
The problem involves a particle of mass $m$ falling under gravity and encountering a resistive force $F_r = -kv$, where $v$ is the velocity and $k$ is a constant. The negative sign indicates the force opposes the velocity (upwards if falling downwards).
Using Newton's second law, the net force $F_{net}$ acting on the particle is:
$F_{net} = F_g + F_r$Taking the downward direction as positive, the gravitational force $F_g = mg$. The equation becomes:
$ma = mg - kv$Since acceleration $a = \frac{dv}{dt}$, we have the differential equation:
$m \frac{dv}{dt} = mg - kv$Rearranging gives:
$\frac{dv}{dt} = g - \frac{k}{m}v$This is a first-order linear differential equation. We can solve it by separation of variables or integrating factors. The initial condition is that the particle starts from rest, so $v(0) = 0$.
The solution to this differential equation with the initial condition $v(0)=0$ is:
$v(t) = \frac{mg}{k} \left( 1 - e^{-\frac{k}{m}t} \right)$Let $V_T = \frac{mg}{k}$. This is the terminal velocity. The equation is:
$v(t) = V_T \left( 1 - e^{-\frac{k}{m}t} \right)$Key characteristics of the velocity-time graph based on the equation $v(t) = V_T (1 - e^{-\frac{k}{m}t})$:
The graph should show velocity increasing from zero, with its rate of increase (slope) decreasing over time, eventually leveling off at the terminal velocity $V_T$. This describes an exponential curve rising from the origin.
Comparing these characteristics with the given options:
Therefore, the graph representing velocity versus time is the one showing exponential increase towards a terminal velocity.
Two circular discs of radius each $10 \text{ cm}$ are joined at their centres by a rod of length $30 \text{ cm}$ and mass $600 \text{ gm}$ as shown in figure.
If the mass of each disc is $600 \text{ gm}$ and applied torque between two discs is $43 \times 10^5 \text{ dyne.cm}$, the angular acceleration of the discs about the given axis $AB$ is________$\text{rad/s}^2$.

Net gravitational force at the center of a square is found to be $F_1$ when four particles having mass $M, 2M, 3M$ and $4M$ are placed at the four corners of the square as shown in figure and it is $F_2$ when the positions of $3M$ and $4M$ are interchanged. The ratio $\frac{F_1}{F_2}$ is $\frac{\alpha}{\sqrt{5}}$. The value of $\alpha$ is _________.

| List-I | List-II |
| A. Spring constant | I. $[\text{M L}^2 \text{ T}^{-2} \text{ K}^{-1}]$ |
| B. Thermal conductivity | II. $[\text{M L}^0 \text{ T}^{-2}]$ |
| C. Boltzmann constant | III. $[\text{M L}^2 \text{ T}^{-3} \text{ A}^{-2}]$ |
| D. Inductive reactance | IV. $[\text{M L T}^{-3} \text{ K}^{-1}]$ |
Two circular discs of radius each $10 \text{ cm}$ are joined at their centres by a rod of length $30 \text{ cm}$ and mass $600 \text{ gm}$ as shown in figure.
If the mass of each disc is $600 \text{ gm}$ and applied torque between two discs is $43 \times 10^5 \text{ dyne.cm}$, the angular acceleration of the discs about the given axis $AB$ is________$\text{rad/s}^2$.
