If the rope of a lift breaks suddenly,what is the tension exerted by the floor of the lift on a person inside? ($a =$ acceleration of the lift)

$A$ coin is dropped in a lift. It takes time $t_1$ to reach the floor when the lift is stationary. It takes time $t_2$ when the lift is moving up with constant acceleration. Then:

$A$ lift is moving downwards with an acceleration equal to the acceleration due to gravity. $A$ body of mass $m$ kept on the floor of the lift is pulled horizontally. If the coefficient of friction is $\mu$,then the frictional resistance offered by the body is:

$A$ smooth rod of length $l$ is kept inside a trolley at an angle $\theta$ as shown in the figure. What should be the acceleration $a$ of the trolley so that the rod remains in equilibrium with respect to it?

Three masses $M = 100 \, kg$,$m_{1} = 10 \, kg$,and $m_{2} = 20 \, kg$ are arranged in a system as shown in the figure. All the surfaces are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. $A$ force $F$ is applied on the system so that the mass $m_{2}$ moves upward with an acceleration of $2 \, m/s^{2}$. The value of $F$ is $...... \, N$. (Take $g = 10 \, m/s^{2}$)

The ratio of the weight of a man in a stationary lift to his weight in a lift moving downwards with an acceleration '$a$' is $3:2$. Then '$a$' = . . . . . . .