

   From $-2a-b-d=0$, $-2(1)-(-1)-d=0$ $-2+1-d=0$ $-1-d=0$ $d=-1$

   Hence, $s \propto \frac{\Delta p}{\rho v f}$

   Taking multiplicative constant as $1$, $s = \frac{\Delta p}{\rho v f}$

4. Substitute the given values

   Given: $\Delta p \approx 10\,\text{Pa}, \quad \rho \approx 1\,\text{kg/m}^3, \quad v \approx 300\,\text{m/s}, \quad f \approx 1000\,\text{Hz}$

   So, $s = \frac{10}{1 \cdot 300 \cdot 1000} \text{ m}$ $s = \frac{10}{3\times 10^5} \text{ m}$ $s = \frac{1}{3\times 10^4} \text{ m}$ $s \approx 3.3 \times 10^{-5} \text{ m}$

5. Convert to mm

   Since $1\,\text{m} = 1000\,\text{mm}$, $s \approx 3.3 \times 10^{-5} \times 1000\,\text{mm}$ $s \approx 3.3 \times 10^{-2}\,\text{mm}$ $s \approx 0.033\,\text{mm}$

   This is of the order of $\frac{3}{100}\,\text{mm}$

6. Match with options

   • A: $1\,\text{mm}$ ✗
   • B: $\dfrac{3}{100}\,\text{mm}$ ✓
   • C: $10\,\text{mm}$ ✗
   • D: $\dfrac{1}{10}\,\text{mm}$ ✗

Therefore, the correct option is B.