# Find the volume of the given rectangular glass plate using vernier calipers and screw gauge.

Formulae:

i) Least count of vernier calipers (L.C) = $\frac{S}{N}$ mm,

S = value of 1 Main scale division , N = Number of vernier divisions.

ii ) Total reading  = Main scale reading (a) mm + ( n*L.C ) mm

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

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

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

P.S.R = Pitch scale reading , n= Corrected Head scale reading , L.C = Least count

vi) Volume of the glass plate V = $l\times{b}\times{h}$ $mm^3$

l = length of the glass plate, b = breadth of glass plate , h = Thickness of glass plate.

Procedure :First we have to determine the least count count of the given vernier calipers.

From the given vernier calipers

S= Length of Main scale division = 1 mm = 0.1 cm,

N = Number of vernier scale divisions = 10 ,

Substitute these values in the formula of Least count L.C = $\frac{S}{N}$ = $\frac{0.1}{10}$ =0.01 cm.

Draw neat diagram of Vernier calipers

Part I : To determine the length ( l )and breadth (b) of the given glass plate with vernier calipers :The given glass plate is held between two jaws of vernier calipers, first to measure its length.Note down the values of the Main scale reading (M.S.R ) and vernier coincidence (VC) in Table-I, take 3set of readings by placing the glass plate in 3 different positions.Each time calculate the total reading by substituting the values of M.S.R and VC in the formula Total reading = M.S.R + ($VC\times L.C$.

Find the average of 3readings and calculate Average Length ( l )of the given glass plate.

Now hold the glass plate between jaws of vernier calipers breadth wise ,repeat the experiment as above , note down the 3 set of readings of M.S.R and VC in Table-II.Calculate average breadth (b) of the glass plate

Part II: To determine thickness(h) of glass plate using Screw gauge:First we have to determine the least count of the given Screw gauge.

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

Draw neat diagram of Screw Gauge

Zero Error :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.

When $S_1$ and $S_2$ are in contact,98 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 = – 2

The given glass plate 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 the glass plate, 3 readings should be taken, and recorded in the table-III. Every time calculate the total thickness (h)of glass plate using equation (1).

Calculate average of 3readings which is average thickness (h) of glass plate.

Table-I: Length (l) of the glass plate :

 S.No M.S.R a cm Vernier Coincidence   (n) Fraction   b=n*L.C Total Reading         (a+b) cm 1 2.5 8 0.01*8=0.08 2.58 2 2.5 9 0.01*9=0.09 2.59 3 2.5 7 0.01*7=0.07 2.57

Average length of glass plate (l) = $\frac{2.58+2.59+2.57}{3}$ = $\frac{7.74}{3}$ = 2.58 cm

Average length of glass plate (l) = 2.58 cm or 25.8mm

Table-II: Breadth (b)of the glass plate :

 S.No M.S.R                   a cm Vernier Coincidence   (n) Fraction   b=n*L.C Total Reading               (a+b) cm 1 1.1 4 0.01*4=0.04 1.14 2 1.1 5 0.01*5=0.05 1.15 3 1.1 5 0.01*5=0.05 1.15

Average Breadth of glass plate (b) = $\frac{1.14+1.15+1.15}{3}$ = $\frac{3.44}{3}$ = 1.15 cm.

Average Breadth of glass plate (b) = 1.15 cm or 11.5 mm.

Table-III: Thickness  (h)of the glass plate :

 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 75 2 75-2=73 73*0.01=0.73 2.73 2 2 74 2 74-2=72 72*0.01=0.72 2.72 3 2 76 2 76-2=74 74*0.01=0.74 2.74

Average Thickness (h) of glass plate (b) = $\frac{2.73+2.72+2.74}{3}$ = $\frac{8.19}{3}$ = 2.73 mm.

Average Thickness  of glass plate (h) = 2.73 mm.

Observations :

i)Average length of glass plate (l) = 2.58 cm or 25.8mm

ii)Average Breadth of glass plate (b) = 1.15 cm or 11.5 mm.

iii)Average Thickness  of glass plate (h) = 2.73 mm.

Calculations : Volume of the given glass plate V = $l\times{b}\times{h}$ $mm^3$

Volume of the given glass plate V = $25.8\times11.5\times2.73$ $mm^3$ =809.99 $mm^3$

Precautions :

1) Take the M.S.R  and vernier coincide every time without parallax error.

