The circular division 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

(D)  1.25 mm

Concept Videos :-

#7 | Vernier Caliper
#8 | Vernier Caliper: Solved Examples 1
#9 | Principle of Vernier Caliper
#10 | Screw Gauge
#11 | Screw Gauge: Solved Examples

Concept Questions :-

Measuring devices

1st figure gives information about zero error as there is nothing in between studs. Zero error is postive as zero of Circular scale is below reference line.

Zero error = 0.01 mm X 5 = 0.05 mm

                                        = + 0.05 mm 

 

L.C. =0.5 mm50 = 0.01 mm

 1 MSD = Pitch = 0.5 mm                                    [Pitch is generally equal to 1 MSD]

 Main scale reading (MSR) = 2 divisions = 2 X 0.5 mm = 1 mm

Circular scale reading (CSR) = 0.01 mm X 25 = 0.25 mm

Observed Reading = 1mm + 0.25 mm

              = 1.25 mm

Correct reading = 1.25 mm - (0.05 mm )

                          = 1.20 mm

Video Solution:

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