$A$ weight is suspended from the ceiling of a lift by a spring balance. When the lift is stationary,the spring balance reads $W$. If the lift suddenly falls freely under gravity,the reading on the spring balance will be

$A$ block can slide on a smooth inclined plane of inclination $\theta$ kept on the floor of a lift. When the lift is descending with a retardation $a$,the acceleration of the block relative to the incline is

If a body of mass $m$ is carried by a lift moving with an upward acceleration $a$,then the forces acting on the body are:
$(i)$ The reaction $R$ on the floor of the lift acting upwards.
$(ii)$ The weight $mg$ of the body acting vertically downwards.
The equation of motion will be given by:

$A$ spring balance and a physical balance are kept in a lift. In these balances,equal masses are placed. If the lift starts moving upwards with constant acceleration,then:

Two wooden blocks of masses $M$ and $m$ are placed on a smooth horizontal surface as shown in the figure. If a force $P$ is applied to the system as shown in the figure such that the mass $m$ remains stationary with respect to the block of mass $M$,then the magnitude of the force $P$ is

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 $g = 10 \, m/s^{2}$)