Measurements and Experimentation — Question 14

(i) As we know,

Pitch = distance moved ahead in 1 revolution

and given,

Distance covered in one revolutions = 1 mm

Hence, Pitch of the screw gauge = 1 mm

(ii) As we know,

$\text{Least count} = \dfrac {\text{Pitch}}{\text{Total no. of div on circular head}}\\[0.5em]$

Here,

Pitch = 1 mm

Number of divisions on circular head = 50

Substituting the values in the formula above we get,

$\text {Least count} = \dfrac{1}{50} \\[0.5em] \text {Least count} = 0.02 \text { mm} \\[0.5em]$

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

(iii) As we know,

Diameter of the wire = main scale reading + circular scale reading     [Equation 1]

and

Circular scale reading = p x L.C.     [Equation 2]

p = 47

L.C. = 0.02 mm

Substituting the values in the Equation 2 we get,

Circular scale reading = 47 x 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.