Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Sup

Question

Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its 10 Vernier Scale Divisions (V.S.D.) are equal to its 9 Main Scale Divisions (M.S.D.). The least division in the M.S. is 0.1 cm and the zero of V.S. is at $x=0.1 \mathrm{~cm}$ when the jaws of Vernier callipers are closed. If the main scale reading for the diameter is $M=5 \mathrm{~cm}$ and the number of coinciding vernier division is 8 , the measured diameter after zero error correction, is

Professor PC Thomas Classes · Answer

(a) : $10 \mathrm{VSD}=9 \mathrm{MSD}$ \ $ \begin{aligned} & 	herefore \quad 1 \mathrm{VSD}=\frac{9}{10} 	imes \mathrm{MSD} \ & 	ext { or } \quad 1 \mathrm{VSD}=\frac{9}{10} 	imes 0.1=0.09 \mathrm{~cm} \ & \begin{aligned} 	ext { LC } & =1 \mathrm{MSD}-1 \mathrm{VSD} \ & =0.1 \mathrm{~cm}-0.09 \mathrm{~cm}=0.01 \mathrm{~cm} \end{aligned} \end{aligned} $ \ $ \begin{aligned} 	ext { Observed reading } & =\mathrm{MSR}+(\mathrm{VSD} 	imes \mathrm{LC}) \ & =5+(8 	imes 0.01)=5.08 \mathrm{~cm} \end{aligned} \ $ Corrected reading $=$ Observed reading - zero error \ $ =5.08-0.1=4.98 \mathrm{~cm} $