In patients suffering from asthma, the thickness of the airway walls increases. This thickening results in a reduction of the local shear modulus of the airways by as much as $50 \%$ compared to the airways of a healthy person. In a particular person, the local shear modulus is reduced by $27 \%$. Assume that the frictional force due to the air flowing through the airways of this person during an asthma attack increases by a faģtor of 2.4 over that of a healthy person. Calculate the ratio $r$ of the shear strain on this person's airway to that of a healthy airway. \[ r= \]

Step 1 ：First, we need to understand the problem. The shear strain ratio $r$ is the ratio of the shear strain on the asthmatic person's airway to that of a healthy airway. The shear strain is inversely proportional to the shear modulus, and directly proportional to the frictional force. Therefore, we can express $r$ as $r = \frac{\text{Shear modulus of healthy airway}}{\text{Shear modulus of asthmatic airway}} \times \frac{\text{Frictional force in asthmatic airway}}{\text{Frictional force in healthy airway}}$

Step 2 ：The shear modulus of the asthmatic airway is reduced by $27\%$ compared to the healthy airway, so the ratio of the shear modulus of the healthy airway to the asthmatic airway is $\frac{1}{1-0.27} = \frac{1}{0.73}$

Step 3 ：The frictional force in the asthmatic airway is increased by a factor of $2.4$ over the healthy airway, so the ratio of the frictional force in the asthmatic airway to the healthy airway is $2.4$

Step 4 ：Substitute these values into the equation for $r$, we get $r = \frac{1}{0.73} \times 2.4 = \frac{2.4}{0.73}$

Step 5 ：To simplify this fraction, we can divide the numerator and the denominator by their greatest common divisor, which is approximately $0.01$, so $r = \frac{2.4/0.01}{0.73/0.01} = \frac{240}{73}$

Step 6 ：Therefore, the ratio $r$ of the shear strain on this person's airway to that of a healthy airway is $\boxed{\frac{240}{73}}$