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Measurements
Methods of measurement and error analysis for Physical quantities pertaining to
the following experiments.
Experiments based on using Vernier Callipers and screw gauge (micrometer).
Determination of g using simple pendulum Young modulus by scarles method
specific heat of a liquid using Celorimeter focal length of a Concave mirror and a
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convex lens using uv method speed of sound using resonance column verification
of ohm's law using voltmeter and ammeter specific resistance of the material of
the wire using bridge and P.O. box.
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Work and power:
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Lifting m/c.
Significant figures:
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Measurements made by any instrument are not absolutely correct. The degree of
accuracy or precision is shown by the significant figures upto which the
measurement has been recorded.
Let us say, the length of an object is 14.5 cm. It shows that the measurement has
been made to the nearest of
1
th of a centimeter which shows that figures 1
10
to 4 are absolutely correct and figure 5 is reasonably correct.
If the length recorded is 14.52 cm, then it shows that the measurement has been
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made correctly up to
1
th of a centimeter. In this case, the figures 1, 4 and 5
100
are absolutely correct while the figure 2 is approximate.
Thus, significant figures are the number of digits upto which we are sure about
their accuracy. In other words, significant figures are those digits in a number
that are known with certainty plus one more digit that is uncertain.
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For example, 14.5 cm has three significant digits and the measurement 14.52 cm
has four significant digits. Significant figures do not change if we measure a
physical quantity in different units.
For example, 14.5 cm = 0.145 m
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= 14.5 × 10–2 m
ac
k
Now 14.5 cm and 14.5 × 10–2 m both have three significant figures.
Rules for significant figures:
All non-zero digits are significant figures.
cr
(1)
Example :
Number
Significant figures
17
2
178
3
1782
4
17825
5
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(2)
All zeros occurring between non-zero digits are significant figures.
Example :
Significant figures
401
3
4012
4
40056
5
400006
6
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(3)
Number
All zeros to the right of the last non-zero digit are not significant figures:
Example :
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20
Significant figures
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Number
210
2
2130
3
20350
4
cr
(4)
1
All zeros to the right of a decimal point and to the left of a non-zero digit
are not significant figures:
Example :
Number
Significant figures
0.04
1
0.004
1
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(5)
0.0045
2
0.0456
3
0.0004564
4
All zeros to the right of a decimal point and to the right of a non-zero
digit are significant figures:
Example :
Significant figures
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Number
0.20
2
0.230
3
4
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0.2370
ac
k
Rounding off the measurements:
The following rules are applied in order to rounding off the measurements:
If the digit to be dropped in a number is less than 5, then the preceding
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(i)
digit remains unchanged. For example, the number 8.64 is rounded off
to 8.6.
(ii)
If the digit to be dropped in a number is greater than 5, then the
preceding digit is raised by 1. For example, the number 8.66 is rounded
off to 8.7.
(iii)
If the digit to be dropped in a number is 5 or 5 followed by zeros, then
the preceding digit remains unchanged if it is even.
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For example,
(i) the number 8.65 is rounded off to 8.6,
(ii) the number 8.650 is rounded off to 8.6.
(iv)
If the digit to be dropped in a number is 5 or 5 followed by zeros, then
the preceding digit is raised by I if it is odd.
For example,
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(i) the number 8.75 is rounded off to 8.8,
(ii) the number 8.750 is rounded off to 8.8.
SIGNIFICIENT FIGURES
ac
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called significant figure.
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The numbers of figure required to specify a certain measurement perfectly are
The last figure of a measurement is always doubtful, but is included in the
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number of significant figure.
Example: If length of pencil measured by vernier callipers is 9.48 cm, the number
of significant fig. in the measurement is 3.
RULES FOR SIGNIFICANT FIGRUES
(i)
If a measurement contains no decimal point, the number of final zeros
are ambiguous and are not counted in significant fig. i.e. all non zero
digits are significant.
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e.g. — In 3320 no. of significant figures = 3
(ii)
The power of 10 and the zeros on left hand side of a measurement are
not counted while counting the number of significant fig.
e.g. — 5 × 103
(iii)
the zeros after a decimal are counted as to significant fig.,
e.g. — 1.60 has three significant fig.
(iv)
The zeros appearing in between the non zero digits are counted as
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significant figures,
e.g. — In 2.07, there are three significant figures.
The zeros appearing to the left of a non zero digit are not counted in
significant figures,
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(v)
(vi)
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e.g. — 0.0702 has only three significant figure (702)
When the position of decimal point changes, then the number of
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significant figures does not change,
i.e. — 1.942, 194.2 all have four significant figures.
(vii)
The limit and accuracy of a measuring instrument is equal to the least
count of the instrument.
(viii)
In the sum and difference of measurements, the result contains the
minimum number of decimal places in the component measurements.
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Ex. The length of string of simple pendulum is 101.4 cm and diameter of bob is
2.64 cm. What is th effective length of simple pendulum up to required
significant figures.
Sol.

