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 _________.
To find the ratio \(\frac{F_1}{F_2}\), we need to calculate the net gravitational force at the center of the square for both configurations.
Let's denote the side of the square by \(a\). The distance from the center to any corner is \(\frac{a}{\sqrt{2}}\).
After simplification, due to symmetry and equal distribution, we find:
Given, \(\frac{F_1}{F_2} = \frac{\alpha}{\sqrt{5}}\)
Comparing both we get, \(\alpha = 3\).
Thus, the correct answer is 3.
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________$\text{rad/s}^2$.

| 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________$\text{rad/s}^2$.
