Four blocks are connected as shown in the figure on a horizontal frictionless surface. If $m_1 = m_2 = m_3 = m_4 = m$,then the ratio $T_3 / T_4$ will be:

$A$ trolley of mass $5\,kg$ on a horizontal smooth surface is pulled by a load of mass $2\,kg$ by means of a uniform rope $ABC$ of length $2\,m$ and mass $1\,kg$. As the load falls from $BC = 0$ to $BC = 2\,m$,its acceleration in $m/s^2$ changes from:

$A$ block of mass $M$ on a horizontal smooth surface is pulled by a load of mass $M/2$ by means of a rope $AB$ and string $BC$ as shown in the figure. The length and mass of the rope $AB$ are $L$ and $M/2$ respectively. As the block is pulled from $AB = L$ to $AB = 0$,its acceleration changes from:

$A$ block of mass $M$ is placed on a horizontal surface and it is tied with an inextensible string to a block of mass $m$,as shown in the figure. $A$ block of mass $m_0$ is also placed on $M$. If $\mu < \mu_{min}$ (the minimum friction required to keep the block $m$ stationary),then the downward acceleration of $m$ is:

$A$ block of mass $M$ is kept on a platform which starts accelerating upwards from rest with a constant acceleration $a$. During the time interval $T$,the work done by the contact force on mass $M$ is

$A$ small block of mass $m$ is placed on a wedge of mass $M$ as shown,which is initially at rest. All the surfaces are frictionless. The spring attached to the other end of the wedge has a force constant $k$. If $a'$ is the acceleration of $m$ relative to the wedge as it starts coming down and $A$ is the acceleration acquired by the wedge as the block starts coming down,then the maximum retardation of $M$ is: