$A$ boy having a mass equal to $40 \, kg$ is standing in an elevator. The force felt by the feet of the boy (apparent weight) will be greatest when the elevator $(g = 9.8 \, m/s^2)$:

$A$ body of mass $1.0 \, kg$ is falling with an acceleration of $10 \, m/s^2$. Its apparent weight will be ......... $kg \, wt$ $(g = 10 \, m/s^2)$.

$A$ person in an elevator accelerating upwards with an acceleration of $2 \, m/s^2$ tosses a coin vertically upwards with a speed of $20 \, m/s$. After how much time will the coin fall back into his hand (in $, s$)? $(g = 10 \, m/s^2)$

$A$ frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. $A$ coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity $\omega$ is an example of a non-inertial frame of reference. The relationship between the force $\vec{F}_{\text{rot}}$ experienced by a particle of mass $m$ moving on the rotating disc and the force $\vec{F}_{\text{in}}$ experienced by the particle in an inertial frame of reference is $\vec{F}_{\text{rot}} = \vec{F}_{\text{in}} + 2m(\vec{v}_{\text{rot}} \times \vec{\omega}) + m(\vec{\omega} \times \vec{r}) \times \vec{\omega}$,where $\vec{v}_{\text{rot}}$ is the velocity of the particle in the rotating frame of reference and $\vec{r}$ is the position vector of the particle with respect to the centre of the disc. Now consider a smooth slot along a diameter of a disc of radius $R$ rotating counter-clockwise with a constant angular speed $\omega$ about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc,the $x$-axis along the slot,the $y$-axis perpendicular to the slot and the $z$-axis along the rotation axis $(\vec{\omega} = \omega \hat{k})$. $A$ small block of mass $m$ is gently placed in the slot at $\vec{r} = (R/2) \hat{i}$ at $t = 0$ and is constrained to move only along the slot.
$(1)$ The distance $r$ of the block at time $t$ is:
$(A)$ $\frac{R}{4}(e^{\omega t} + e^{-\omega t})$
$(B)$ $\frac{R}{2} \cos \omega t$
$(C)$ $\frac{R}{4}(e^{2\omega t} + e^{-2\omega t})$
$(D)$ $\frac{R}{2} \cos 2\omega t$
$(2)$ The net reaction of the disc on the block is:
$(A)$ $\frac{1}{2} m \omega^2 R(e^{\omega t} - e^{-\omega t}) \hat{j} + mg \hat{k}$
$(B)$ $\frac{1}{2} m \omega^2 R(e^{\omega t} + e^{-\omega t}) \hat{j} + mg \hat{k}$
$(C)$ $-m \omega^2 R \cos \omega t \hat{j} - mg \hat{k}$
$(D)$ $m \omega^2 R \sin \omega t \hat{j} - mg \hat{k}$

The weight of a man in a lift moving with the same acceleration upwards is $608 \ N$,while the weight of the same man in the lift moving downwards with the same acceleration is $368 \ N$. His normal weight is ............ $N$.

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