Two thin wires, Wire-1 of diameter 0.650 mm and Wire-2 of unknown diameter d are given. To obtain the value of d , the diameters of the two wires are measured with a screw gauge. The screw gauge has a pitch of 0.5 mm

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Correct answer: 0

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

\text{LC} = \dfrac{\text{Pitch}}{\text{Number of divisions on CS}} = \dfrac{0.5}{100} = 0.005 \text{ mm}

For Wire-1, the measured reading is:

\text{Reading}_1 = \text{LS} + \text{CS} \times \text{LC}

\text{Reading}_1 = 0.5 + 42 \times 0.005 = 0.5 + 0.210 = 0.710 \text{ mm}

The actual diameter of Wire-1 is

0.650 \text{ mm}

. The zero error of the screw gauge is:

\text{Zero Error} = \text{Measured Reading} - \text{Actual Value} = 0.710 - 0.650 = +0.060 \text{ mm}

For Wire-2, the measured reading is:

\text{Reading}_2 = \text{LS} + \text{CS} \times \text{LC}

\text{Reading}_2 = 1.5 + 95 \times 0.005 = 1.5 + 0.475 = 1.975 \text{ mm}

The actual diameter

d

of Wire-2 is obtained by subtracting the zero error from its measured reading:

d = \text{Reading}_2 - \text{Zero Error} = 1.975 - 0.060 = 1.915 \text{ mm}

Converting the diameter into micrometers (

\mu

m):

d = 1.915 \times 1000 \text{ } \mu\text{m} = 1915 \text{ } \mu\text{m}

Answer:

1915

	Readings
	LS (mm)	CS
Wire-1	$0.5$	$42$
Wire-2	$1.5$	$95$

Two thin wires, Wire-1 of diameter 0.650 mm and Wire-2 of unknown...

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