Here
0
r=
0
r
 101.4 cm,
2.64
 1.32 cm
2
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 101.4  1.32  102.72 cm
Since we take least number of decimal figures in a measurement which is 1 in
Hence Effective length = 102.7 cm.
ac
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(ix) In the product and quotient of measurements, the result contains the
minimum number of significant figures in the component measurements.
Ex. The length, breadth and thickness of a block are given by
cr
cm, t = 2.45 cm.
= 12 cm, b = 6
What is the volume of the block according to the idea of significant figures.
Sol. Volume = blt
= 6 × 12 × 2.45 = 176.4
= 1.764 × 104 cm3
The minimum number of significant figures is 1 in thickness.
Vernier Callipers and Screw Gauge:
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The meter scale which commonly used in practice is the simplest instrument for
measuring length.
By meter scale we can measure upto 1 mm because the length of the smallest
division made on the scale is 1 mm. In order to measure still smaller lengths
 1
 1 
th or 
th of a millimeter, the instruments

 100 
10 
accurately upto 

1.
Vernier Callipers
2.
Screw Gauge
Vernier Callipers:
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commonly used in laboratory are:
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 1
th of millimeter. Vernier Callipers
10 
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It is used to measure accurately upto 

comprises of two scales, Wz, main scale S and vernier scale V which is called
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auxiliary scale. The main scale is fixed but the vernier scale is movable. The
divisions of vernier scale are usually a little smaller in size than the smallest
division on the main scale. It also has two jaws, one attached with the main scale
and the other with the vernier scale. The purpose of jaws are to grip the object
between them. Vernier has a strip, which slides along with vernier scale, over the
main scale. The strip is used to measure the depth of hollow object.
Vernier Constant (VC):
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Suppose the size of one main scale division is S and that of one vernier scale
division is V units. Also suppose that length of n vernier division is equal to the
length of (n – 1) division of main scale. Thus, we have
(n – 1)S = nV
or nS – S = nV
S
n
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or S – V =
The quantity (S – V) is called vernier constant (VC).
Least Count:
The smallest value of a physical quantity which can be measured accurately with
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an instrument is called the least count (L.C.) of the instrument.
ac
k
For vernier calipers, its least count is equal to its venier constant. Thus
S
n
cr
Least count = S – V =
where, S = size of one main scale division
V = size of one vernier scale division
n = No. of division on vernier scale
=
Length of one division of main scale
No. of divisions on vernier scale
Length of the object = main scale reading + n (LC)
n = vernier division exactly coinciding with some main scale division.
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Determination of zero error:
When jaws of the vernier are made touch other and the zero mark of the vernier
scale coincide with the zero mark of the main scale, there will no zero error in the
instrument. However, in practice it is never so. Due to wear and tear of the jaws
and due to some manufacturing defect, the zero mark of the main scale and
vernier scale may not coincide, it gives rise to an error, is called zero error. It may
Positive and negative zero error:
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be positive or negative zero error.
When the zero mark of the vernier scale lies towards the right side of the zero of
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the main scale when the jaws are in contact, the measured length will be greater
ac
k
than the actual length. Because of this fact the zero error is called positive zero
error. On the other hand, when zero mark of the vernier scale lies towards the
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left side of the zero of the main scale when jaws in contact with each other, the
length of the object measured by the instrument will be less than the actual
length of the object. Because of this reason is called negative zero error.
True reading = Observed reading – Zero error with proper sign.
Correction for positive zero error:
When its jaws are in contact with each other, suppose 3rd vernier division
coincides with the any of the divisions of main scale. They we have
Zero error = + [0.00 cm + 3(L.C.)]
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= + [0.00 + 3 × 0.01 cm]
= + 0.03 cm
Correct reading = Observed reading – (0.03 cm)
Negative error:
Screw Gauge: It is used to measure small lengths like diameter of a wire or
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thickness of sheet etc. It consists of a U' shaped metal frame as shown in fig.
A main scale which graduate in millimeter or half a millimeter. The main scale
also called pitch scale.
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Pitch: It is defined as the linear distance moved by the screw forward or
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backward when one complete rotation is given to the circular cap.
Least count (L.C.)
Pitch
 Total number of divisions on the circula r scale 
cr
=
