Class 11 NEET Physics Physical World, Units and Measurements

$\text{Least count} = 1\ \text{MSD} - 1\ \text{VSD}$$= 1\ \text{MSD} - \frac{9}{10}\ \text{MSD} = \frac{1}{10}\ \text{MSD}$$= \frac{1}{10} \times 0.1\ \text{cm} = 0.01\ \text{cm}$When the jaws are closed, the zero of the V.S. lies at $x = 0.1\ \text{cm}$, i.e. it is shifted $0.1\ \text{cm}$ to the right of the main scale zero, so the instrument reads $+0.1\ \text{cm}$ with nothing between the jaws.Hence the zero error is $+0.1\ \text{cm}$ (positive), so the zero correction is $-0.1\ \text{cm}$ and must be subtracted.Observed reading $= M + (\text{coinciding V.S.D.}) \times \text{Least count}$$= 5\ \text{cm} + 8 \times 0.01\ \text{cm} = 5.08\ \text{cm}$Measured diameter after zero error correction:$= 5.08\ \text{cm} - 0.1\ \text{cm} = 4.98\ \text{cm}$Thus the correct answer is $4.98\ \text{cm}$.