In the system shown in the figure, $m_{1} > m_{2}$ . System is held at rest by thread BC. Just after the thread BC is burnt:

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

Acceleration of $m_{2}$ will be upwards

Acceleration of $m_{1}$ will be equal to zero

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

answer is A, C.

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Detailed Solution

Before the thread BC is burnt.
From FBD of mass $m_{1}$
Spring force $= m_{1} g$
Spring force does not change just after cutting the thread
Just after thread BC is burnt, forces on $m_{2}$ are: spring force $({=m}_{1} g)$ upwards and weight $(m_{2} g)$ downwards.
Since, $m_{1} {>m}_{2}$ , acceleration of $m_{2}$ will be upwards. Option (a) is correct.
For $m_{2}$ : acceleration, $a = \frac{m_{1} g - m_{2} g}{m_{1} + m_{2}} = \frac{(m_{1} - m_{2})}{m_{1} + m_{2}} g$
Forces on $m_{1}$ are: spring force $(m_{1} g)$ upwards and weight $(m_{1} g)$ downwards. Both are equal and opposite. Hence, acceleration of $m_{1}$ is equal to zero. Option (b) is wrong. Option (c) is correct. Option (d) is wrong.