# 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 +$n\times L.C$ ,

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 )    $mm^3$

# Vernier Calipers

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 called Vernier 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.No M.S.R                acm Vernier Coincidence   (n) Fraction   b=(n-x)*L.C Total Reading (a+b) cm 1. 2. 3. 4. 5.

Average length of the cylinder l = ……… cm.

b) Diameter of the cylinder :

 S.No M.S.R                acm Vernier Coincidence   (n) Fraction   b=(n-x)*L.C Total 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.No M.S.R                acm Vernier Coincidence   (n) Fraction   b=(n-x)*L.C Total 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 + $8 \times L.C$

# What are Units

Metric system

In my previous post, I explained with example how to solve the In problems in physics. To solve problems and to under stand the basics of the Physics it is very important to know what is a physical quantity, types of physical quantities, what is a unit, what are the units of different physical quantities, types of units, symbols of units.

There is one and only branch of science which measures a physical quantity, that branch of science is “Physics”. Measurements have an important role not only in physics but also in every branch of science and everywhere in our day-to-day life.

To measure physical quantities we need units. Let’s try to understand necessity of measurements and units of measurement in Physics.

The information about a physical quantity, by description of its external properties like color, taste etc is incomplete with out knowing its temperature, size (dimensions), which depends on measurement

, i.e. with out measurements it is impossible to know about the external properties of any object. So, it becomes necessary to measure it.

To measure a physical quantity we require a unit. Different physical quantities will have different units.

What is unit? A standard reference of the same physical quantity is essential to measure any physical quantity. That standard which we use to measure a physical quantity is called unit.

Let me put it this way, if we want to measure length of a table, we have to select a standard length (length of our hand), and by comparing the table’s length with the standard length we can measure the length of the table. If the table is 3.5 times that of standard length, i.e. length of our hand then we can write the result as “length of table = 3.5 times the length of our hand. In this example length of hand is taken as standard length to measure the table’s length.

Like that we can define any convenient standard or unit to measure a physical quantity.

But, if we choose standards as in the above example which are not consistent, and can not be reproduced then  errors and confusion in measurements will creep in. To avoid such confusion, instead of taking any undefined reference as a standard, well-defined and universal standards are used. Such a reference taken a standard is generally called a well defined unit or unit. Measurement of every physical quantity will have two parts a number (n) followed by a unit (u).

There fore n u = constant.

Ex: If the length of a table is 1.2 meters.In this measurement number n= 1.2 and unit is meter.

→ length (L)= $n_1u_1$ = 1.2 meters

→ length (L)= )= $n_2u_2$ = 120 centimeters

→ length (L)= $n_3u_3$ = 1200 millimeters

From the above data we can understand that

i)we can measure a physical quantity in different units.what ever may be the unit it’s value is same.

→ L =$n_1u_1$ = $n_2u_2$=$n_3u_3$

If theunit cosen is smaller, the multiple number will be greater.

$u_1$>$u_2$>$u_3$ = $n_1$ <$n_2$<$n_3$

iii)The units(u) of a physical quantity will be reciprocal to the multiple (n)

nu=constant$\Rightarrow$ $n_1u_1$=$n_2u_2$

u$\propto\frac{1}{n}$ or n$\propto\frac{1}{u}$

$\frac{n_1}{ n_2}$= $\frac{u_2}{ u_1}$