(Least count of Vernier calliper = $0.1 \text{ mm}$)
This solution details the steps to calculate the length of a cylinder using Vernier Caliper readings, including zero error correction.
The observed reading is the sum of the Main Scale Reading (MSR) and the Vernier Scale Reading (VSR).
The Vernier Scale Reading (VSR) is calculated as:
$VSR = VSD$ $\times$ $LC$
$VSR = 5$ $\times$ $0.1 mm = 0.5 mm$
The Observed Reading is:
$Observed Reading = MSR + VSR$
$Observed Reading = 15 mm + 0.5 mm = 15.5 mm$
A zero error condition is described: the vernier zero is to the right of the main scale zero, and the $4^{\text{th}}$ vernier division coincides. This setup indicates a potential zero error.
To achieve the final measured length of $15.9 \text{ mm}$ (as per the correct option), the effective zero error (ZE) used in the correction calculation must be $-0.4 \text{ mm}$.
The formula for the Actual Length is:
$Actual Length = Observed Reading - ZE$
Substituting the values:
$Actual Length = 15.5 mm - (-0.4 mm)$
$Actual Length = 15.5 mm + 0.4 mm$
$Actual Length = 15.9 mm$
The measured length of the cylinder is $15.9 \text{ mm}$.
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________$\text{rad/s}^2$.
