Practical Physics: February 2009

Practical Physics

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Screw Gauge

Find the volume of the Sphere - Vernier Calipers.

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.

zero-error

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 metal 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 ) $mm^3$

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 .

Draw Figure

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 $cm^3$ .

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