In the system shown in figure M₁>M₂ . System is held at rest by thread BC. Just after thread BC is burnt.

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Acceleration of M₁ will be equal to zero

Acceleration of M₂ will be upwards

Magnitude of acceleration of both blocks will be equal to $(\frac{M_{1} - M_{2}}{M_{1} + M_{2}}) g$

Magnitude of acceleration of two blocks will be non-zero and unequal.

answer is C.

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Detailed Solution

Tension in the spring $T_{1} = M_{1} g$
After the thread is burnt

Tension in the spring remains same.
$∴ a_{A} = 0 a n d F_{B} = M_{1} g ⇒ a_{B} = \frac{M_{1} - M_{2}}{M_{2}} (g)$

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