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Question

A screwgauage has pitch 1.5 mm and there is no zero error. Linear scale has marking at MSD=1mm and there are 100 equal division of circular scale. When diameter of a sphere is measured with instrument, main scale is having 2 mm mark visible on linear scale, but 3mm mark is not visible, 76th division of circular scale is in line with linear scale. What is the diameter of sphere.

A
2.64 mm
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B
3.14 mm
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C
1.14 mm
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D
2.76 mm
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Solution

The correct option is B 3.14 mm
pitch =1.5 mm

LC
= pitch no of div on circular scale=1.5100=0.015 mm


MSR =2 mm Final reading =2+CSR×LC=2+76×0.015=2+1.140=3.14 mm



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