$A$ man has a mass of $80 \text{ kg}$. He stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of $6 \text{ m/s}^2$. What would be his apparent weight in $\text{kg}$? (Take $g = 10 \text{ m/s}^2$)

$A$ car is moving with a uniform speed on a level road. Inside the car,there is a balloon filled with helium and attached to a piece of string tied to the floor. The string is observed to be vertical. The car now takes a left turn while maintaining the same speed on the level road. The balloon in the car will:

$A$ 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:

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

$A$ vehicle is moving with a constant speed of $10 \ m/s$ on a circular horizontal track of radius $20 \ m$. $A$ bob is suspended from the roof of the vehicle by a massless string. The angle made by the string with the vertical will be (acceleration due to gravity,$g = 10 \ m/s^2$).

The readings shown in two extended spring balances,$S_1$ and $S_2$,are $60\,kg$ and $30\,kg$ respectively. The lift is accelerating upward with an acceleration of $10\,m/s^2$. The mass $m$ is stationary with respect to the lift. Then the mass of the block will be ......... $kg$. (Take $g = 10\,m/s^2$)