$A$ block of mass $m$ is resting on a wedge of angle $\theta$ as shown in the figure. The wedge is given a horizontal acceleration $a$. What is the minimum value of $a$ so that the mass $m$ falls freely?

$A$ person is standing in an elevator. In which situation does he experience weight loss?

$A$ man weighing $75 \,kg$ is standing in a lift. The weight of the man standing on a weighing machine kept in the lift when the lift is moving downwards freely under gravity is

Two blocks of masses $m_1$ and $m_2$ are connected with a massless unstretched spring and placed over a plank moving with an acceleration '$a$' as shown in the figure. The coefficient of friction between the blocks and the platform is $\mu$.

$A$ mass $M$ is placed on a smooth wedge of angle $\theta$ resting on a frictionless surface. To keep the mass $M$ at rest relative to the wedge,the wedge must be given a horizontal acceleration $a$ equal to:

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