Two full turns of the circular scale of a screw gauge cover a distance of $1 m m$ on its main scale. The total number of divisions on the circular scale is $50$ . Further, the screw gauge has a zero error of $- 0.03 m m$ . If main scale reading is $3 m m$ and circular scale reading is $35$ during the measurement of diameter of a wire. Then diameter of wire is $(3 + \frac{x}{100}) m m$ . The value of $x$ is_____.

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Solution

Least count of screw gauge
Reading = [main scale reading + circular scale Reading $\times L . C .)$ – (zero error)
$L . C = \frac{1}{50 \times 2} = 0.01 m m$
Diameter $= 3 + 35 \times 0.01 + 0.03$
$= 3.38 m m$
$= (3 + \frac{38}{100}) m m$

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Q. Two full turns of the circular scale of a screw gauge covers a distance of

1 m m

on its main scale. The total number of divisions on the circular scale is

50

. Further, it is found that the screw gauge has a zero error of

- 0.03 m m

. While measuring the diameter of a thin wire, a student notes the main scale reading of

3 m m

and the number of circular scale divisions in line with the main scale as

35

. The diameter of the wire is:

Q. Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale. The total number of divisions on the circular scale is 50. Further, it is found that the screw gauge has a zero error of -0.03 mm while measuring the diameter of a thin wire, a student notes the main scale reading of 3 mm and the number of circular scale divisions in line with the main scale as 35. The diameter of the wire is. [AIEEE-2008]

Two full turns of the circular scale of a screw gauge cover a distance of $1 m m$ on its main scale. The total number of divisions of $1 m m$ on its main scale. The total number of divisions on the circular scale is $50$ . Further, it is found that the screw guage has a zero error of $- 0.03 m m$ . While measuring the diameter of a thin wire, a student notes the main scale reading of $3 m m$ and the number of circular scale divisions in line with the main scale as $35$ . The diameter of the wire is

The pitch of a screw gauge is 1 mm and there are 100 division on the circular scale. A students measures a diameter of a wire using this screw gauge and he gets main scale reading as 5 mm and circular divisions as 25. If the screw gauge has positive zero error $0.03 m m$ , the correct value of diameter will be

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Least count of screw gauge Reading = [main scale reading + circular scale Reading ×L.C.) – (zero error) L.C=150×2=0.01mm Diameter =3+35×0.01+0.03 =3.38 mm =(3+38100)mm

Least count of screw gauge
Reading = [main scale reading + circular scale Reading $\times L . C .)$ – (zero error)
$L . C = \frac{1}{50 \times 2} = 0.01 m m$
Diameter $= 3 + 35 \times 0.01 + 0.03$
$= 3.38 m m$
$= (3 + \frac{38}{100}) m m$