B

Given a convex mirror with a focal length (f) of +10 cm. According to the mirror formula, we have:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Here, the distance behind the mirror (x) acts as the object distance (u), and is taken as negative for virtual images in the sign convention. Therefore, the object distance for different cases is:
(i) For x = 6 cm, u = -6 cm:
\[ \frac{1}{10} = \frac{1}{v} - \frac{1}{6} \Rightarrow \frac{1}{v} = \frac{1}{10} + \frac{1}{6} = \frac{3 + 5}{30} = \frac{8}{30} \Rightarrow v = 3.75 \text{ cm (virtual)}\]
Therefore, (i) corresponds to [B] Virtual.
(ii) For x = 11 cm, u = -11 cm:
\[ \frac{1}{10} = \frac{1}{v} - \frac{1}{11} \Rightarrow \frac{1}{v} = \frac{1}{10} + \frac{1}{11} = \frac{11 + 10}{110} = \frac{21}{110} \Rightarrow v = 5.238 \text{ cm (virtual)}\]
Therefore, (ii) again corresponds to [B] Virtual.
(iii) For x = 16 cm, u = -16 cm:
\[ \frac{1}{10} = \frac{1}{v} - \frac{1}{16} \Rightarrow \frac{1}{v} = \frac{1}{10} + \frac{1}{16} = \frac{16 + 10}{160} = \frac{26}{160} \Rightarrow v = 6.154 \text{ cm (virtual)}\]
Therefore, (iii) corresponds to [B] Virtual.
(iv) For x = 30 cm, u = -30 cm:
\[ \frac{1}{10} = \frac{1}{v} - \frac{1}{30} \Rightarrow \frac{1}{v} = \frac{1}{10} + \frac{1}{30} = \frac{3 + 1}{30} = \frac{4}{30} \Rightarrow v = 7.5 \text{ cm (virtual)}\]
Therefore, (iv) corresponds to [B] Virtual.
As all cases demonstrate that the image is virtual, therefore, the answer is (ii) which corresponds to [B] Virtual.