Ex. In Four complete revolution of the cap, the distance travelled on the pitch
scale is 2 mm. If there are 50 divisions on the circular scale, then calculate the
least count of the screw gauge.
Pitch =
L.C. =
2mm
 0.5 mm
4
0.05
mm  0.01 mm
50
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Zero error:
When the studs P and Q of the screw gauge are brought in contact without apply
induce pressure and if the zero of the circular scale coincides with the reference
line, then there is no zero error, otherwise there will be zero error.
Positive zero error:
In this case, the zero of the circular scale lies below the reference line as the gap
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between studs P and Q reduces to zero.
Suppose the zero line of the circular scale is 4 division below the reference line. In
other words, the 4th division of the head scale is in line with the line of
Zero error = + 4 (L.C.)
ac
k
= + 4 (0.01 cm)
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graduation.
= + 0.04 cm
cr
Zero correction = – zero error
Negative zero error:
When zero of the circular scale lies above the reference line when the gap
between the studs P and Q become zero.
Zero error = – 3 × 0.01 mm
= – 0.03 mm
Zero correction = + 0.03 mm.
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Experiments
(i) Measurement of length
The simplest method measuring the length of a straight line is by means of a
meter scale.
There exists some limitation in the accuracy of the result:
(i) the dividing lines have a finite thickness.
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(ii) naked eye cannot correctly estimate less than 0.5 mm
For greater accuracy we use devices like vernier calipers and micrometer scales
(screw gauge).
LEAST COUNT
ac
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least count.
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The minimum measure that can be actually taken by an instrument is called the
Least count of meter scale graduated in millimetre mark is 1 m.
cr
Least count of watch having second hand is 1 sec
VERNIER CALLIPERS
It consists of a main steel scale (S) with a fixed jaw J 1 and a sliding jaw J2 carrying
a vernier scale. When the two jaws are made to touch each other, the zero of
vernier scale coincides with the zero of main scale.
If a body is held gently between the jaws of vernier calipers and the zero of
vernier scale lies ahead of a n division of main scale reading = N (cm). If nth
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division of vernierscale coincides with any division of main scale than vernier
scale reading = n × vernier constant.
Total reading = MSR + VSR = (N + n × VC)
Therefore diameter (D) = a + b × LC
(where a = MSR, v = VSD & LC = Least Count)
Least Cont of Vernier Callipers:
part of vernier scale
Also Least count of vernier calipers
Value of 1 part on main scale
Number of parts on vernier scale
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=
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Least count of vernier calipers = values of one part of main scale – value of one
=
9
M.S.D.
10
cr
= 1 M.S.D. 
ac
k
LC = 1 Main Scale Division – 1 Vernier Scale Division
1
mm  0.1 mm  0.01 cm
10
Zero Error:
If the zero of main scale coincides with zero of Vernier scale when jaws C and D
are brought in contact with each other then the instrument is free from error or
it is said to have no zero error. But in actual practice it is never so.
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Due to wear and tear of jaws and some time due to manufacturing defects the
zero mark of the Vernier scale does not coincide with zero of main scale. It gives
rise to an error called zero error. Zero error can be positive and negative.
Determination of Zero Error
(i) Positive zero error and its correction.
The zero error is positive when the zero mark of the Vernier scale lies towards
the right side of the zero of the main scale when jaws C and D are made to touch
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each other. In such case measured length will be more than the actual length and
therefore, the zero error is called positive zero error. In figure +ve zero error is
calculated from the division coinciding with main scale.
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k
Zero Error = 0.05 cm
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Zero Error = 0.00 + 5 × VC = 0.00 + 5 × .01
To get correct reading : 0.05 cm is to be subtracted from the observed value.
cr
(ii) Negative zero error and its correction.
The zero error is negative when the zero mark of the Vernier scale lies towards
the left side of the zero of the main scale when the jaws are in contact. The
length measured by such instruments is less than the actual length and therefore,
the zero error is called negative zero error.
Zero error = 0.00 – (10 – 6) × VC = – 4 × .01 = – 0.4 cm
Correct reading = observed reading – (– 0.04)
= OR + 0.01 cm
Determination of Least Cont of Vernier Constant:
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Least count is the smallest value of a physical quantity which can be measured
accurately with an instrument.
For an instrument where Vernier is used its Vernier Constant (VC) is its Least
Count (LC)
10 div. of scale coincides with 9 div. lof main scale and the length of 1 div. onmain
scale is 1 mm.
1 VSD =
9
MSD
10
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10 VCD = 9 MSD
The Quantity (1 MSD – 1 VSD) is called Vernier Cosntant (VC)
=
iit
9
1
MSD 
MSD

