Two blocks are in contact on a frictionless table. One has mass $m$ and the other $2\,m$. $A$ force $F$ is applied on $2\,m$ as shown in the figure. Now,the same force $F$ is applied from the right on $m$. In the two cases respectively,what is the ratio of the contact force between the two blocks?

$A$ lift of mass $m$ is ascending with an acceleration $a$ $(a < g)$. The tension in the cable of the lift is $(g = \text{acceleration due to gravity})$

$A$ 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:

$A$ block of mass $m$ slides on a wooden wedge,which in turn slides backward on a horizontal surface. The acceleration of the block with respect to the wedge is: Given $m = 8 \, kg, M = 16 \, kg$. Assume all surfaces shown in the figure to be frictionless.

Three blocks $A$,$B$,and $C$ of masses $5 \text{ kg}$,$3 \text{ kg}$,and $2 \text{ kg}$ respectively are pulled on a horizontal smooth surface by a force of $80 \text{ N}$ as shown in the figure. The tensions $T_1$ and $T_2$ in the strings are respectively:

$A$ block of mass $m_1$ rests on a horizontal table. $A$ string tied to the block is passed over a frictionless pulley fixed at the end of the table,and to the other end of the string is hung another block of mass $m_2$. The acceleration of the system is