Question: A student measured the diameter of a small steel ball using a screw gauge of least count $0.001\ \text{cm}$. The main scale reading is $5\ \text{mm}$ and the zero of the circular scale coincides with 25 divisions above the reference level. If the screw gauge has a zero error of $-0.004\ \text{cm}$, the correct diameter of the ball is: [PMT/NEET-2018]
a. $0.521\ \text{cm}$
b. $0.525\ \text{cm}$
c. $0.053\ \text{cm}$
d. $0.529\ \text{cm}$

Answer. d. $0.529\ \text{cm}$

Solution. $= MSR + CSR \times (\text{Least count}) – \text{Zero error}$
$= 5\ \text{mm} + 25 \times 0.001\ \text{cm} – (0.004)\ \text{cm}$
$= 0.5\ \text{cm} + 25 \times 0.001\ \text{cm} – (0.004)\ \text{cm} = 0.529\ \text{cm}.$