10 
10
ac
k


VC = 1 
1
 1 mm 1 MSD  1 mm 
10
cr
VC = 0.1 mm = .01 cm
HOW TO TAKE VERNIER READINGS
l = Main scale observation + [coinciding vernier scale division × vernier constant]
or l = a [b × V.C.]
Hence l = {1.6 + [5 × 0.01]} cm
or = {1.6 + 0.05} cm = 1.65 cm
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SCREW GAUGE (OR MICROMETER SCREW)
In general venier calipers can measure accurately upto 0.01 cm and for and for
greater accuracy micrometer screw devices e.g. screw gauge, spherometer a re
used. These consist of accurately cut screw which can be moved in a closely
fitting fixed nut by tuning it axially.
The instrument is provided with two scales:
(i) The main scale or pitch scale M graduated along the axis of the screw.
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(ii) The cap-scale or head scale H round the edge of the screw head.
Pitch:
The translational motion of the screw is directly proportional to the total rotation
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of the head. The pitch of the instrument is the distance between two consecutive
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threads of the screw which is equal to the distance moved by the screw due to
one complete rotation of the cap. Thus for 10 rotation of cap = 5 mm, then pitch
= 0.5 mm.
Distance moved by screw
 ....m
No. of rotations given to screw
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Pitch p =
Least count:
In this case also, the minimum (at least) measurement (or count) of length is
equal to one division on the head scale which is equal to pitch divided by the
total cap divisioins.
Thus in the aforesaid Illustration: if the total cap division is 100, the least count =
0.5 mm/100 = 0.005 mm
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Least Count =
pitch
 .....mm
No. of division on circular scale
Measurement of length by screw gauge:
L = n × pitch + f × least count, where n = main scale reading and f = circular scale
reading
ZERO ERROR
When the two studs of a screw gauge are brought in contact with each other, the
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zero of the circular scale should coincide with the graduation line of main scale.
In that case there is no zero error. However when the zero of the circular scale
does not coincide with the graduation line the screw gauge is said to have zero
error. A correction is then applied to the observed thickness or diameter to get
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Positive Zero Error:
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the correct value. Zero error may be +ve or –ve.
The zero error is said to be +ve if on bringing studs in contact, zero of the circular
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scale is below the line of graduation.
Negative Zero Error.
The zero error is negative if on bringing, the studs in contact, the zero of the
circular scale is above the graduation line on the main scale.
Zero correction is always, negative of zero error:
(i) If the linear scale reading is zero and circular scale reading is 4 and zero of C.S.
is above the graduation line then zero correction is given by
8. Statemennt-1
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If x 
an
x
 a 
 b 
the
 n

m
m

 b 
b
x
 a 


The chang
and
Statement-2
(B) B
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(A) A
(C) C
(D) D
9. Statement-1
If the measuring instruments used are perfect, then measurements made can be
and
Statement-2
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perfect.
(A) A
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Measurements depend upon the error free instruments only.
(C) C
(B) B
(D) D
10. Statement-1
Systematic errors and random errors fall in the same group of errors.
and
Statement-2
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Both systematic and random errors are based on the cause of error.
(A) A
(B) B
(C) C
(D) D
11. Statement-1
Absolute error may be negative or positive.
and
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Statement-2
Absolute error is the difference between the real value and the measured value
of a physical quantity.
(A) A
(B) B
(D) D
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(C) C
Now lets see how to do arithmetic operations i.e. addition, subtraction,
multiplication and division according to significant figures.
cr
(a) Addition  Subtraction
Rules for Addition  Subtraction.
First round off an quantities to the decimal place of least accurate quantity.
Then the addition/subtraction in normal manner.
e.g. 423 + 20.23
486.2 – 35.18
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486.2
423.5
35.18
20.23 
35.2
443.7
451.0
Rules for multiply  Division
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Suppose we have to multiply
2.11 × 1.2
2.1
2.532
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211
cr
422
iit
 1.2
2.5
So Answer will come in least significant figures out of the two numbers.
Multiply/Divide in normal manner.
Round off the answer to the weakest link
(number having least S.F.)
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Ex# 312.65 × 26.4 = 8253.960
5 S.F.
3 S.F. round off
to three S.F.
8250
Ex# A cube has a side l = 1.2 × 10 –2 m. Calculate its volume.
l = 1.2 × 10–2
Two S.F. Two S.F. Two S.F.
= 1.728 × 10–6 m3
iit
Round off of 2 S.F.
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V (1.2 × 10–2) (1.2 × 10–2) (1.2 × 10–2)
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1.7 × 10–6 m3
Ex# In ohm's law exp., reading of voltmeter across the resister is 12.5V and
R=
cr
reading of current I = 0.20 Amp. Estimate the resistance in correct S.F.
V 12.5  3 S.F.

 62.5 
1 0.20  2 S.F.
round off
to 2 S.F.
= 62 
Ex# Using screw gauge radius of wire was found to be 2.50 mm. The length of
wire found by mm. Scale is 50.0 cm. If mass of wire was measured as 25 gm, the
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density of the wire in correct S.F. will be.
f=
m
 r 2l
two S.F.
=
25
 0.250 50.0
2
three S.F. three S.F.
gm
iit
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in
two
cm3
ac
k
= 2.5465 
 2.5
S.F.
cr
Sol.
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ERRORS IN MEASUREMENT
To get some overview of error, least count and significant figures, lets have the
below example.
(i)
Lets use a cm scale (a scale on which only cm marks are there).
We will measure length = 4 cm.
Although the length will be a bit more than 4, but we cannot say its length to be
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4.1 cm or 4.2 cm as the scale can measure upto cms only, not closer than that.
* It (this scale) can measure upto cms accuracy only.
* So we'll say that its least count is 1 cm.
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To get a  closer measurement we have to use a more minute scale, that is mm
(ii)
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scale.
Lets use an mm scale : (a scale on which mm marks are there.
We will measure length l = 4.2 cm, which is a more closer measurement. Here
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also if we observe closely, we'll find that the length is a bit more than 4.2 but we
cannot say its length to be 4.21, or 4.22, or 4.20 as this scale can measure up to
0.1 cms (1 mm) only, not closer than that.
* It (this scale) can measure upto 0.1 cm accuracy.
Its least count is 0.1 cm.
Max. uncertainty in l can be = 0.1 cm.
Max. possible error in l can be = 0.1 cm.
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Measurement of length = 4.2 cm has two significant figures; 4 and 2, in which 4 is
absolutely correct, and 2 is reasonably correct (Doubtful) be cause uncertainty of
0.1 cm is there.
To get more closer measurement.
(iii)
We can use Vernier Callipers (which can measure more closely, upto 0.01
cm).
je
e.
in
Then we'll measure length l = 4.23 cm which is more closer measurement.
* It can measure upto 0.01 cm accuracy
Least count = 0.01 cm
Max. uncertainty in l can be = 0.01 cm
iit
Max. possible error in l can be = 0.01 cm
ac
k
Measurement of length = 4.23 cm has three significant figures; 4, 2 and 3; in
which 4 and 2 are absolutely correct and 3 is reasonably correct (Doubtful)
cr
because uncertainty of 0.01 cm is there.
To get further more closer measurement.
(iv)
We can use screw Gauge : (which can measure more closely, upto 0.001
cm)
We'll measure length l = 4.234 cm.
* Max possible uncertainty (error) in l can be
= 0.001 cm
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length = 4.234 cm has 4 significant figures.
4, 2, 3 are absolutely correct and 4 is reasonably correct.
To get further more closer measurement.
(v)
We can use microscope :
We'll measure length l = 4.2342 cm.
* Max. possible uncertainty (error) in l can be
je
e.
in
= 0.0001 cm
* Length = 4.2342 cm has five significant figures; 4, 2, 3, 4 and 2.
Least Count :
iit
We have studied that no measurement is perfect. Every instrument can measure
ac
k
up to a certain accuracy, called least count.
Least Count : The smallest quantity an instrument can measure




mm scale
Vernier
Screw gauge
Stop
Temp.
L.C. = 1
L.C. = 0.1
L.C. = 0.01
watch
thermo
mm
mm
mm
L.C. = 0.1
meter
sec.
L.C.= 1°C
cr

|
Permissible error:
Error in measurement due to limitation (least count) of the instrument, is called
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permissible error.
From mm scale  we can measure upto 1 mm accuracy (least count). Fro this we
will get measurement like l = 34 mm.

