In a complete rotation- spindle of a screw gauge advances by 1-2mm- There are 50 divisions on circular scale- The main scale has 1-2 mm marks - is graduated to 1-2mm or has least count - 1-2 mm or has least count - 1-2 mm If a wire is put between the jaws- 3 main scale divisions are clearly visible- and 20 division of circular scale co-inside with the reference line- Find diameter of wire in correct S-F-

The correct option is A 1.70 mmDiameter of wire (3 × 1/2 mm) + (20) ( 1/2 mm/50 ) = 1.5 + 0.20 = 1.70 mm ...

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Byju's Answer

Standard XII

Physics

Vernier Caliper

In a complete...

Question

In a complete rotation, spindle of a screw gauge advances by $\frac{1}{2}$ mm. There are 50 divisions on circular scale. The main scale has $\frac{1}{2}$ mm marks $\to$ (is graduated to $\frac{1}{2}$ mm or has least count = $\frac{1}{2}$ mm or has least count = $\frac{1}{2}$ mm) If a wire is put between the jaws, 3 main scale divisions are clearly visible, and 20 division of circular scale co-inside with the reference line. Find diameter of wire in correct S.F.

1.70 mm

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1.71 mm

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1.72 mm

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1.73 mm

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Solution

The correct option is A 1.70 mm
Diameter of wire $(3 \times \frac{1}{2} m m) + (20) (\frac{1 / 2 m m}{50})$ = 1.5 + 0.20 = 1.70 mm

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The correct option is A 1.70 mmDiameter of wire (3×12mm)+(20)(1/2mm50) = 1.5 + 0.20 = 1.70 mm

The correct option is A 1.70 mm
Diameter of wire $(3 \times \frac{1}{2} m m) + (20) (\frac{1 / 2 m m}{50})$ = 1.5 + 0.20 = 1.70 mm