A

To determine the friction coefficient between the record and the mosquito, we can use the concepts of circular motion and friction.
Step 1: Calculate the angular speed of the disc in radians per second. Since the disc is rotating at 33 revolutions per minute (rpm), we convert that to radians per second:
\[ \omega = 33 \times \frac{2\pi \text{ rad}}{1 ext{ rev}} \times \frac{1 ext{ min}}{60 ext{ s}} = \frac{33 \times 2\pi}{60} = \frac{11\pi}{10} \text{ rad/s} \]
Step 2: Now, we calculate the centripetal acceleration \(a_c\) of the mosquito at a distance of 10 cm (0.1 m) from the center of the disc using the formula:
\[ a_c = r\omega^2 \]
where \(r = 0.1 \text{ m}\).

So,
\[ a_c = 0.1 \times \left(\frac{11\pi}{10}\right)^2 = 0.1 \times \frac{121\pi^2}{100} = \frac{121\pi^2}{1000} \text{ m/s}^2 \]
Step 3: The frictional force is required to provide this centripetal acceleration; hence, \[ f = ma_c. \]
where \(m\) is the mass of the mosquito (assume a small mass, e.g., 0.001 kg for calculation).

Thus,
\[ f = m \times a_c = 0.001 \times \frac{121\pi^2}{1000} = \frac{121\pi^2}{1000000} ext{ N} \]
Step 4: The normal force \(N\) acting on the mosquito is equal to its weight: \[ N = mg = 0.001 \times 10 = 0.01 ext{ N} \]
Step 5: The coefficient of friction \(\mu\) is given by:
\[ \mu = \frac{f}{N} = \frac{\frac{121\pi^2}{1000000}}{0.01} = \frac{121\pi^2}{10000} \]
Step 6: Now substituting \(\pi\) (approximately 3.14):
\[ \mu = \frac{121(3.14)^2}{10000} \approx \frac{121 \times 9.86}{10000} \approx \frac{1198.66}{10000} \approx 0.1199 \]
Now, to conclude that this is greater than \(\frac{2}{81} \) (which is approximately 0.02469):
Therefore, \(\mu > \frac{2}{81}\) holds true.
Thus, we have shown that the friction coefficient between the record and the mosquito is greater than 2/81. Therefore, the answer is A.