A

Step 1: We start by using the centripetal force requirement for the motorcyclist to maintain motion on the vertical wall of the well. For a body moving in a circle of radius R with speed v, the centripetal force is given by:
$$ F_c = \frac{mv^2}{R} $$
Step 2: The gravitational force acting on the motorcyclist is:
$$ F_g = mg $$
Step 3: For the motorcyclist to stay on the wall of the well, the frictional force should provide the necessary centripetal force. Thus, we have:
$$ F_f = F_c $$
where the frictional force is given by:
$$ F_f = \mu N $$
Here, N is the normal force. In this scenario, the normal force will also equal the gravitational force:
$$ N = mg $$
Therefore, the frictional force becomes:
$$ F_f = \mu mg $$
Step 4: Setting these equal gives us:
$$ \mu mg = \frac{mv^2}{R} $$
Step 5: By canceling mass (m) from both sides, we find:
$$ \mu g = \frac{v^2}{R} $$
Step 6: Solving for the coefficient of friction (\mu), we have:
$$ \mu = \frac{v^2}{gR} $$
Step 7: Substituting the known values: R = 5 m, g = 10 m/s², and v = \sqrt{gR} = \sqrt{10\times5} = \sqrt{50} \approx 7.07 m/s.
Therefore:
$$ \mu = \frac{(7.07)^2}{10 \times 5} = \frac{50}{50} = 1 $$
Hence, the minimum value of the coefficient of friction is 1.