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.
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.
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.NoM.S.R
a cm		Vernier Coincidence   (n)Fraction   b=n*L.CTotal Reading         (a+b) cm
12.580.01*8=0.082.58
22.590.01*9=0.092.59
32.570.01*7=0.072.57
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.NoM.S.R                   a cmVernier Coincidence   (n)Fraction   b=n*L.CTotal Reading               (a+b) cm
11.140.01*4=0.041.14
21.150.01*5=0.051.15
31.150.01*5=0.051.15
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
275275-2=7373*0.01=0.732.73
2274274-2=7272*0.01=0.722.72
3276276-2=7474*0.01=0.742.74
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

04/03/2009

Compare the radii of given three wires using Screw Gauge.

Filed under: Experiments,Screw Gauge — gyaunnrraje @ 5:30 pm 
Tags: , 
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.
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.
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
1.145345-3=4242*.001=0.421.42
2.146346-3=4343*0.01=0.431.43
3.146346-3=4343*0.01=0.431.43
4.147347-3=4444*.001=0.441.44
5.146346-3=4343*0.01=0.431.43
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
1.212312-3=99*0.01=0.092.09
2.213313-3=1010*0.01=0.102.10
3.214314-3=1111*0.01=0.112.11
4.212312-3=99*0.01=0.092.09
5.214314-3=1111*0.01=0.112.11
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
1.185385-3=8282*0.01=0.821.82
2.184384-3=8181*0.01=0.811.81
3.184384-3=8181*0.01=0.811.81
4.186386-3=8383*0.01=0.831.83
5.186386-3=8383*0.01=0.831.83
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

18/02/2009

Screw Gauge

Filed under: Experiments,Screw Gauge — gyaunnrraje @ 9:44 am 
Tags: , 
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.
photo-8-001
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_1and 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.NoPitch Scale Reading (P.S.R) ammObserved H.S.R    (n’)Correction   (x)Corrected H.S.R  n=n’-xFraction  b=n*L.CTotal 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.NoPitch Scale Reading (P.S.R) ammObserved H.S.R    (n’)Correction   (x)Corrected  H.S.R n=n’-xFraction  b=n*L.CTotal 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
Watch this video to get idea about Screw Gauge :

13/02/2009

Find the volume of the Sphere – Vernier Calipers.

Filed under: Experiments,Vernier calipers — gyaunnrraje @ 9:02 am 
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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.NoM.S.R                acmVernier Coincidence   (n)Fraction   b=n*L.CTotal Reading (a+b) cm
1.1.970.051.97
2.1.960.041.96
3.1.960.061.96
4.1.970.051.97
5.1.970.061.97
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.

12/02/2009

Find the volume of cylinder – Vernier calipers.

Filed under: Experiments,Vernier calipers — gyaunnrraje @ 7:25 pm 
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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.NoM.S.R                acmVernier Coincidence   (n)Fraction   b=n*L.CTotal Reading (a+b) cm
1.2.690.092.69
2.2.710.012.71
3.2.720.022.72
4.2.720.022.72
5.2.680.082.68
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.NoM.S.R                acmVernier Coincidence   (n)Fraction   b=n*L.CTotal Reading (a+b) cm
1.1.450.051.45
2.1.440.041.44
3.1.460.061.46
4.1.450.051.45
5.1.460.061.46
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.

08/02/2009

Practical Exams Question bank.