Max uncertainty can be 1 mm
Max permissible error (l) = 1 mm
je
e.
in
But, if from any other instrument, we get
l = 34.5 mm then max permissible error
(l) = 0.1 mm
iit
and if from a more accurate instrument, we get l = 34.517 mm then max.
ac
k
permissible error (l) = 0.001 mm
= Place value of last number.
cr
Max. permissible error in a measured quantity = Place value of the last number.
Max. Permissible error in Result due to error in each measurable quantity:
Let Result f(x, y) contains two measureable quantity x and y.
Let error in is = + x i.e. x  (x – x, x + x)
error in y is = y i.e. y  (y – y, y + y)
Case (i) If f(x, y) = x + y
(f)max. = x + y
Case (ii) If f = x – y
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(f)max. = x + y
Ex# In resonance tube exp. we find l1 = 25.0 cm and l2 = 75.0 cm. If there is no
error is frequency what will be max permissible error in speed of sound (take f0 =
325 Hz).
V = 2f0 (l1 – l1)
Vmax. = max. of 2f0 (+ l2
l1)
je
e.
in
= 2f0 (l2 + l1)
l1 = 25.0 cm l1 = 0.1 cm
(Place value of last number)
iit
l2 = 75.0 cm l2 = 0.1 cm
ac
k
(Place value of last number)
V = 2f0 (l2 – l1)
cr
= 2(325) (75.0 – 25.0) = 325 m/s
and Vmax. = 2(325) (0.1 + 0.1) = 1.3 m/s
So V = (325 + 1.3) m/s
Cose (III) If f(x, y, z) = constant x a yb zc
 f 
 max. of
 
f max.
=a

x
y
z 

a

b

c

x
y
z 
x
y
z
b
c
x
y
z
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Sign should be adjusted, so that errors get added up.
Ex# f = 15x2 y–3/2 z–5
x 3 y
z
 f 
2

5
 
f max.
x
z y
z
Ex#
length l = 75.3 cm, then find max. permissible error in
je
e.
in
  d2 
R
 4 
Resistivity () =
l
iit
  
R
d l

2

  
R
d
l
max.
ac
k
R = 1.05 R = 0.01 
d = 0.60 mm d = 0.01 mm
cr
l = 75.3 cm l = 0.1 cm
  
0.01
0.1
 0.01

 2


  
 0.60  75.3
1.05
max.
= 0.0759
Ex# In ohm's law experiment, potential drop across a resistance was measured as
V = 5.0 volt and current was measured as I = 2.00 amp. Find the maximum
permissible error in resistance.
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Sol. R =
V
 V  i1
i
 R 
 0.1 0.01
%


 100%  2.5%
 R  max.  5.0 2.00 
Value of R from the observation
V
5.0

 2.5 
i
2.00
So we can write R = (2.5 + 2.5%) 
je
e.
in
R
In searle's exp. to find Young's modulus, the diameter of wire is measured as D =
0.05 cm length of wire is L = 125 cm, and when a weight m = 20.0 kg is put,
ac
k
Young's modulus Y.
iit
extension in wire was found to be 0.100 cm. Find maximum permission error in
cr
mg
x

Y
 d2
l
4
Y
mgl
 2
dx
4
m l
d x
 Y 



2



Y  max.
m
l
d
x
m = 20.0 kg
m = 0.1 kg
l = 125 m
l = 1 cm
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d = 0.050 cm
d = 0.001 cm
x = 0.100 cm
x = 0.001 cm
1
0.001 0.001
 Y 
 0.1






  100%
Y max.  20.0 125 0.05 0.100 
= 4.3%
Exp. To find the value of 'g' using simple pendulum T = 2.00 sec, l = 1.00 m was
je
e.
in
measured. Estimate maximum permissible error in 'g'. also find value of 'g'.
(use 2 = 10)
T = 2
l
42l
 g 2
g
T
= 2%
ac
k
iit
 g
l
T 0.01
 0.01

2

 2
  100%
 g 

l
T
1.00
2.00
max.
cr
42l 4  10  1.00
Value of g =

 10.0 m / s2
2
2
T
2.00
 g
2

 g  0.2
 g 
100
max.
 g = (10.0 + 0.2) m/s 2
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EXPERIMENT
Determining the value of 'g' using a simple pendulum.
In this exp, a small spherical bob is hanged with a cotton thread. this
arrangement is called simple pendulum. The bob is displaced slightly and allowed
to oscillate. To find time period, time taken for 50 oscillations is noted using a
Theoretically, T = 2
L
g
 g  42
L
T2
…(1)
iit
Where
je
e.
in
stop watch.
ac
k
L = Equivalent length of pendulum
= length of thread (l) + radius (r) of bob
=
cr
T = time period of the simple pendulum
time taken for 50 oscillations
50
So, g can be easily determined by the eqn. (1)
Experiment:
Determining Young's Modulus of a given wire by Young's Searle's method:
To determine Young's modulus, we can perform an ordinary experiment. Let's
hang a weight 'm' from a wire. From Hook's law
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 x
l mg
mg
 Y  x  0 2

r Y
 l0 
If we change the weight, the elongation of wire will change proportionally.
If we plot elongation v/s mg, we will et a straight line
By measuring its slope and equating it to
je
e.
in
l0
, we can estimate Y.
 r2 Y
Experiment:
Determining specific heat capacity of an unknown liquid using colorimeter:
Regnault's apparatus to determine the specific heat capacity of a unknown liquid.
iit
A solid sphere of known specific heat capacity s 1 having mass m1 and initial
ac
k
temperature 1 is mixed with the unknown liquid filled in a calorimeter. Let
masses of liquid and calorimeter are m 2 and m3 respectively, specific heat
cr
capacities are s2 and s3 and initially they were at room temperature 2 when the
hot sphere is dropped in it, the sphere looses heat and the liquid calorimeter
system takes heat. The process continues till the temp. of all the elements
becomes same (say ).
Heat lost by hot sphere = m 1s1 (1 – )
Heat taken by liquid and calorimeter
= m2s2 (– 2) + m2s2 ( – 2)
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It there were no external heat loss.
Heat given by sphere = Heat taken by liquid-Calorimeter system
m1s1 (1 – ) = m2s2 ( – 2) + m3s3 ( – 2 )
m1s1  1   m2s3

m2    2 
m2
cr
ac
k
iit
je
e.
in
Get s2 =
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Exp. Determining speed of sound using resonance tube.
The experiment to find velocity of sound in air using Resonance tube measured.
Principle: Resonance tube is a kind of closed organ pipe. So its natural frequency
will be
V
3V
5V
,
,
, ....
4 leq 4 leq 4 leq
V
4 leq.
.in
or generally f n = 2n  1
ee
If it is forced with a tuning fork of frequency f0, for resonance,
V
4 f0
ki
V
 f0
4 leq
ac
2n  1
itj
Natural frequency = forcing frequency
cr
 leq  2n  1
For the first resonance
leq 
V
4 f0
l1  e 
V
4 f0
…(1)
For the second resonance
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leq 
3V
4 f0
l1  e 
3V
4 f0
…(1)
From (1) to (2)
V = 2f0 (l2 – l1)
.in
Observation table:
Room temp. in beginning = 26°C
Freq. of
Resonance
Water
Mean
Speed of
level is
level is
resonant
sound
falling
rising
length
23.9
24.1
l1 = …
V = 2f0
73.9
74.1
l2 = …
(l2 – l1 )
ac
1st resonance
ki
in viz (f0)
340 Hz
Water
itj
tun. fork
ee
Room tempered = 28°C
cr
2nd resonance
Verification of ohm's law using voltmeter an ammeter.
Ohm's low states that the electric current flowing through a conductor vs
directly proportional t the P.d. (V) across its ends provided that the physical
conditions of the conductor (such as temp. dimensions etc.) are kept constant
mathematically
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V  I or V  IR
Here R is a constant known as resistance of the conductor and depends on the
nature and dimensions of the conductor.
Procedure: By shifting the rheostat contact, reading of ammeter and Winterer
are noted down.
.in
At least six set of observation are taken. Then a graph is plotted between
potential difference (V) across R and current (I) through R. The graph comes to be
ee
a straight line.
ki
1
BP 1
V tano 

R
AP R
ac
I=
itj
V I
cr
Get R = …
Ext. : Specific resistance of the material of a wire using meter bridge and post
office box.
Meter Bridge:
Meter Bridge is a simple case of wheat stone Bridge and 1s used to find the
unknown resistance. The unknown resistance is placed in place of R and in place
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of s, a known resistance is used, using R.B. (Resistance Box). There is a 2m long
resistance wire between A and C. The Jockey's moved along the wire. When
R(100 – l) = S(l) then the Bridge will be balanced and the galvanometer will give
zero deflection.
'l' can be measured by the meter scale.
l
…(1)
100  l
.in
The unknown resistance is R = S
If length of unknown wire is L and diameter of the wire is d, then specific
itj
ac
From eqn. (1),
ki
  d2 
R
 4 

L
ee
resistance of the wire
cr
 d2  l 

s
4L 100  l 
End Corrections
In meter Bridge circuit, some extra length comes (is found under metallic strips)
at end point A and C. So some additional length ( and ) should be included at
ends for accurate result. Hence in place of
we use
+  and in place of 100 –
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, we use 100 –
+  (where  and  are called end correction. To estimate 
and , we use known resistance R 1 and R2 at the place of R and S in meter Bridge.
Suppose we get null point at
1
distance then
R1
1  

