C

Step 1: The initial distance of B from the mirror is given as 42.0 cm.
Step 2: The acceleration of the elevator is 2.00 m/s² upwards. Therefore, the effective acceleration of block B, when released, is g - a = 10 m/s² - 2 m/s² = 8 m/s² downwards.
Step 3: Use the equation of motion to find the distance fallen by the block after t = 0.200 s.
The equation is: \( s = ut + \frac{1}{2} a t^2 \)
Where:
\( u = 0 \) (initial velocity), \( a = 8 \, \text{m/s}^2 \), \( t = 0.200 \text{ s} \)
Thus, \( s = 0 + \frac{1}{2} \cdot 8 \cdot (0.200)^2 \) = \( 0.5 \cdot 8 \cdot 0.04 \) = 0.16 m = 16 cm.
Step 4: The new distance of block B from the mirror after 0.200 s is: 42.0 cm - 16 cm = 26.0 cm.
Step 5: The distance between the image of block B and the mirror:
The image distance \( v \) can be calculated using the mirror formula, \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \). Here, \( u = -26 \) cm (object distance is taken negative in mirror formula), and \( f = 12.0 \) cm.
Rearranging the formula, we get \( \frac{1}{v} = \frac{1}{12} + \frac{1}{-26} = \frac{13 - 6}{78} = \frac{7}{78}. \)
Thus, \( v = \frac{78}{7} \approx 11.14 \) cm.
The image distance from the mirror will be the absolute value since it will be in the same direction as the object relative to the mirror.
Step 6: Final distance between the image of block B and the mirror is approximately 11.14 cm.