Question 4

Step 1: Determine the Least Count (LC) from Figure 1 From the first figure, we can see the main scale is calibrated such that 10 divisions make up 1 cm. Therefore, 1 Main Scale Division ($1\text{ MSD}$) = $\frac{1}{10}\text{ cm} = 0.1\text{ cm}$.

Looking at the alignment, 10 Vernier Scale Divisions (VSD) exactly coincide with 7 Main Scale Divisions (MSD). $10\text{ VSD} = 7\text{ MSD}$ $1\text{ VSD} = \frac{7}{10}\text{ MSD} = 0.7 \times 0.1\text{ cm} = 0.07\text{ cm}$.

The Least Count (LC) of the Vernier caliper is given by: $\text{LC} = 1\text{ MSD} - 1\text{ VSD}$ $\text{LC} = 0.1\text{ cm} - 0.07\text{ cm} = 0.03\text{ cm}$.

(Note: Figure 1 also shows that the zero of the Vernier scale aligns perfectly with the zero of the main scale, meaning there is no zero error).

Step 2: Read the measurement from Figure 2 In the second figure, the zero mark of the Vernier scale has crossed the first mark of the main scale ($0.1\text{ cm}$). So, the Main Scale Reading ($\text{MSR}$) = $0.1\text{ cm}$.

Next, we look for the Vernier Scale Coinciding Division ($\text{VCD}$). The $1^{\text{st}}$ division of the Vernier scale perfectly aligns with a division on the main scale. So, $\text{VCD} = 1$.

Step 3: Calculate the Final Measured Value (D) The formula for the total reading is: $\text{Reading} = \text{MSR} + (\text{VCD} \times \text{LC})$ $D = 0.1\text{ cm} + (1 \times 0.03\text{ cm})$ $D = 0.1\text{ cm} + 0.03\text{ cm} = 0.13\text{ cm}$.

Final Answer: The measured value of $D$ is $0.13\text{ cm}$.