Filed under: Experiments — gyaunnrraje @ 11:16 am 
Tags: 
Physics Practical Examination
( Board of Intermediate Education, Andhrapradesh)
Question Bank
SECTION – A
5.Find the volume of given cylinder bt measuring the length with vernier calipers and radius with screw gauge.
6.Find the thickness of given glass plate and radius of curvature of given curved surface (watch glass) using spherometer.
7.Find the thickness of glass plate with spherometer and verify the result with a screw gauge.
8.Find the focal length of given sphrecal mirror with spherometer.
9.Find the mass of the given body correct to milligram using a common balance.
SECTION – B
10.Find the value of ‘g’ at a given place using a simple pendulum.Also find the percentage error in measurement.
11.Find the acceleration due to gravity at your place using simple pendulum from l -T^2 graph.Find length of seconds pendulum also.
12.Find length of seconds pendulum from l -T^2 graph.
13.Verify parallelogram law of forces.Find weight of given stone in air.
14.Verify the triangle law of forces take 3 readings).Find relative density of given stone.
15.Find the volume of the given stone using parallelogram law of forces.Take 3 observations.
16.Find the volume of the stone applying the triangle law of forces.
17.Verify Boyle’s law and draw p-l graph ( Take 8 observations).
SECTION – C
18.Verify Boyle’s law. Draw  h-\frac{1}{l} graph.(take 6 observations)
19.Using Boyle’s law apparatus,draw h-\frac{1}{l} graph.Hence find the atmospheric pressure.
20. Find the co-efficient of apparent expansion of given liquid using the specific gravity bottle.
21.Find the co-efficient of real or absolute expansion of the given liquid using specific gravity bottle.
22.Find the specific heat of solid by method of mixtures.
23.Find the surface tension of water by capillary rise method.
24.Determine the radius of the bore of given capillary tube by capillary rise method.(T = 72 dyne/cm)
SECTION – D
25.Find the focal length of given concave mirror by u – v method.Draw u – v graph and hence find its focal length.Compare these two values.
26.Find the focal length of given concave mirror by u , v method.Verify the result from \frac{1}{u}-\frac{1}{v} graph method.
27.Find the focal length of given convex lens by u – v method.Verify the result from u – v graph.
28.Find the focal length of given convex lens from u – v graph.Verify result from \frac{1}{u}-\frac{1}{v} graph.
29.Find the focal length of given convex lens by u – v method.Verify the results from \frac{1}{u}-\frac{1}{v} graph.
30.Find the focal length of yhe given convex lens by lens displacement  or conjugate foci  method.
31.For a prism , assuming the angle of the Prism as 60^0,draw I-d curve and hence determine the refractive index of material of the prism.
32.Draw I-d curve and determine the angle of the prism assuming refractive index of glass as 1.5 .
SECTION – E
33. Find velocity of sound in air at room temperature.(using 3 tuning forks) and estimate its value at 0^0 C.
34.Compare the frequencies of the given two tuning forks using resonance apparatus.
35.Find the frequencies of the two given tuning forks using resonating air column apparatus.(Velocity of sound in air = 330 m/sec^2
36.Draw the lines of force around a short bar magnet placed with its North pole pointing the North in the earths magnetic field and locate null or neutral points, and find the magnetic moment of the given magnet.
37.Draw the lines of force around a short magnet placed with the  south pole pointing the North in the earth’s magnetic field and locate null points and find pole strength of the magnet.
38.Verify Inverse square law using deflection magneto-meter.Take 3 readings.
39.Compare the magnetic moments of given two magnets using a deflection magnetometer in Tan – B position by i)equal distance method and compare results with null method.Take 2 observations in each case.
40.Compare the magnetic moments of given two magnets using a deflection magnetometer in Tan - A position by i)equal distance method and compare results with null method.Take 2 observations in each case.
41.Measure the current flowing through a circuit using a tangent galvanometer.Verify the results using an ammeter in the circuit.
42.Find the reduction factor of the tangent galvanometer using an ammeter in the circuit.Verify the result from theory.
SECTION – F
43.Draw R – Cos\theta graph and hence verify the Ohm’s law.
44.Draw R – Cos\theta graph and hence find the resistance of the tangent galvanometer and the battery.
45.Find the resistance and specific resistance of the given wire using a meter bridge.
46.Using mere bridge , find the resistance of given wires by connecting them in series and in parallel.
47.Compare the resistance’s of the given two wires using meter bridge.
48. Draw voltage – current ( V-I) characteristics of junction diode.

