Flash Education

Question

The figure below shows the reading obtained while measuring the diameter of a wire with a screw gauge. The screw advances by 1 division on the main scale when the circular head is rotated once. Find — (i) pitch of the screw gauge, (ii) least count of the screw gauge, and (iii) the diameter of the wire.

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Answer

The figure below shows the reading obtained while

(i) Pitch = distance moved ahead in 1 revolution

and given,

Distance covered in one revolution = 1 mm

Hence, Pitch of the screw gauge = 1 mm

(ii) Least count = Pitch / Total no. of div on circular scale

Pitch = 1 mm

Number of divisions on circular head = 50

Substituting the values in the formula above we get,

Least count = 1/50 mm

Hence, the least count of the screw gauge = 0.02 mm

(iii) Diameter of the wire = main scale reading + circular scale reading     [Eq 1]

and

Circular scale reading = p × L.C.     [Eq 2]

p = 47

L.C. = 0.02 mm

Substituting the values in Eq 2 we get,

Circular scale reading = 47 × 0.02 = 0.94 mm

Hence, circular scale reading = 0.94 mm and main scale reading = 4 mm.

Using Equation 1 we get,

Diameter of the wire = 4 mm + 0.94 mm = 4.94 mm

Hence, the diameter of the wire is 4.94 mm.

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