2)Record all the reading in same system preferably in C.G.S system.

3) Do not apply excess pressure on the body held between the jaws.

4) Check for the ZERO error.When the two jaws of the vernier are in contact,if the zero division of the main scale coincides with the zero of the vernier scale no ZERO error will be there.If not ZERO error will be there, apply correction.

5) Pitch scale reading (P.S.R) should be taken carefully without parallax error

6) Head scale reading (H.S.R) should be taken carefully without parallax error

7)Screw must be rotated by holding the safety device ‘D’

8 ) Do not apply excess pressure on the object held between the surfaces $S_1$ and $S_2$.

Result : Volume of the given glass plate is V= 809.99 $mm^3$

# 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 +$n\times L.C$ ,

P.S.R = Pitch scale reading , n= Corrected Head scale reading , L.C = Least count

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

# Screw Gauge

Experiment No :2

Aim: To measure the i)Thickness of the Glass Plate ii) Diameter of the metal wire iii) Volume of the given Glass Plate.

Apparatus : Screw Gauge , Glass Plate and Metal wire .

Description : Screw Gauge consists of  U shaped metallic frame.To one side of this U frame a long hallow cylindrical tube with a nut inside it, the inner side of cylindrical nut contains a uniform thread cut in it.On the other side of U frame a fixed stud $S_1$ with a plane face is attached.

A screw $S_2$ is fitted in the cylindrical nut.One side of the screw $S_2$ has a plane face similar to that of stud $S_1$. The faces of $S_1$ and $S_2$ are plane and parallel to one another. The other end of the screw $S_2$ carries a milled head ‘H’ attached to a cap ‘C’ with a sloping edge. When the head H is rotated, the screw moves ”to and fro” in the nut.The milled head H is provided with a safety device ‘D’ to rotate the head H.When the object is held between the stud $S_1$ and  screw $S_2$  and the head H is rotated using the safety device (D), it produces crackling sound when optimum pressure is applied on the object.

The outer surface of long cylindrical nut consists of a thick horizontal line ‘P’ parallel to the axis of cylindrical tube.This line ‘P’ is called Index line. Along the index line a scale is graduated in millimeters.This scale is called Pitch Scale.On the sloping edge of the cap ‘C’ a circular scale is graduated, which consists of 100 equal divisions, this scale is called Head scale.

Theory : The screw gauge works on the principle of screw.

When we rotate the head ‘H’ by means of safety device ‘D’ through one complete rotation, the distance moved by the screw for every complete rotation is constant. This constant distance moved by the screw for one complete rotation of head ‘ H ‘ is called Pitch of the screw.If the head scale has 100 equal divisions, then the distance moved by the screw for even 1/100 of a complete rotation can be measured accurately,this is called Least count of screw gauge.

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

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.

a) Determine the thickness of glass plate : The given object glass plate 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).

Thickness of the Glass plate = Total reading = P.S.R +$n\times L.C$ – – –  – – (1)

Changing the position of glass plate , 5 readings should be taken, and recorded in the table-1. Every time calculate the total thickness of the glass plate using equation (1).

Average of the 5 readings  of the glass plate should be calculated, to get the average thickness(t) of the given glass plate.

b) Determine the radius(r) of the given metal wire :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 the given wire = Total reading = P.S.R +$n\times L.C$

Changing the position of metal wire, 5 readings should be taken, and recorded in the table-2. 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 given metal wire.

Radius (r) of the metal wire =$\frac{d}{2}$ 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.

Observations : i ) Zero error =

ii) Zero correction =                   mm

iii ) Distance moved by the head for 5 complete revolutions =                         mm

iv ) Number of head scale divisions =

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

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

Table -1 ( Thickness of glass plate ) :

 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. 3. 4. 5.

Average thickness of the glass plate (t) =        mm

Table – 2 (Diameter of the metal 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. 3. 4. 5.

Average diameter  d =       mm

Average radius r = $\frac{d}{2}$ =         mm .

c ) Volume of Glass plate (v) : The length ( l ) , breadth ( b) are determined using vernier calipers and thickness ( t ) of the glass plate is determined using screw gauge. The values of l ,b and t are substituted in the equation of volume  V = ( l )( b )( t )    $mm^3$