Screw Gauge: Practice Problem & Solution
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 zero of circular scale division coincides with \(25 \text{divisions}\) above the reference level. If screw gauge has a zero error of \(-0.004 \text{cm}\), the correct diameter of the ball is : [NEET 2018]
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Screw Gauge. Understanding the underlying formula is key to arriving at the correct answer below:
Least Count (LC) = \(0.001 \text{cm}\). Main Scale Reading (MSR) = \(5 \text{mm} = 0.5 \text{cm}\). Circular Scale Reading (CSR) = \(25 \text{divisions}\). Observed Reading = MSR + (CSR \(\times\) LC) = \(0.5 \text{cm} + (25 \times 0.001 \text{cm}) = 0.525 \text{cm}\). Correct Reading = Observed Reading - Zero Error = \(0.525 \text{cm} - (-0.004 \text{cm}) = 0.525 \text{cm} + 0.004 \text{cm} = 0.529 \text{cm}\).
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