Class – 11 Physics, Chapter – 2 : Units and Measurement
Physical quantity : A quantity that can be measured is called as physical quantity.
It has two types according to magnitude and direction , one is vector quantity and other is scalar quantity.
Vector quantity can be defined as the physical quantity having both magnitude and direction. Force, velocity, momentum and acceleration are the physical quantity as they have both magnitude and direction.
Scalar quantity is the physical quantity having only magnitude and has no direction. Speed, work, mass, volume and current are the scalar quantities as they have only magnitude.
NOTE : The topic “vector” will become a chapter in the coming unit. There, we will discuss this topic ( vector ) in details.
On the basis of units, the physical quantities can be classified in two categories. one is Fundamental and other is Derived Physical Quantity.
Fundamental Physical Quantity : The physical quantities which can not be defined in terms of other physical quantities are called as fundamental quantities. Mass , length and time are the basic or fundamental quantities.
Derived Physical Quantities : The physical quantities that can be defined in terms of other physical quantities, are termed as derived physical quantities.
speed=distance/time
Speed can be defined in terms of distance and time (fundamental physical quantities ). Hence speed is the derived physical quantity.
Area = length * breadth → (in terms of one basic physical quantity i.e length )
Density = mass / volume → (in terms of two basic physical quantities i.e mass and length )
Momentum = mass * velocity
Momentum = mass * (distance/time) → (in terms of three fundamental physical quantities i.e mass, length and time )
Unit : The unit can be defined as the reference standard used to measure a physical quantity.
Characteristics of Units :
→ It should be of convenient size
→ It should be well defined
→ It should be easily available so that as many laboratories as possible can duplicate
→ It should not change with the change in physical condition like temperature, pressure etc
→ It should not change with time and space
→ It should be universally agreed upon so that result agreed in different countries are comparable
Types of unit :
There are two types of unit one is fundamental unit and derived unit.
Fundamental Units : Fundamental unit are the units of fundamental physical quantities like cm, m, sec, hour, min, gm, kg
Derived Units : Derived units are the units of derived physical quantities like cm/sec, m/sec, m/sec^{2}, newton etc.
System of Units :
There are three system of units : –
→ International System of Units or S.I System
→ C.G.S System
→ F.P.S System
Physical Quantities S.I Unit C.G.S Unit F.P.S Unit
Mass kg gm Pound
Length m cm foot or feet
Time sec sec sec
Significant Figures : The digits which tell us the no of units we are reasonably sure of having counted in making a measurement are called as significant figure.
Rules for determining the no of significant figures :
 All non zero digits are significant
132.73 contains 5 significant figures
111.111 contains 6 significant figures
1.1 contains 2 significant figures  All zeros between two non zero digits are significant
1007.001 contains 7 significant figure
207.009 contains 6 significant figure
1.0002 contains 5 significant figure  If there is no decimal point, zeros to the right of the right most non zero digits are significant if they come from a measurement
400m contains 3 significant figure
While 307000 contains only three significant figure i.e All zeros two the left of an understood decimal point but to the right of a non zero digits are not significant
 All zeros to the left of an understood decimal point but to the right of a non zero digit are significant if they come from measurement .
 All zeros to the left of an expressed decimal point and to the right of a non zero digits are significant
307000m contains 6 significant figure
4000m contains 4 significant figure  All zeros to the right of decimal point but to the left of non zero digits are not
significant provided there is only a zero to the left of the decimal point .
0.000345 m contains only 3 significant figure .
Here , single zero , left of the decimal point is never significant .  All zeros to the right of the decimal point and to the right of non zero digits are significant
0.0320 contains 4 significant figures  The no of significant figures does not vary with the choice of different units
Problem 1 : State the no of significant figures in 600900.
Answers : The no .of significant figure is 4 in 600900
Explanation : a) All non zero digits are significant .
b) All zeros between two non zero digits are significant
c) All zeros to the left of an understood decimal point but to the right of a non
Zero digits are not significant, that’s why last two zero of the given no (60900) are not considered as significant.
Problem 2 : Find the number of significant figure in 400m.
Answer : The number of significant figure is 3 in 400m.
Explanations : a) All non zero digits are significant.
b) All zeros to the left of an understood decimal point but to the right of non zero digit are significant if they come from measurement.
Problem 3: What is the number of significant figure in 978.850?
Answer : The number of significant figure is 6 in 978.850
Explanation : a) All non zero digits are significant
b) All zeros to the right of a decimal point but to the left of a non zero digit are significant
Problem 4 :There are how many significant figures in 70?
Answer : The number of significant figure is 2
Explanation : a) All non zero digits are significant.
b) All zeros to the left of an expressed decimal point and to the right of a non zero digit are significant
Problem 5 : Find the no . of significant figure in 0.007
Answer : The number of significant figure is 1
Explanations : a) All non zero digits are significant
b) All zeros to the right of a decimal point but left to the non zero digit are not significant provided there is only a zero to the left of a decimal point
Problem 6 : How many significant figures are there in 9.1*1031?
