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Two blocks each of mass m lie on a smooth table. They are attached to two other masses as shown in the figure. The pulleys and strings are light. An object O is kept at rest on the table. The sides AB & CD of the two blocks are made reflecting. The acceleration of two images formed in those two reflecting surfaces w.r.t. each other is:

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5g/6

5g/3

g/3

17g/6

answer is D.

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Detailed Solution

T - 3 mg = - 3 ma \to (1)

$T = ma$

$ma - 3 mg = - 3 ma$

$4ma = 3 mg$

$a = \frac{3g}{4} (To wards left)$

$T - 2 mg = - 2 {ma}^{|}$

$T = {ma}^{|}$

${ma}^{|} - 2 mg = - 2 {ma}^{|}$

$3 {ma}^{|} = 2 mg$

$a^{|} = \frac{2 g}{3} (Towards right)$

Acceleration of mirror AB w,r,to CD (or) CD w.r.to AB is

$a_{m} = \frac{3g}{H} + \frac{2g}{3}$

$a_{m} = \frac{17g}{12}$

$a_{I} = {2a}_{m} - a_{0}$

$a_{I} = 2 \times \frac{17g}{12} = 0$

$a = \frac{17g}{6}$

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