In a space of width d , the absolute temperature of air linearly...

The speed of sound at an absolute temperature $T$ is given by:

$v_T = v \sqrt{\dfrac{T}{273}}$

Since the temperature increases linearly with distance $x$ from $0$ to $d$, the temperature at a distance $x$ is:

$T(x) = T_1 + \dfrac{T_2 - T_1}{d} x$

Differentiating with respect to $x$ gives:

$dT = \dfrac{T_2 - T_1}{d} dx \Rightarrow dx = \dfrac{d}{T_2 - T_1} dT$

The time taken by the sound wave to travel a small distance $dx$ is $dt = \dfrac{dx}{v_T}$. Integrating from $x = 0$ to $x = d$ (which corresponds to $T = T_1$ to $T = T_2$):

$t = \int_{T_1}^{T_2} \dfrac{d}{T_2 - T_1} \dfrac{\sqrt{273}}{v \sqrt{T}} dT$

$t = \dfrac{d \sqrt{273}}{v (T_2 - T_1)} \int_{T_1}^{T_2} T^{-1/2} dT$

$t = \dfrac{d \sqrt{273}}{v (T_2 - T_1)} \left[ 2\sqrt{T} \right]_{T_1}^{T_2}$

$t = \dfrac{2d \sqrt{273}}{v (T_2 - T_1)} (\sqrt{T_2} - \sqrt{T_1})$

Using the identity $T_2 - T_1 = (\sqrt{T_2} - \sqrt{T_1})(\sqrt{T_2} + \sqrt{T_1})$, we get:

$t = \dfrac{2d \sqrt{273}}{v (\sqrt{T_1} + \sqrt{T_2})}$

Substituting the given values $d = 33 \text{ m}$, $v = 330 \text{ m s}^{-1}$, $T_1 = 280 \text{ K}$, and $T_2 = 310 \text{ K}$:

$t = \dfrac{2 \times 33 \times \sqrt{273}}{330 (\sqrt{280} + \sqrt{310})}$

$t = \dfrac{66 \times 16.52}{330 (16.73 + 17.61)}$

$t = \dfrac{1090.32}{330 \times 34.34} \approx 0.096 \text{ s} = 96 \text{ ms}$