Answer : The number of significant figure is 2
Explanation : a) All non zero digits are significant.
Problem 7 : How many significant figures are there in 0.045?
Answer : The number of significant figure is 2
Explanation: Same as explanation of question no – (5)
Problem 8 : State the number of significant figures in 25300
Answer : The number of significant figure is 3
Explanations : Same as explanation of question no – (1)
Problem 9 : State the number of significant figures in 7785
Answer : The no .of significant figure is 4
Explanation : All non zero digits are significant
Problem 10 : State the number of significant figures in 984.06
Answer : The number of significant figure is 5
Explanations : a) All non zero digits are significant
b) All zeros between nonzero digits are significant .
Problem 11 : State the number of significant figures in 6.032
Answer : The number of significant figure is 4
Explanation : SSame as explanation of question no – (10)
Problem 12 : How many significant figures are there in 0.0050?
Answer : The number of significant figure is 2
Explanations : a) All non zero digits are significant
b) All zeros to the right of a decimal point but to the left of a non zero digit are significant
c) All zeros to the right of decimal point but to the right of a non zero digit are significant
Problem 13 : State the number of significant figures in 2.64*1024
Answer : The number of significant figure is 3
Explanations : Same as the explanation of question no – (1)
Problem 14 : How many significant figures are there in 0.0006032?
Answer : The number of significant figure is 4
Explanation : Same as the explanation of question no – (7)
Problem 15: How many significant figures are there in 648700?
Answer : The number of significant figure is 4
Explanations : Same as the explanation of question no – (1)
Problem 16: How many significant figures are there in 5212.0?
Answer : The number of significant figure is 5
Explanations : a) All non zero digits are significant
b) All zeros to the right of a decimal point are significant.
Problem 17: How many significant figures are there in 426.071?
Answers: The number of significant figure is 6
Explanations : Same as the explanation of question no – (10)
Problem 18 : How many significant figures are there in 0.0631?
Answer : The number of significant figure is 3
Explanation : Same as the explanation of question no – (12)
Problem 19 : How many significant figures are there in 6.320?
Answer : The number of significant figure is 4
Explanations : a) All zero digits are significant
b) All zeros to the right of decimal point and right of a non zero digit are significant
Problem 20: How many significant figures are there in 0.2370?
Answer : The number of significant figure is 4
Explanation : Same as the explanation of question no – (5)
Rounding off Rules :
 If the digit to be dropped is smaller than 5 ,the preceding digit should be left unchanged . eg. 7.34 is rounded off to 7.3
 If the digit to be dropped is greater than 5 , the preceding digit is added by 1 . As for example 9.58 is rounded off to 9.6
 If the digit to be dropped is 5 , followed by zero , then the preceding digit is left unchanged if it is even and is added by 1 if it is odd . As for example 7.250is rounded off to 7.2 and 8.45 is rounded off to 8.4 where as 6.350 is rounded off to 6.4 and 7.75 is rounded off to 7.8.
Problem : Solve the following with due regard to significant figures and rounding off.
A —> 5.8 + 0.125
Solution : 5.800 + 0.125 = 5.925
Required answer is 5.9
Note : Second decimal place is less than 5 so first decimal place is left as it is
B —> 9.15 + 3.8
Solution : 9.15 + 3.80 = 12.95
Required answer is 13.0
Note : Second decimal place is equal to 5 , hence 1 is added to first decimal place
C —> 3.5 + 2.51
Solution : 3.50 – 2.51 = 0.99
Required answer is 1.0
Note : Here second decimal place is greater than 5 so 1 is added to first decimal place
D —> (3.5 – 3.31)½
Solution: (3.503.31)½ = (0.19)½
=0.4358 ( by using scientific calculator )
Required answer is 0.4
Note : Here second decimal place is less than 5 so first decimal place is left as it is
E —> 3.9 * 10^{5}– 2.5 * 10^{4}
Solution:
3.9 * 10^{5 }– 2.5 * 104 = 39 * 10^{4 }– 2.5 * 10^{4}
=(39.0 – 2.5) *10^{4}
=36.5*10^{4}
=3.65*10^{5}
=3.6*10^{5}
Errors in Measurement:
The uncertainty in measurement is called error . It is the difference between the measured value and true value of a physical quantity .
Types of error :
There are many types of error . Some of them are in the following
Constant Error : If the error is repeated every time in a series of observation , the error is said to be constant error . It is due to faulty calibration of measuring instrument . So measurement must be made with all possible different methods . The mean value , so obtained may be regarded as true value.
Systematic Error : systematic errors are those errors that occur according to certain pattern or a system these errors are due to known reasons .It can be classified into following four main categories .
