The problem describes a spherometer used to measure the depth of a hole in a metal piece. The pitch of the screw is 1 mm, and the circular scale is divided into 100 parts. We need to find the reading of the spherometer as shown in the figure.

1. Problem Description

The problem describes a spherometer used to measure the depth of a hole in a metal piece. The pitch of the screw is 1 mm, and the circular scale is divided into 100 parts. We need to find the reading of the spherometer as shown in the figure.

2. Solution Steps

The main scale reading (MSR) is the value on the linear scale that is just below the circular scale. From the diagram, the MSR is 4 mm.

The least count (LC) of the spherometer is given by:

$LC = \frac{Pitch}{Number\, of\, divisions\, on\, circular\, scale}$

$LC = \frac{1\, mm}{100} = 0.01\, mm$

The circular scale reading (CSR) is the division on the circular scale that coincides with the main scale line. From the diagram, the CSR is

2

0. The total reading is given by:

$Total\, Reading = MSR + (CSR \times LC)$

$Total\, Reading = 4\, mm + (20 \times 0.01\, mm)$

$Total\, Reading = 4\, mm + 0.20\, mm$

$Total\, Reading = 4.20\, mm$

3. Final Answer

4.20 mm