03/01/2009

Vernier Calipers

Filed under: Experiments,Vernier calipers — gyaunnrraje @ 3:13 am 
Tags: , 
Aim: To determine i) The volume of the given cylinder by measuring its length and diameter
ii ) The volume of the given sphere by measuring its diameter.
Apparatus : Vernier Calipers,Cylinder and sphere.
Description of Vernier Calipers: A Vernier calipers consists of mainly two parts i) A 2cm wide 15cm long rectangular metal strip .The left end bottom side of this strip consists of a fixed jaw 1 (A) and at the same end jaw 2(C) at the top of this strip. On the strip a scale (5)is graduated in Inches along the upper edge and another scale(4) is graduated in Centimeters along the lower edge. This is called Main Scale ‘S’ .
ii) A metal frame V called vernier slides over the Main Scale ‘S’ . At the bottom of this frame V a button 8(P) is attached,which helps to fix this vernier at any desired place on the main scale.This verier frame consists of jaw1 (B) at the bottom and a jaw 2(D) at the top .Two scales are graduated on this frame corresponding to two scales on the Main Scale ‘S’. The two scales 6 and 7 on the vernier are calledVernier scale.Vernier scale consists of equal number of divisions. When we move vernier frame over the main scale, a thin strip (3) will be projected out.The projection will be exactly equal to the distance between Jaws 1(AB) i.e the thickness of the object between jaws.
The lower jaws 1,1(AB) are used to measure the thickness or external diameter of the tubes,cylinders or spheres.
The upper jaws 2,2 (CD) are used to measure the inner diameters of hallow bodies like tubes or holes.
The thin strip ( 3) is used to measure the depth of the objects like test tubes.
Theory :  Principle of vernier calipers – N divisions on the vernier scale is equal to (N-1) divisions on the main scale.
N V.S.D = (N-1) M.S.D
1 V.S.D = \frac{(N-1)}{N} M.S.D
Least count (L.C) of vernier calipers : Minimum length or thickness measurable with the vernier calipers is called its least count.
Least count (L.C) = 1 M.S.D – 1 V.S.D
L.C = 1 M.S.D – \frac{(N-1)}{N} M.S.D
L.C = 1 M.S.D [ 1-\frac{(N-1)}{N}]
L.C = \frac{1 M.S.D}{N}  = \frac{S}{N}
Where S is the value of  one Main scale division and N is the number of equal divisions on the vernier scale.
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.
d)To determine the diameter of the sphere : The given sphere is held firmly between jaws of the vernier calipers, in such a way the points where the jaws are in contact with sphere should be the two extremes of the chord of the cylinder.Post the values of the 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=2 r ) of the sphere.
e)To determine the volume of the sphere :Calculate the radius of the sphere  r = d/2 .Substitute the value of  mean radius (r)  in the formula of the volume of the sphere V = \frac{4}{3} \pi r^3 cm^3 .
Precautions to be taken while doing the experiment : 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.
Observations:
I) Least count of Vernier calipers :
a) Value of 1 Main scale division  = 1 M.S.D = S = ……..  cm,
b) Number of divisions on the vernier scale   N= ……… cm,
Least count              L.C = \frac{S}{N}   = ………. cm.
c) Zero error =x ( positive error)
II) Volume of the Cylinder :
a)Length of the cylinder :
S.NoM.S.R                acmVernier Coincidence   (n)Fraction   b=(n-x)*L.CTotal Reading (a+b) cm
1.
2.
3.
4.
5.
Average length of the cylinder l = ……… cm.
b) Diameter of the cylinder :
S.NoM.S.R                acmVernier Coincidence   (n)Fraction   b=(n-x)*L.CTotal Reading (a+b) cm
1.
2.
3.
4.
5.
Average diameter of the cylinder  d = 2r = …………. cm,
Mean radius of the cylinder           r = d/2 = ………….. cm,
Volume of the cylinder                   V = \pi r^2 l cm^3.
III ) Volume of the sphere :
a) Diameter of the sphere :
S.NoM.S.R                acmVernier Coincidence   (n)Fraction   b=(n-x)*L.CTotal Reading (a+b) cm
1.
2.
3.
4.
5.
Average diameter of the sphere  d = 2r = …………. cm,
Mean radius of the sphere           r = d/2 = ………….. cm,
Volume of the sphere                   V = \frac{4}{3} \pi r^3 cm^3.
Result : 1. Volume of the given cylinder  V= ……….. cm^3
2.Volume of the given sphere      V= ……….. cm^3.
Watch this vedio :
Review Questions – Vernier calipers.
During practical examinations  the examiner can shoot any question related to the experiment you are doing.Here I will try to answer few Frequently asked questions by   examiner.
1Q: What is the advantage of vernier calipers over a regular  scale graduated in millimeters?
Ans: We can measure up to a millimeter with regular scale which is graduated in millimeters, Where as  with a vernier calipers with 10 equal divisions on the vernier scale we can measure up to 1/10 (0.1) mm accurately.
2Q:What is the principle of vernier ?
Ans: (N-1) main scale divisions  = N divisions on vernier scale,
(N-1) M.S.D = N V.S.D   is the principle of vernier.
3Q:What do you mean by least count of vernier calipers?
Ans: The difference  of 1 M.S.D  and 1 V.S.D is called least count of a vernier calipers. (or) The minimum
length which can be measured by a vernier calipers is called its least count.
4Q:What is the use of lower jaws of a vernier calipers?
Ans: The lower jaws or used to measure the  thickness of object,outer diameters of tubes,spheres   and  cylinders.
5Q:What is the use of upper jaws?
Ans: The upper jaws of the vernier calipers are used to measure the inner diameters of rings,tubes and inner diameters of hallow cyliders , hallow spheres.
6Q:What is the use of the thin strip moving behind the main scale?
Ans: The strip is used to measure  the depths of the tubes and level of the liquid inside tubes or jars.
7Q:Two different vernier calipers  have different number of equal divisions on their vernier scales i) 10 equal divisions ii) 50 equal divisions . Which can measure more accurately.
Ans: The least count of the I vernier calipers  = S/N = \frac{1mm}{10} =0.1mm
The least count of the II vernier calipers  = S/N = \frac{1mm}{50} =0.02mm.
That is the II vernier calipers with 0.02mm of L.C can measure up to 0.02mm accurately.Hence the accuracy of II vernier calipers in more than I vernier calipers.
8Q:What is the ZERO error of a vernier calipers?
Ans:When two lower jaws of the vernier calipers are in contact, if the zero division of main scale does not coincide with the zeroth division of vernier scale, such vernier calipers will have zero error.
9Q: What is the formula of least count?
Ans: Least count L.C = \frac{S}{N}.
10Q:What is the formula to calculate the volume of the cylinder?
Ans: Volume of the cylinder V = \pi r^2 l. , where r = radius of the cylinder and l = length  of the cylinder.
11Q:What is the formula to calculate the volume of a Sphere?
Ans: Volume of the Sphere V = \frac{4}{3} \pi r^3 , where r = radius of  the sphere.
12Q:What is the formula to calculate total reading with a vernier calipers?
Ans: Total reading = M.S.R + n\times L.C.
13.How many types of Zero errors will be possible in vernier calipers?
Ans: Two types of Zero errors are possible in vernier calipers they are i) Negative Zero error ii) Positive Zero error
14. What is Positive zero error,how do we correct such an error in a vernier calipers?
Ans:When two lower jaws of the vernier calipers are in contact, if the zero division of main scale does not coincide with the zeroth division of vernier scale and the zero of the vernier is to the right of zero of main scale such an error is called Positive Zero error.So,the zero correction should be subtracted from the reading which is measured.
Ex:When two lower jaws of the vernier calipers are in contact, if the zero division of the vernier is to the right of  zeroth division of main scale and if the vernier coincidence is 1.Then the correction is subtraction i.e Total Reading – 1 \times L.C
15.What is Negative zero error,how do we correct such an error in a vernier calipers?
Ans:When two lower jaws of the vernier calipers are in contact, if the zero division of main scale does not coincide with the zeroth division of vernier scale and the zero of the vernier is to the left of zero of main scale such an error is called negative Zero error.So,the zero correction should be added to the reading which is measured.
Ex:When two lower jaws of the vernier calipers are in contact, if the zero division of the vernier is to the left of  zeroth division of main scale and if the vernier coincidence is 8.Then the correction is addition i.e Total Reading + 8 \times L.C
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