Question

10 rotations of the cap of screw gauge is equivalent to 5mm. The cap has 100 divisions. Find the least count. A reading taken for the diameter of wire with the screw gauge shoes four complete rotations and 35 divisions on the circular scale. Find the diameter of the wire.

Answer 1

The pitch of the screw is the distance moved by the spindle per revolution. To find this, the distance advanced by the head scale over the pitch scale for a definite number of complete rotations of the screw is determined. And the Least count is the distance moved by the tip of the screw, when the screw is turned through 1 division of the head scale.

Complete step-by-step solution:
Following these steps, we can calculate the diameter of wire:
->Fix the given wire between the stud and spindle by turning the ratchet in the clockwise direction.
->Note the main scale reading (M.S.R) (up to complete division visible on the horizontal line just beyond the circular scale.
->Note the circular scale division says ‘n’ which is exactly in front of the index line or reference line.
->Multiply ‘n’ with the least count (L.C) to get a fraction of the circular scale reading (C.S.R) to be added into the main scale reading (M.S.R).
->The diameter of the wire will be the diameter of wire= (MSR+ C.S.R) + zero correction.
->Repeat the experiment at least three times by gipping the wire at different places to obtain its average diameter.
First, we have to find the least count.
10 rotations are equal to 5mm
Therefore, 1 rotation will be equal to $\dfrac{5}{{10}}mm$
$LC = \dfrac{{reading\\_in\\_one\\_rotation}}{{total\\_divisions}}$
$LC = \dfrac{{0.5}}{{100}} = 0..005 mm$
We have rotated screw gauge’s shoe 4 times completely,
On one rotation we have got 0.5mm
Thus on 4 rotations it will : $4 \times 0.5mm = 2mm$ (M.S.R.)
Now after 4 rotations its M.S.R. is matching with ${35^{th}}$ division so circular scale reading(C.S.R.) will be : $LC \times 35 = 0.005 \times 35 = 0.175mm$
Now the diameter of the wire will be = M.S.R + C.S.R. + zero correction.
Since there is no zero correction in our case we’ll skip that part.
So, the diameter will be: M.S.R + C.S.R
Diameter = $(4 \times 0.5 + 35 \times 0.005)mm$

Note:-
Certain terms related to screw gauge:
Pitch: The distance between two nearest threads along the axis of the screw is known as the pitch of the screw.
Positive Zero Error: If the zero of the circular scale lies above the reference line, provided that the fixed and movable studs are in contact.
Range: If the zero of the circular scale lies above the reference line, provided that the fixed and movable studs are in contact.