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:

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

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.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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.