Consider the shown arrangement. Assume all surfaces to be smooth. If $N$ represents magnitude of normal reaction between block and wedge then acceleration of $M$ along horizontal equals.

 

A cylinder rests in a supporting carriage as shown. The side $AB$ of carriage makes an angle $30°$ with the horizontal and side $BC$ is vertical. The carriage lies on a fixed horizontal surface and is being pulled towards left with an horizontal acceleration $a$. The magnitude of normal reactions exerted by sides $AB$ and $BC$ of carriage on the cylinder be $N_{AB}$ and $N_{BC}$, respectively. Neglect friction everywhere. Then as the magnitude of acceleration $a$ of the carriage is increased, pick up the correct statement.

Two blocks $A$ and $B$ each of mass $m$ are placed on a smooth horizontal surface. Two horizontal forces $F$ and $2F$ are applied on the two blocks $A$ and $B$, respectively as shown in figure. The block $A$ does not slide on block $B$. Then the normal reaction acting between the two blocks is

Two masses $\mathrm{m}$ and $\mathrm{M}$ are attached with strings as shown. For the system to be in equilibrium we have

Three blocks $\mathrm{A} , \mathrm{B}$ and $\mathrm{C}$ are suspended as shown in the figure. Mass of each block $\mathrm{A}$ and $\mathrm{C}$ is $\mathrm{m}$. If system is in equilibrium and mass of $\mathrm{B}$ is $\mathrm{M}$, then :

In the arrangement shown in figure, pulleys are massless and frictionless and threads are light and inextensible. Block of mass ${\mathrm{m}}_1$ will remain at rest if :

A uniform rope of length $L$ and mass $M$ is placed on a smooth fixed wedge as shown. Both ends of rope are at same horizontal level. The rope is initially released from rest, then the magnitude of initial acceleration of rope is

Block $B$ moves to the right with a constant velocity $v_0$. The velocity of block $A$ relative to $B$ is

System is shown in the figure and man is pulling the rope from both sides with constant speed $u$. Then the speed of the block will be ($M$ moves vertical)