The circular divisions of shown screw gauge are 50. It moves 0.5 mm on main scale in one rotation. The diameter of the ball is

- (A)2.25 mm
- (B)2.20 mm
- (C)1.20 mm✓ Correct
- (D)1.25 mm
Answer: (C)
Solution
Least count of the screw gauge = mm
From the figure, the zero error = mm (positive zero error).
Reading from the screw gauge:
Actual measurement = Observed reading - Zero error
= mm
Therefore, the answer is (C).
From the figure, the zero error = mm (positive zero error).
Reading from the screw gauge:
- Main scale reading (MSR) = mm
- Circular scale reading (CSR) = mm
- Observed reading = MSR + CSR = mm
Actual measurement = Observed reading - Zero error
= mm
Therefore, the answer is (C).
JEE Advanced · 2006 · 09 Apr 2006 · Single correct option