B

Step 1: Understand the relationship between coefficient of friction and maximum speed in a circular path. The maximum speed ($v_{max}$) can be obtained using the formula:

$$ v_{max} = ext{sqrt}(g imes r imes ext{µ}) $$
where $g$ is the acceleration due to gravity (approximately $9.81 ext{ m/s}^2$) and $r$ is the radius of the circular path.

Step 2: Given that the maximum speed on a dry road is $10 ext{ m/s}$ with a coefficient of friction $ ext{µ}$, we can express this as:

$$ 10 = ext{sqrt}(g imes r imes ext{µ}) $$

Step 3: Now let's address the wet road condition. The coefficient of friction on a wet road is diminished. Assuming the new coefficient of friction is given by:

Let this new coefficient of friction be expressed as \( ext{µ}_{wet} = k \times ext{µ} \) for some constant \( k < 1 \).

Step 4: Substitute the new coefficient of friction into the maximum speed equation:

$$ v_{max, wet} = ext{sqrt}(g imes r imes (k \times \text{µ})) $$

Step 5: Since we want to find the new maximum speed in relation to the original speed, we can express it in terms of the original speed:

$$ v_{max, wet} = ext{sqrt}(k) imes v_{max} $$

where \( v_{max} = 10 ext{ m/s} \).

Step 6: If given conditions state that the wet coefficient of friction is half the dry coefficient (assumption), then \( k = 0.5 \). Thus:

$$ v_{max, wet} = ext{sqrt}(0.5) imes 10 \approx 7.07 ext{ m/s} $$

Therefore, the maximum speed permitted on the wet road is approximately \(7.07 ext{ m/s} \). This is the superior option from the given choices.