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

$A$ body of mass $M$ tied to a string is lowered at a constant acceleration of $(g/4)$ through a vertical distance $h$. The work done by the string will be..............

Two masses $M$ and $m$ are connected by a weightless string. They are pulled by a force $F$ on a frictionless horizontal surface as shown in the figure. The tension in the string will be

$A$ block of mass $1 \, kg$ is suspended by a string of mass $1 \, kg$ and length $1 \, m$ as shown in the figure. Taking $g = 10 \, m/s^2$,calculate the force exerted by the support on the string. (in $, N$)

$A$ rod of length $L$ and mass $M$ is acted on by two unequal forces $F_1$ and $F_2$ $(F_2 < F_1)$ as shown in the figure. The tension in the rod at a distance $y$ from the end $A$ is given by:

$A$ 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 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