Question

The circular scale of a screw gauge has 50 divisions and pitch of 0.5 mm. Find the diameter of the sphere. Main scale reading is 2.
A. 1.2 mm
B. 1.25 mm
C. 2.20 mm
D. 2.25 mm

Answer 1

From the pitch and number of divisions on the scale we have to find the least count of the screw gauge. We have to find the main scale reading and the circular scale reading and then add those values of those two readings to get the final reading.

Complete step-by-step answer:
We know that, the pitch of the circular scale is,
P = 0.5 mm(given)
We also know that, the number of divisions on circular scale is,
N=50

Therefore the least count of screw gauge is,
L.C = P/N ​=0.5/50​ mm
L.C = 0.01 mm

Now, we know that the main scale reading is given as,
M.S.R =2 mm
On the circular scale the marking is,
n=25

Thus, diameter of sphere
d=M.S.R + n×L.C
⟹ d=2 mm+25×0.01 mm
d=2.25 mm

Therefore option D is the correct option.

Additional Information:
Least count is the smallest amount that can be calculated by a scale.
A screw gauge is a device that is used to measure the length of a circular or irregular object.
Pitch is the measurement increased when the circular gauge completes one revolution.
Divisions are the markings on a scale used to recognize the difference between two distances.

Note: Main scale and circular scale is different.
Students forget to do the addition of the main scale and circular scale.
Least count recognition is necessary.