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  • A precision laboratory experiment requires measuring the thickness of a very thin film using a screw gauge. The main scale reading is 0.8 mm, and the circular scale is divided into 100 divisions. T

A precision laboratory experiment requires measuring the thickness of a very thin film using a screw gauge. The main scale reading is 0.8 mm, and the circular scale is divided into 100 divisions. The pitch of the screw gauge (p) is 0.125 mm. The screw gauge has a negative zero error of −0.03 mm.

Calculate the actual thickness of the thin film.

Option: 1

160mm


Option: 2

27.46mm


Option: 3

13.33 mm


Option: 4

20 mm


Answers (1)

Given: Main scale reading = 0.8 mm

Number of circular scale divisions = 100

Pitch of screw gauge (p) = 0.125 mm

Negative zero error (Z.E.) = −0.03 mm

The actual reading on the circular scale (C.S.R.) is given by:

\text { C.S.R. }=\frac{\text { Number of circular scale divisions }}{\text { Number of main scale divisions }} \times \text { Pitch }Plugging in the values:

\text { C.S.R. }=\frac{10}{1} \times 0.5 \mathrm{~mm}=5 \mathrm{~mm}

The corrected reading is the sum of the circular scale reading and the absolute value of the zero error:

Corrected reading = C.S.R. + |Z.E.|

Plugging in the values:

Corrected reading = 12.5 mm + 0.03 mm = 12.53 mm

The total reading is the sum of the main scale reading and the corrected reading:

Total reading = Main scale reading + Corrected reading Plugging in the values:

Total reading = 0.8 mm + 12.53 mm = 13.33 mm

Therefore, the actual thickness of the thin film is 13.33 mm. Therefore, the correct option is 3.

Posted by

Kshitij

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