Question

For the given figure List - I gives acceleration of blocks of masses $m_1,m_2,m_3\&m_4$ and List - II gives their magnitudes. Here $m_1=m_2=2m$ and $m_3=m_4=m$

$\begin{array}{cc}\text{List - I}& \text{List - II}\\ \text{(I)}{a}_1& \text{(P)}2g\\ \text{(II)}{a}_2& \text{(Q)}g\\ \text{(III)}{a}_3& \text{(R)}\frac{g}{2}\\ \text{(IV)}{a}_4& \text{(S)}\frac{g}{4}\\ & \text{(T)}0\\ & \text{(U)}\text{None}\end{array}$
Just after cutting string connecting $m_1$ and $m_2$ correct match for accelerations of blocks is

• A. $I\to Q;II\to Q;III\to P;IV\to Q$
• B. $I\to Q;II\to P;III\to R;IV\to T$
• C. $I\to Q;II\to Q;III\to T;IV\to T$
• D. $I\to P;II\to Q;III\to R;IV\to U$

Answer 1

The correct option is C $I\to Q;II\to Q;III\to T;IV\to T$

$I\to Q;II\to Q;III\to T;IV\to T$
$m_1$ falls freely after the cut, so its acceleration is $g$.
The tension in the right side spring before cutting is $(m_1+m_2)g=4mg$. After cutting, the net force on $m_2$ is $4mg-m_{2}g=2mg$. So, its acceleration is $\frac{2mg}{m_2}=g$.

Just after cutting, all the springs still maintain the tension in them. Therefore, the blocks $m_3$ and $m_4$ remain in equilibrium.