A) 3.32 mm
B) 3.73 mm
C) 3.67 mm
D) 3.38 mm
Correct Answer: D
Solution :
[d]: Pitch of screw gauge \[=\frac{1}{2}mm=0.5mm\] Least count of the screw gauge \[=\frac{0.5}{50}mm=0.01mm\] Zero error = -0.03 mm Zero correction = + 0.03 mm Main scale reading = 3 mm Circular scale reading = 35 Observed diameter of the wire \[=3\text{ }mm+35\times \left( 0.01 \right)mm=3.35\text{ }mm\] Corrected diameter of the wire = (3.35 + 0.03) mm = 3.38 mmYou need to login to perform this action.
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