In a screw gauge, when the two studs are in contact, the main-scale reference line coincides with the 5th division of the circular scale. The circular scale has 100 divisions and the pitch is 0.1 mm. While measuring the diameter of a sphere, the main-scale reading is 5 mm and the 50th circular-scale division coincides with the reference line. What is the corrected diameter of the sphere?
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Question

In a screw gauge, when the two studs are in contact, the main-scale reference line coincides with the 5th division of the circular scale. The circular scale has 100 divisions and the pitch is 0.1 mm. While measuring the diameter of a sphere, the main-scale reading is 5 mm and the 50th circular-scale division coincides with the reference line. What is the corrected diameter of the sphere?

Options

G1 Physics · Accepted Answer

The least count is LC = pitch ÷ number of circular-scale divisions = 0.1 mm ÷ 100 = 0.001 mm. When the studs are in contact, the instrument reads zero error = 5×0.001 = +0.005 mm. Therefore the zero correction is -0.005 mm. The observed reading for the sphere is reading(observed) = main-scale reading + circular-scale reading = 5 + 50×0.001 = 5.050 mm. Hence the corrected diameter is diameter = 5.050-0.005 = 5.045 mm. Correct option: A.