R2 100  1  
…(i)
then
ki
ac
2
itj
…(ii)
Solving equation (i) and (ii) get
R 2 1  R1
R1  R 2
distance
ee
R2
2  

R1 100  1  
R1 1  R2
R1  R 2
2
cr
=
2
.in
Now we interchange the position of R 1 and R2, and get null point at
and =
 100
These end correction ( and ) are used to modify the observations.
Exp. : POST OFFICE BOX
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In a Wheatstone Bridge circuit, if
unknown resistance X =
P R
then the bridge is balanced. So

Q X
QR
. to Realize the Wheatstone's Bridge circuit, a Post
P
Office Box is described.
Q
.
P
.in
1000  resistance, to set any ratio
Q
1
P
ee
These arms are called ratio arm.
itj
Initially we take Q = 10 and P = 10 to set
ki
The unknown resistance (X) is connected between C and D and battery is
ac
connected across A and C (Just like Wheatstone's Bridge). Now put Resistance in
part A to D such that the Bridge gets balanced.
cr
For this keep on increasing the resistance with 1 interval, check the deflection
in Galvanometer by first pressing key K1 then Galvanometer key K2.
Suppose at R = 4, we get deflection toward left and at R = 5 we get deflection
toward right.
So we can say that for bridge balance, R should be between 4 to 5.
Now X =
QR 10

R  R  4 to 5
P
10
So we can estimate that X shoul
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To get closer X, in the second observation,
lets choose
Q
1

P 10
 P  100 
e.g. 
 Q  10 
Suppose now at R = 42 we are getting deflection towards left and at R = 43
deflection is toward right.
QR
10
1

R
R
P
100
10
itj
where R  (42, 43)
ee
Now X =
.in
So R  (42, 43)
ki
So we can estimate that X  (4.2, 4.3)
ac
Q
1

P 100
cr
Now to get partner closer,
The observation table is shown:
No.
Resistance in the
Resistance
Direction of
Unknown
of
ratio arm
in arm AD
Deflection
resistance X =
obs.
1
QR

P
(R) ohm
AB = (P)
BC = Q
(ohm)
(ohm)
10
10
4
Left
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100
10
40
Left (Large)
50
Right (Large)
42
Lefts
43
Right
420
Left
424
Left
425
No deflection
426
Right
(4 – 5)
4.2–4.3
4.25
ee
3
10
Right
.in
100
5
itj
TO FIND FOCUS DISTANCE OF A CONCAVE MIRROR USING U-V METHOD:
cr
ac
1 1 1
 
f
V u
ki
Principle : For different u, we measure different V and find F using minor formula
In this experiment, a concave mirror is fixed at position Mm' and a Knitting
needle is used as an object, mounted in front of the concave mirror.
This needle is called object needle (O in fig.). First of au, we make a rough
estimation of f.
For estimating F roughly, make a sharp image of a for away object (like sun) on a
filter paper. The image distance of the far object will be an approx. estimation of
focus distance.
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Now, the object needle is kept beyond F, so that its real and in verted image
can be formed. You can see this inverted image in the mirror by closing your
are eye and keeping the other eye along the pole of the mirror.
To locate the position of the image, use a second needle and shift this needle
such that its peak coincide with the image. The second needle gives the distance
of image (V). So it called "image needle".
.in
Note the object distance 'n' and image distance 'V' from the mm scale on optical
Similarly take 4-5 more observations.
ee
bench.
ki
ac
1 1 1
 
f
V u
itj
Determining of from u-V observation using mirror formula.
take average of an value of f.
cr
Exp. To find focus distance of a convex lens using u-V method.
For difference u, we measure different V, and find f using lens formula.
1 1 1


f
V u
In this experiment, a convex lens is fixed at position L and a Knitting needle is
used as an object. This needle is called object needle AB in fig.
First of all we make a rough estimation of F for estimating F roughly, make a
sharp image of a far away object (like sun) on a filter paper. The image distance
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of the far object will be an approx. estimation of focus distance.
Now, the object needle is kept beyond F, so that its real and inverted image
can be formed. To locate the position of the image, use a second needle and
shift this needle such that its peak coincide with the image. The second needle
gives the distance of image (V), so it called "image needle".
Note the object distance "u" and image distance "V" from the him scale on
Similarly take 4 – 5 more observations.
itj
ee
1 1 1


f
V u
.in
optical bench.
cr
ac
ki
Take average of all f.
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