# 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 +$n\times L.C$ ,

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 $mm^3$

# Find the volume of the Sphere – Vernier Calipers.

2 Q : Find the volume of the given sphere using vernier calipers.

Ans:

Formula :

1. Volume of the Sphere V =  $\frac{4}{3} \pi r^3 cm^3$,

V= volume of Sphere, r = radius of  Sphere
.

2.Least count of vernier calipers L.C = $\frac{S}{N}$ cm,

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

3.Length (or) diameter  of Cylinder = Main scale reading (a) cm + ( n*L.C ) cm.

n = vernier coincidence .

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

To determine the volume of the  Sphere we have to determine the radius  (r) of the cylinder and substituting this value in the equation for the volume of the Sphere we can calculate it.

a) To determine the diameter of the Sphere : Given Sphere is held gently between jaws 1,1 of the vernier calipers.The reading on the main scale just before the zero of the vernier is noted.This is called Main scale reading (M.S.R).The number of division (n) on the vernier which coincides perfectly with any one of the main scale divisions is noted.This is called vernier coincidence (V.C).The vernier coincidence (V.C=n) is multiplied by least count to get the fraction of a main scale division.This is added to the main scale reading (M.S.R) to total reading or total diameter of the sphere.

Total reading = M.S.R + ($V.C\times L.C$)

Take the readings,keeping the Sphere between jaws 1,1 at different positions.Post the values of M.S.R and vernier coincidence (n) in the table.Take at least 5 readings, get the average of these 5 readings which is mean diameter (d)of the Sphere.

Place the Sphere diametrically between the jaws 1,1 of the vernier calipers, post the values of M.S.R and vernier coincidence (n) in the table. Take at least 5 readings, calculate the average of these readings which gives the mean diameter ( d=2r ) of the Sphere.

c) To determine the volume of the Sphere :Substituting the value mean radius ( r) of the sphere which is already determined, in the formula V = $\frac{4}{3} \pi r^3 cm^3$,

Determine Least count of 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.

Table for  Diameter of the Sphere :

 S.No M.S.R                acm Vernier Coincidence   (n) Fraction   b=n*L.C Total Reading (a+b) cm 1. 1.9 7 0.05 1.97 2. 1.9 6 0.04 1.96 3. 1.9 6 0.06 1.96 4. 1.9 7 0.05 1.97 5. 1.9 7 0.06 1.97

Average diameter of the sphere  d = 2r =  $\frac{(1.97+1.96+1.96+1.97+1.97)}{5}$ cm = $\frac{9.83}{5}$

Average radius of the sphere r =$\frac{d}{2}$ = $\frac{1.966}{2}$cm = 0.98 cm.

Observations :

Average radius of the cylinder r = 0.98 cm.

Calculations : Volume of the sphere V = $\frac{4}{3} \pi r^3 cm^3$ = $\frac{4}{3}\times\frac{22}{7}\times(0.98)^3$ $cm^3$

=3.94 $cm^3$

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.

Result and Units : Volume of the sphere V = 3.94 $cm^3$.

# Find the volume of cylinder – Vernier calipers.

1 Q : Find the volume of the  given cylinder  using vernier calipers.

Ans :

Formula :

1. Volume of the cylinder V = $\pi r^2 l$ $cm^3$,

V= volume of cylinder, r = radius of cylinder  l = length of cylinder.

2.Least count of vernier calipers L.C = $\frac{S}{N}$ cm,

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

3.Length (or) diameter  of Cylinder = Main scale reading (a) cm + ( n*L.C ) cm.

n = vernier coincidence .

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

To determine the volume of the  cylinder we have to determine a)the length of the cylinder and b) radius of the cylinder and substituting these values in the equation for the volume of the cylinder we can calculate it.

a) To determine the length of the cylinder : Given cylinder is held gently between jaws 1,1 of the vernier calipers.The reading on the main scale just before the zero of the vernier is noted.This is called Main scale reading (M.S.R).The number of division (n) on the vernier which coincides perfectly with any one of the main scale divisions is noted.This is called vernier coincidence (V.C).The vernier coincidence (V.C=n) is multiplied by least count to get the fraction of a main scale division.This is added to the main scale reading (M.S.R) to total reading or total length of the cylinder.

Total reading = M.S.R + ($V.C\times L.C$)

Take the readings,keeping the cylinders between jaws 1,1 at different positions.Post the values of M.S.R and vernier coincidence (n) in the table.Take at least 5 readings, get the average of these 5 readings which is mean length(l )of the cylinder.

b)To determine the diameter of the cylinder : Place the cylinder diametrically between the jaws 1,1 of the vernier calipers, as in the above case post the values of M.S.R and vernier coincidence (n) in the table. Take at least 5 readings, calculate the average of these readings which gives the mean diameter ( d=2r ) of the cylinder.

c) To determine the volume of the cylinder :Substituting the values of mean length (l ) of the cylinder and mean diameter ( r) of the cylinder which is already determined, in the formula V = $\pi r^2 l cm^3$.

Determine Least count of 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.

Table – Length of the cylinder :

 S.No M.S.R                acm Vernier Coincidence   (n) Fraction   b=n*L.C Total Reading (a+b) cm 1. 2.6 9 0.09 2.69 2. 2.7 1 0.01 2.71 3. 2.7 2 0.02 2.72 4. 2.7 2 0.02 2.72 5. 2.6 8 0.08 2.68

Average length of the cylinder l = $\frac{2.69+2.71+2.72+2.72+2.68}{5}$ cm=$\frac{13.52}{5}$ = 2.70 cm

b) Diameter of the cylinder :

 S.No M.S.R                acm Vernier Coincidence   (n) Fraction   b=n*L.C Total Reading (a+b) cm 1. 1.4 5 0.05 1.45 2. 1.4 4 0.04 1.44 3. 1.4 6 0.06 1.46 4. 1.4 5 0.05 1.45 5. 1.4 6 0.06 1.46

Average diameter of the cylinder  d = 2r = $\frac{1.45+1.44+1.46+1.45+1.46}{5}$ cm=$\frac{7.26}{5}$ = 1.45 cm,

Average radius of the cylinder r =$\frac{d}{2}$ =$\frac{1.45}{2}$ cm = 0.73 cm.

Observations :

Average length of the cylinder l = 2.70 cm,

Average radius of the cylinder r = 0.73 cm.

Calculations : Volume of the cylinder V = $\pi r^2 l$ $cm^3$ = $\frac{22}{7}\times(0.73)^2\times2.70$ $cm^3$ = 4.52 $cm^3$

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.

Result and Units : Volume of the cylinder V = 4.52 $cm^3$.