Step 1: Relationship between main scale and vernier scale Given that:
\[ n \, \text{MSD} = (n + 1) \, \text{VSD}. \]
From this:
\[ 1 \, \text{VSD} = \frac{n}{n + 1} \, \text{MSD}. \]
Step 2: Least count formula The least count (L.C.) of a vernier caliper is given by:
\[ \text{L.C.} = 1 \, \text{MSD} - 1 \, \text{VSD}. \]
Substitute \( 1 \, \text{VSD} \) from Step 1:
\[ \text{L.C.} = m - m \left( \frac{n}{n + 1} \right). \]
Simplify:
\[ \text{L.C.} = m \left[ 1 - \frac{n}{n + 1} \right]. \]
\[ \text{L.C.} = m \left( \frac{n + 1 - n}{n + 1} \right). \]
\[ \text{L.C.} = \frac{m}{n + 1}. \]
Final Answer: \( \frac{m}{n + 1} \).
Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows the configuration corresponding to the measurement of diameter $ D $ of a tube. The measured value of $ D $ is:
The velocity-time graph of an object moving along a straight line is shown in the figure. What is the distance covered by the object between \( t = 0 \) to \( t = 4s \)?
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $