Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows th

Question

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

Professor PC Thomas Classes · Answer

(c) : Main scale division (MSD) $=7$ \Vernier scale division (VSD) $=10$ 
$ \begin{aligned} & 7 \mathrm{MSD}=10 \mathrm{VSD} \Rightarrow 1 \mathrm{VSD}=\frac{7}{10} \mathrm{MSD} \ & 1 \mathrm{MSD}=\frac{1}{10} \mathrm{~cm}=1 \mathrm{~mm} \end{aligned}$
L.C. $=1 \mathrm{MSD}-1 \mathrm{VSD}$ 
$ =\left(1-\frac{7}{10}ight) \mathrm{MSD}=\frac{3}{10} \mathrm{MSD}=0.3 \mathrm{~mm} $
$ \begin{aligned} 	ext { Reading } & =\mathrm{MSR}+\mathrm{L} . \mathrm{C} . 	imes \mathrm{VSD} 	ext { (matching) } \ & =1 \mathrm{~mm}+1 	imes 0.3 \mathrm{~mm}=1.3 \mathrm{~mm} \ & =0.13 \mathrm{~cm} \end{aligned} $