To measure the diameter of a sphere using a screw gauge

[Amitkumarr]

Homework Statement

A screw gauge has pitch 1.5 mm and there is no zero error. Linear scale has marking at MSD = 1 mm 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 3 mm mark is not visible, 76th division of circular scale is in line with linear scale. What is the diameter of sphere?

Relevant Equations

Diameter=M.S.R + L.C×C.S.R
(there is no zero error)
Where M.S.R=Main scale reading
L.C=Least count of screw gauge
C.S.R=Circular scale reading
Least count of screw gauge= Pitch÷No. of divisions on circular scale

Least count of the screw gauge = Pitch÷No. of divisions on circular scale=1.5÷100 mm =0.015mm
According to me,in this case the main scale reading should be taken as 2 mm because it is the one which is visible and circular scale reading should be 76.
So, Diameter=2 mm + 0.015×76 mm
= 2 mm + 1.14 mm= 3.14 mm
But none of the given options matches my answer and the correct answer is given as 2.64 mm which is option A(in the attached question).
According to the solution manual, the Main scale reading should be 1× pitch=1.5 mm and the diameter should be 1.5 mm + (1.5÷100)×76 mm = 2.64 mm ,but why?
Where am I going wrong?

 

[TSny]

Suppose you start with the gauge reading zero; i.e., both the main scale and the circular scale read zero. After one full rotation of the screw, where will you be on the main scale?

Suppose the scale is measuring a distance of exactly 2 mm. Would the circular scale read zero?

 

[Amitkumarr]

Suppose you start with the gauge reading zero; i.e., both the main scale and the circular scale read zero. After one full rotation of the screw, where will you be on the main scale?

Suppose the scale is measuring a distance of exactly 2 mm. Would the circular scale read zero?

When we start with the gauge reading zero,after one full rotation,we should be at a distance equal to 1.5 mm(equal to pitch) on the main scale.If the scale is measuring a distance of exactly 2mm,the circular scale would not read zero.

But,how we have to use the information that the 2mm mark is visible but the 3mm mark is not visible, in getting the main scale reading?

 

[TSny]

When we start with the gauge reading zero,after one full rotation,we should be at a distance equal to 1.5 mm(equal to pitch) on the main scale.If the scale is measuring a distance of exactly 2mm,the circular scale would not read zero.

Right. The circular scale does not read zero when the main scale is exactly 2 mm.

But, your solution appears to assume that the circular scale does read zero when the main scale is exactly 2 mm. You wrote: Diameter=2 mm + 0.015×76 mm. So, you assumed that the additional amount beyond 2 mm equals the distance associated with 76 tick marks on the circular dial.

But, how we have to use the information that the 2mm mark is visible but the 3mm mark is not visible, in getting the main scale reading?

This tells you that the diameter of the sphere lies between 2 mm and 3 mm.

 

[Amitkumarr]

Right. The circular scale does not read zero when the main scale is exactly 2 mm.

But, your solution appears to assume that the circular scale does read zero when the main scale is exactly 2 mm. You wrote: Diameter=2 mm + 0.015×76 mm. So, you assumed that the additional amount beyond 2 mm equals the distance associated with 76 tick marks on the circular dial.

This tells you that the diameter of the sphere lies between 2 mm and 3 mm.

Thanks a lot.I can now understand.