A mosquito rests on an L.P. record disc that is rotating on a turn...

The angular velocity of the turn table is given by $\omega = 33\dfrac{1}{3} \text{ rev/min}$.

Converting to radians per second:

$\omega = \dfrac{100}{3} \times \dfrac{2\pi}{60} = \dfrac{10\pi}{9} \text{ rad/s}$

The distance of the mosquito from the centre is $r = 10 \text{ cm} = 0.1 \text{ m}$.

For the mosquito to not slip, the static friction must provide the necessary centripetal force:

$\mu m g \ge m \omega^2 r$

$\mu \ge \dfrac{\omega^2 r}{g}$

Substituting the values:

$\mu \ge \dfrac{\left(\dfrac{10\pi}{9}\right)^2 \times 0.1}{10}$

$\mu \ge \dfrac{100\pi^2}{81} \times \dfrac{1}{100}$

$\mu \ge \dfrac{\pi^2}{81}$

The minimum friction coefficient required is $\dfrac{\pi^2}{81}$.