Lab Practical of Vernier Caliper and Screw Gauge.

EXPERIMENT 1 (A)

OBJECTIVE:

To Measure the dimensions of the given sample using Vernier Caliper. 

APPARATUS:

Vernier Caliper, Given Sample.

DESCRIPTION:

A Vernier caliper consists of mainly two parts.

        i.            A 2cm wide 15cm long rectangular metal strip. The left end bottom side of this strip consists of fixed jaw and at the same end jaw at the top of the strip. On the strip a scale is graduated in inches along the upper edge and another scale is graduated in centimeters along the lower edge. This is called main scale.

      ii.            A metal frame V called vernier slides over the Main scale. At the bottom of this frame V a button is attached, which helps to fix this vernier at any desired place on the main scale. This vernier frame jaws at the bottom and a jaw at the top. Two scale are graduated on this frame corresponding to two scales on the main scale. The two scales on the verneir are called Verneir  scales. Verneir scales consists of equal number of divisions.

The lower jaws are used to measure the thickness or external diameter of the tubes , cylinders or spheres.

The upper jaws are used to measure the inner diameter of hollow bodies like tubes or holes. The thin strip is used to measure the depth of the objects like the tubes.

THEORY:

Principle of Vernier Caliper-N division on the Vernier scale is equal to division on the main scale.

N (V.S.D) = (N-1) M.S.D

1 V.S.D = (N-1/N) M.S.D

Least count (L.C) of Vernier calipers: Minimum length or thickness measurable with the Vernier caliper is called its least count.

Least count (L.C) = 1 M.S.D – 1 V.S.D

L.C = 1 M.S.D – (N-1/N)M.S.D

L.C = 1 M.S.D [1- N-1/N]

L.C = 1M.S.D/N = S/N

Where S is the value of one Main scale division and N is the number of equal division on the Vernier scale.

PROCEDURE:

        i.            First we have to determine the least count of the given Vernier caliper.

      ii.            To determine the volume of cylinder we have to determine

a)      The length of the cylinder

b)      Radius of 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:

Give cylinder is held gently b/w jaws of Vernier caliper. The reading on the main scale just before the zero of the Vernier is noted. This is called main scale reading. The number of division on the Vernier which coincides perfectly with any one of the main scale divisions is noted. This is called the Vernier coincidence. The Vernier coincidence (V.C= n) is multiply by least count to get the fraction of a main scale division. This is added to the main scale reading (M.S.D) to total reading or total length of the cylinder.

Total reading = M.S.R + (V.C*L.C)

Take the reading, keeping the cylinders b/w jaws at different positions. Post the values of  M.S.R & Vernier coincidence in the table.

B.     To determine the DIAMETER of the CYLINDER:

Place the cylinder diametrically b/w the jaws of vernier, as in above case post the values of M.S.R &vernier coincidence 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 the mean length of the cylinder &mean diameter of the cylinder which is already determined, in the formula V = πr²Lcm3

OBSERVATIONS:

·         Least count of the vernier caliper:

1)      Value of 1 Main scale division = 1M.S.D = S= ………cm

2)      Number of division on the vernier scale N =………….cm

Least count            L.C =S/N =……..cm

3)      Zero error = x (positive error)

·         Volume of the cylinder:

o   Length of the cylinder

S.NO.	M.S.R (cm)	Vernier Coincidence	Fraction B= (n-x)*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 = 2.70 cm

o   Diameter of the cylinder

S.NO.	M.S.R (cm)	Vernier Coincidence	Fraction B= (n-x)*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 = 2r =  1.45 cm

Mean radius of the cylinder       r = d/2

Volume of the cylinder              V = πr²Lcm³

EXPERIMENT 1 (A)

OBJECTIVE:

To Measure the dimensions of the given sample using Screw Gauge.

APPARATUS:

Screw Gauge, Given Sample.

DESCRIPTION:

The screw gauge is an instrument used for measuring accurately the diameter of a thin wire or the thickness of a sheet of metal.  It consists of a U-shaped frame fitted with a screwed spindle which is attached to a thimble.Parallel to the axis of the thimble, a scale graduated in mm is engraved. This is called pitch scale. A sleeve is attached to the head of the screw.

The head of the screw has a ratchet which avoids undue tightening of the screw. On the thimble there is a circular scale known as head scale which is divided into 50 or 100 equal parts. When the screw is worked, the sleeve moves over the pitch scale.

A stud with a plane end surface called the anvil is fixed on the ‘U’ frame exactly opposite to the tip of the screw. When the tip of the screw is in contact with the anvil, usually, the zero of the head scale coincides with the zero of the pitch scale.

Zero Error and Zero Correction

To get the correct measurement, the zero error must be taken into account. For this purpose, the screw is rotated forward till the screw just touches the anvil and the edge of cap is on the zero mark of the pitch scale. The Screw gauge is held keeping the pitch scale vertical with its zero down wards.

When this is done, anyone of the following three situations can arise:
 

1. The zero mark of the circular scale comes on the reference line. In this case, the zero error and the zero correction, both are nil.
2. The zero mark of the circular scale remains above the reference line and does not cross it.  In this case, the zero error is positive and the zero correction is negative depending on how many divisions it is above the reference line.
3. The zero mark of the head scale is below the reference line.  In this case, the zero error is negative and the zero correction is positive depending on how many divisions it is below the reference line.

 

PROCEDURE:

·         Determine the pitch and least count of the screw gauge using the equations (1) and (2) respectively..

·         Bring the anvil and screw in contact with each other and find the zero error. Do it three times and record them. If there is no zero error, then record ‘zero error nil’.

·         Move the screw away from the anvil and place the lead shot and move the screw towards the anvil using the ratchet head. Stop when the ratchet slips without moving the screw.

·         Note the number of divisions on the pitch scale that is visible and uncovered by the edge of the cap. The reading N is called the pitch scale reading(PSR)

·         Note the number (n) of the division of the circular scale lying over the reference line.

·         Repeat steps 4 and 5 after rotating the lead shot by 900 for measuring the diameter in a perpendicular direction. Record the observations in the tabular column.

·         Find total reading using the equation 3 and apply zero correction in each case.

Ø To find the thickness of the glass plate:

The glass plate is gripped between the tip of the screw and the anvil. The PSR and HSR are noted as before.

The thickness of the glass plate is;

t = PSR+ corrected HSR = N + (n x L.C)

Ø To find the diameter of the wire:

Place the wire between the anvil and the screw and note down the PSR and HSR as before.

The diameter of the wire is given by;

T.R = PSR+ (corrected HSR x L.C) = N + (n x L.C)

If r is radius of the wire, and l be the mean length of the wire.

Then, volume of the wire,

THICKNESS OF THE GLASS PLATE

S.NO.	Pitch Scale reading (P.S.R)	Observed H.S.R	Correction	Corrected H.S.R	Fraction B= n*L.C	Total reading
1	2	75	2	73	0.73	2.73
2	2	74	2	72	0.72	2.72
3	2	76	2	74	0.74	2.74

Average thickness of glass plate = 2.73 mm