$A$ boy whose mass is $50\,kg$ stands on a spring balance inside a lift. The lift starts to ascend with an acceleration of $2\,m/s^2$. The reading of the machine or balance $(g = 10\,m/s^2)$ is ........ $kg$.

$A$ 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:
$(A)$ $\frac{1}{2} m \omega^2 R(e^{\omega t} - e^{-\omega t}) \hat{j} + mg \hat{k}$
$(B)$ $\frac{1}{2} m \omega^2 R(e^{\omega t} + e^{-\omega t}) \hat{j} + mg \hat{k}$
$(C)$ $-m \omega^2 R \cos \omega t \hat{j} - mg \hat{k}$
$(D)$ $m \omega^2 R \sin \omega t \hat{j} - mg \hat{k}$

In the given arrangement,all surfaces are smooth. What acceleration $a$ should be given to the system (block $M$) so that the block $m_2$ does not slide down?

$A$ wedge $Y$ with mass of $10 \text{ kg}$ has all frictionless surfaces, and the inclined surface makes an angle of $37^{\circ}$ with the horizontal. $A$ block $X$ with mass $2 \text{ kg}$ is placed at the highest point of the wedge as shown in the figure and is at rest. At $t=0$, the wedge $Y$ is pulled toward the right with a constant force $f$ of $24 \text{ N}$. Taking the block $X$ at rest at $t=0$, the time taken by it to slide down $8.8 \text{ m}$ on the slope, while $Y$ is in motion, is . . . . . . s. (Take $\tan(37^{\circ}) = 3/4$ and $g = 10 \text{ m/s}^2$)

$A$ block of mass $m$ is in contact with the cart $C$ as shown in the figure. The coefficient of static friction between the block and the cart is $\mu$. The acceleration $\alpha$ of the cart that will prevent the block from falling satisfies:

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 ............ $N$.