# Compare the radii of given three wires using Screw Gauge.

Formulae :

i ) Pitch of the screw =   $\frac{Distance moved by sloped edge over the pitch scale}{Number of rotations of the screw}$.

ii ) Least count (L.C) = $\frac{Pitch of the screw}{Number of divisions on Head scale}$,

iii ) Total Reading = P.S.R +$n\times L.C$ ,

iv ) Ratio of radii of the three given wires  is $r_1$:$r_2$:$r_3$,

where $r_1$ = Average radius of first wire,

$r_2$ = Average radius of second wire,

$r_3$ = Average radius of third wire.

Draw figure

Procedure :First we have to determine the least count of the given Screw gauge.

To determine the least count of the screw gauge, the head ‘H’ is rotated through certain (say 5) number of complete rotations.The distance moved by the sloped  edge over the pitch scale is measured.

Now substitute these values in the formula of pitch of the screw = $\frac{Distance moved by sloped edge over the pitch scale}{Number of rotations of the screw}$.

Least count L.C =  $\frac{Pitch of the screw}{Number of divisions on Head scale}$ .

Now check whether the given screw gauge has any ZERO ERROR or not. To determine the ZERO ERROR, the head H is rotated until the flat end of the screw  $S_2$ touches the plane surface of the stud $S_1$ (do not apply excess pressure) i.e we have to rotate the head only by means of safety device ‘D’ only.

When $S_1$ and $S_2$ are in contact,the zero of the head scale perfectly coincides with the index line as in Fig-(a). In such case there will be no ZERO ERROR and no correction is required.

When $S_1$ and $S_2$ are in contact,the zero of the head scale is below the index line as in Fig(b), such ZERO ERROR is called positive ZERO ERROR, and the correction is negative.

When $S_1$ and $S_2$ are in contact,the zero of the head scale is above the index line as in Fig(c) , such ZERO ERROR is called negative ZERO ERROR, and the correction is positive.

Determine the radii  of the given  metal wires :The given object metal wire is held between the two parallel surfaces of fix stud $S_1$ and screw tip $S_2$. Note the completed number of divisions on pitch scale, which is called PITCH SCALE READING (P.S.R). The number of the head scale division coinciding with the index line is noted, which is called OBSERVED HEAD SCALE READING  n’. If the given screw gauge has ZERO ERROR (x) the correction is made by adding or subtracting the ZERO ERROR (x) from the  OBSERVED HEAD SCALE READING  n’.The corrected value (n’-x) or (n’+x)  is called the HEAD SCALE READING (H.S.R) n.

To calculate the fraction the H.S.R (n) is multiplied by the least count (L.C).

Diameter of first wire = Total reading = P.S.R +$n\times L.C$ –  –  –  –   –  – (1)

Changing the position of metal wire, 5 readings should be taken, and recorded in the table-1. Every time calculate the total diameter (d) of the metal wire using equation (1).

Average of the 5 diameter of the metal wire should be calculated, to get the average diameter(d) of the first metal wire.

Radius ($r_1$) of the first metal wire =$\frac{d_1}{2}$ mm.

The diameters of 2nd  and 3rd wires are also  measured following the above procedure. From diameters of 2nd  and 3rd wires we can calculate their radii $r_2$  and $r_3$ .

Calculation of least count:

Number of complete rotations of the screw = 5

Distance moved by  sloped edge  over the pitch scale = 5mm

Pitch of the screw = $\frac{Distance moved by sloped edge over the pitch scale}{Number of rotations of the screw}$ = $\frac{5mm}{5}$ =1mm.

Number of divisions on the head scale = 100

Least count (L.C) = $\frac{Pitch of the screw}{Number of divisions on Head scale}$ = $\frac{1mm}{100}$ =0.01mm

Zero Error :

When $S_1$ and $S_2$ are in contact,97 th division of head scale is coinciding with index line i.e the zero of the head scale is 3 divisions below the index line as in Fig(b), such ZERO ERROR is called positive ZERO ERROR, and the correction is negative.

