In the figure,the reading of the spring balance will be $......\,N$ $\left[ g = 10\,m/s^2 \right]$.

$A$ plumb bob is hanging from the ceiling of a car. If the car moves with an acceleration $a$,the angle made by the string with the vertical is

$A$ monkey is sitting on the pan of a spring balance which is placed on an elevator. The maximum reading of the spring balance will be when

$A$ block of mass $m$ is placed on a smooth wedge of inclination $\theta$. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block ($g$ is acceleration due to gravity) will be

$A$ 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 apparent weight of a girl of mass $30 \ kg$ when she is in a lift moving vertically upwards with an acceleration of $2 \ m \ s^{-2}$ is (Acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $N$)