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(D) For the left block of mass $m$,the net pulling force is $3mg$ and the total mass of the system is $(m + 3m) = 4m$. Thus,the acceleration of the le

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$17g / 6$

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(D) For the left block of mass $m$,the net pulling force is $3mg$ and the total mass of the system is $(m + 3m) = 4m$. Thus,the acceleration of the left block is $a_1 = \frac{3mg}{4m} = \frac{3}{4}g$.Since the mirror (side $AB$) moves with acceleration $a_1$ towards the object $O$,the image formed by this mirror moves with acceleration $a_{I_1} = 2a_1 = 2 	imes \frac{3}{4}g = \frac{3}{2}g$ towards the object.For the right block of mass $m$,the net pulling force is $2mg$ and the total mass of the system is $(m + 2m) = 3m$. Thus,the acceleration of the right block is $a_2 = \frac{2mg}{3m} = \frac{2}{3}g$.Since the mirror (side $CD$) moves with acceleration $a_2$ away from the object $O$,the image formed by this mirror moves with acceleration $a_{I_2} = 2a_2 = 2 	imes \frac{2}{3}g = \frac{4}{3}g$ away from the object.Since the two images are moving in opposite directions relative to the object,their relative acceleration is $a_{rel} = a_{I_1} + a_{I_2} = \frac{3}{2}g + \frac{4}{3}g = \frac{9g + 8g}{6} = \frac{17}{6}g$.

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