 Instrumental Error : If an instrument is faulty and inaccurate then the error occurred due to this inaccuracy in a measurement can be regarded as instrumental error .
 Personal Error : The errors occurred due to the individual qualities of experimenter himself are known as personal error. These error may arise due to the lack of intention , bad sight , habits and peculiarities of the observer . This error can be minimized by performing the measurement by different experimenter.
 The errors due to external source : The errors cause due to change in external condition like pressure ,temperature, wind, humidity etc. are termed as errors due to external source .
 Errors due to internal sources or due to imperfections: These errors are due to limitation of experimental arrangement .
Gross Errors : These errors are due to following reason

 Improper setting of instrument
 Recording observation wrongly
 Not to take into account the source of error and precautions.
 Using some wrong values in calculations
Random Errors : In a common experience that the repeated measurements of quality give values which are slightly different from each other . These error have no set pattern . They take place in a random manner . These errors depend on the errors in measuring process and also on the individual observer.
A_{mean} =(A_{1}+A_{2}+A_{3}+…………………….+A_{n})/n
Where n is no of observation
A_{mean}=1/n ∑A_{i}
How to express an error ?
There are three ways to express an error
Absolute Error :
Absolute Error = ( value of the quantity – individual measurement value )
As we are not sure about the correct or true value of quantity ,then their arithmetic mean value will be considered as the true or correct value.
Mathematical treatment of the absolute error may be discussed in the following way
The absolute error is given by
ΔA_{1 }= A – A_{1}
ΔA_{2 }= A – A_{2}
ΔA_{3 }= A – A_{3}
……………..
ΔA_{n }= A – A_{n}
Where A represents the arithmetic mean
n is the no of observation
Mean absolute error is given by
ΔA_{ }= (ΔA_{1} + ΔA_{2} + ΔA_{3 }+………………..+ ΔA_{n})/n = 1/n∑  A_{i} 
From these discussion we can conclude that
A – ΔA ≤ A_{i} ≤ A + ΔA
Relative Error :
It may be defined as the ratio of mean absolute error to the value of the quantity to be measured ( which may be taken as the arithmetic mean of the values of the quantity )
Relative error = ΔA ÷ A
Where ΔA is absolute error and A is arithmetic mean of the values of the quantity
Percentage Error :
Percentage Error = (ΔA ÷ A )* 100%
Question (1): The density of the material of a cylindrical rod was determined by the formula d=m / (πr^{2}l ) the percentage error in m, r and l are 2 %,1.5 % and 0.8 % respectively. Calculate the maximum possible percentage error in the determination of density.
Solution : Given :
d=m / (πr^{2}l ) —————— (1)
Percentage error in m = △m / m = 2%
Percentage error in r = ∆r / r = 1.5%
Percentage error in l = ∆l / l = 0.8%
Now from equation (1) , we have
d=m / (πr^{2}l )
∆d / d = (∆m/m) + 2* (∆r / r) + ( ∆l / l)
= (2+1.5+0.8) %
=5.8%
Therefore , the maximum percentage error in density is 5.8 %.
Question 2 : The centripetal force is given by F = mv^{2 }/ r , the mass , velocity and radius of the circular path of an object are 0.5 kg, 10 m/sec,.4 m respectively. Find the percentage error in the force. Given m, v and r are measured to accuracy of 0.005kg, 0.01 m/s and 0.01 m respectively.
Solution : Given
F = mv^{2 }/ r —————— (1)
m = 0.5kg & ∆m = 0.005kg
v = 10 m/s & ∆v=0.01m/s
r = 0.4 m & ∆r=0.01m
From equation (1) we have
∆F / F = (∆m / m) + (∆v / v) + (∆r / r)
=(0.005 / 0.5) + (0.01 / 10) + (0.01 / 0.4)
= 0.01 + 0.002 + 0.025
=0.037
(∆F / F)* 100% = 0.037*100%
=3.7%
Therefore , the percentage error in force is 3.7%
Question 3 : Given : specific resistance , ρ = πr^{2}l
Where r = radius of the wire
l = length of the wire
R = resistance of the wire
Calculate the percentage error in p , if ρ = (64 ± 2) ohm,
l = (156.0 ± 0.1) cm
r = (0.26 ± 0.02) cm
Solution :
Given : ρ = πr^{2}l ———————————–(1)
R = 64 ohm & ∆R = 2 ohm
l = 156 cm & ∆l = 0.1cm
r = 0.26 cm & ∆r = 0.02 cm
From equation (1)
ρ = πr^{2}l
∆ρ / ρ = 2* (∆r / r) + (∆R / R) + (∆l / l)
= 2*(0.02 / 0.26) + (2 / 64) + (0.1 / 156)
= 0.1538 + 0.0312 + 0.00064
=0.18564
= 0.186 (by rounding off )
(∆ρ / ρ)*100% = 0.186 * 100%
=18.6%
Therefore ,the percentage error in p is 18.6%