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A laboratory technician needs to measure the thickness of a very thin metal plate using a screw gauge. The main scale reading is $1.5\ mm$ , and the circular scale reading is $42$ . The pitch of the screw gauge is $0.25\ mm$ , and the least count of the instrument is $0.02\ mm$ . Due to the delicate nature of the plate, it is important to apply very gentle pressure while measuring its thickness.

Determine the thickness of the metal plate.
Option: 1
$1.452\ mm$

Option: 2
$0.541\ mm$

Option: 3
$0.585\ mm$

Option: 4
$0.894\ mm$

Answers (1)

best_answer

Given values:
• Main scale reading (M) $=1.5\ mm$
• Circular scale reading (C) $=42$
• Pitch of the screw gauge (P) $=0.25\ mm$
• Least count (LC) $=0.02\ mm$
The total reading of the screw gauge can be calculated by adding the main scale
reading to the product of the circular scale reading and the least count:

Total reading $=M\times(C+LC)$
Step 1: Calculate the total reading of the screw gauge:
Total reading $=1.5 \mathrm{~mm}+(42 \times 0.02 \mathrm{~mm})=1.5 \mathrm{~mm}+0.84 \mathrm{~mm}=2.34 \mathrm{~mm}$

The thickness of the metal plate can be calculated using the formula:

Thickness = Total reading $\times$ P
Step 2: Calculate the thickness of the metal plate:

Thickness $=2.34 \mathrm{~mm} \times 0.25 \mathrm{~mm}=0.585 \mathrm{~mm}$

The thickness of the metal plate is $0.585\ mm$
Therefore, the correct option is (C).

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rishi.raj

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