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  • A laboratory experiment involves measuring the thickness of a thin plastic sheet using a screw gauge. The main scale reading is 1.5 mm, and the circular scale is divided into 50 divisions. The pitc

A laboratory experiment involves measuring the thickness of a thin plastic sheet using a screw gauge. The main scale reading is 1.5 mm, and the circular scale is divided into 50 divisions. The pitch of the screw gauge (p) is 0.2 mm. The screw gauge has a negative zero error of −0.01 mm.

Calculate the actual thickness of the plastic sheet.

Option: 1

11.51mm


Option: 2

27.46mm


Option: 3

200 mm


Option: 4

20 mm


Answers (1)

best_answer

Given: Main scale reading = 1.5 mm

Number of circular scale divisions = 50

Pitch of screw gauge (p) = 0.2 mm

Negative zero error (Z.E.) = −0.01 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 = 10 mm + 0.01 mm = 10.01 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 = 1.5 mm + 10.01 mm = 11.51 mm

Therefore, the actual thickness of the plastic sheet is 11.51 mm.

Therefore, the correct option is 1.

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Nehul

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