The Zero correction for the given screw gauge = – 3

Table -1 ( Diameter of the 1st wire) :

 S.No Pitch Scale Reading (P.S.R) amm Observed H.S.R    (n’) Correction   (x) Corrected H.S.R  n=n’(+/-)x Fraction  b=n*L.C Total reading  (a+b) mm 1. 1 45 3 45-3=42 42*.001=0.42 1.42 2. 1 46 3 46-3=43 43*0.01=0.43 1.43 3. 1 46 3 46-3=43 43*0.01=0.43 1.43 4. 1 47 3 47-3=44 44*.001=0.44 1.44 5. 1 46 3 46-3=43 43*0.01=0.43 1.43

Average diameter of the 1st wire ($d_1$) = $\frac{1.42+1.43+1.43+1.44+1.43}{5}$ =$\frac{7.15}{5}$ mm.

Average diameter of the 1st wire ($d_1$) =    1.43 mm

Average radius of 1st wire ($r_1$) =$\frac{d_1}{2}$ = $\frac{1.43}{2}$=0.72 mm

Table – 2 (Diameter of the 2nd wire):

 S.No Pitch Scale Reading (P.S.R) amm Observed H.S.R    (n’) Correction   (x) Corrected H.S.R  n=n’(+/-)x Fraction  b=n*L.C Total reading  (a+b) mm 1. 2 12 3 12-3=9 9*0.01=0.09 2.09 2. 2 13 3 13-3=10 10*0.01=0.10 2.10 3. 2 14 3 14-3=11 11*0.01=0.11 2.11 4. 2 12 3 12-3=9 9*0.01=0.09 2.09 5. 2 14 3 14-3=11 11*0.01=0.11 2.11

Average diameter of the 1st wire ($d_2$) = $\frac{2.09+2.10+2.11+2.09+2.11}{5}$ =$\frac{10.5}{5}$ mm.

Average diameter of the 2nd wire ($d_2$) = 2.10  mm

Average radius of 2nd wire ($r_2$) =$\frac{d_2}{2}$ = $\frac{2.10}{2}$=1.05 mm

Table -3 ( Diameter of the 3rd wire) :

 S.No Pitch Scale Reading (P.S.R) amm Observed H.S.R    (n’) Correction   (x) Corrected H.S.R  n=n’(+/-)x Fraction  b=n*L.C Total reading  (a+b) mm 1. 1 85 3 85-3=82 82*0.01=0.82 1.82 2. 1 84 3 84-3=81 81*0.01=0.81 1.81 3. 1 84 3 84-3=81 81*0.01=0.81 1.81 4. 1 86 3 86-3=83 83*0.01=0.83 1.83 5. 1 86 3 86-3=83 83*0.01=0.83 1.83

Average diameter of the 1st wire ($d_3$) = $\frac{1.82+1.81+1.81+1.83+1.83}{5}$ =$\frac{9.10}{5}$ mm.

Average diameter of the 3rd wire ($d_3$) = 1.82 mm

Average radius of 1st wire ($r_3$) =$\frac{d_3}{2}$ = $\frac{1.82}{2}$=0.91 mm

Observations : i)Average radius of 1st wire ($r_1$) = 0.72 mm,

ii)Average radius of 2nd wire ($r_2$) = 1.05 mm,

iii )Average radius of 3rd wire ($r_3$) = 0.91 mm.

Precautions : i ) Pitch scale reading (P.S.R) should be taken carefully without parallax error ii ) Head scale reading (H.S.R) should be taken carefully without parallax error iii )Screw must be rotated by holding the safety device ‘D’ iv ) Do not apply excess pressure on the object held between the surfaces $S_1$ and $S_2$.

v ) The screw is rotated in only one direction either clock wise or anti-clock wise to avoid the back lash error.

Result : Ratio of radii of the given wires is $r_1$:$r_2$:$r_3$ = 0.72 : 1.05 : 0.91

## 2 thoughts on “Compare the radii of given three wires using Screw Gauge.”

1. harsimranjot singh

It’s a very good demonstration but you should also give real picture of screw gauge so that students can understand relation between the scales provided on instrument.

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