3 Units & Dimensions

Tunneling microscope:
1 Measurement of Length Mass
& time Uses quantum tunneling for atomic-level
objects.
Parallax Used to measure large Electron microscope:
Method distance
Uses electrons for high-resolution imaging.
p
Measurements:
Pa r a l l a x θ
x b Astronomical Unit (AU):
Angle =
x
 Used to measure distances within our

b solar system.

Where, 1AU = 1.496 × 10

11
m
 θ = The angle formed between two lines of Light year (Ly):
sight to an object when observed from two  The distance that light travels in one year.
distinct positions. 15
1Ly = 9.46 × 10 m
 b = The distance between the two observation
Parsec:
points
 Another unit used in astronomy to
 x = The actual distance to the object being
measure large distances.
measured
1parsec = 3.08 × 1016m
Note:
+ –15
 Size of proton (P ) = 10 m (1 fermi)
The conversion between degrees, minutes,
–10
and seconds to radians helps in precise  Size of an atom, = 10 m (1A°)
calculation and understanding of the small  Radius of earth, REarth = 10 m
7

angles involved in parallax measurements.

 Distance of boundary of
= 1026 m
 1º = 60' = 1.745 × 10
-2
rad Observable Universe
-4
 1΄ = 60' = 2.91 × 10 rad
2 Conversion of Units
-6
 1˝ = 4.85 × 10 rad
 nu = constant, n = measure value of Physical
Measuring Very Small Sizes:
quantity, u = unit of that Physical quantity.
Optical microscope: 1
n ∝
For small objects visible with light. u
 n nu 3 SI System (International
System of Units)

The SI (International System of Units) is a

modern version of the metric system, which
u u
is used worldwide for measuring different
n things. It is based on seven fundamental
units, called base units. Each base unit has a
specific symbol and represents a fundamental
quantity in science and everyday life. Here
1/u is a simple explanation of these base units,
their symbols, and what they measure:
→ Only use to find value of physical quantity
in new system of unit, if value is know in 7 Base/Fundamental Units:-
one unit.
No. Quantity Unit Symbol
Measurement of Mass & Time
1. Length Meter m
1 12
 1 amu =
12 Mass of C atom 2. Mass Kilogram kg

 1 amu = 1.66 × 10
-27
kg 3. Time Second S
–
 e mass = 9.1 × 10-31 kg 4. Temperature Kelvin K
25
 Earth mass:- 10 kg 5. Elec. Current Ampere A
55
 Observable Universe =10 kg. 6. Luminous Candela cd
Intensity
Q.1. Convert 18 km/hr in m/s. s
7. Amt. of Sub Mole Mol
Sol. n1u1 = n2u2
2 supplementary Units:-
18 km/hr = n2 m/s
No. Quantity Unit Symbol
18 × 10 3 m
= n 2 m/s 1. Plane Angle radian rad
60 × 60s
5 2. Solid Angle Steradian sr
n 2 = 18 × =5
18 Plane Angle
Q.2. If unit of length is y m in new system of It is the ratio of the arc length (ds) to the
unit then find value of x m2 area in new radius (r) of the circle. It is measured in
system of unit. radians, which is a dimensionless unit.
Sol. un = cost
r ds
n1u1 = n2u2
dθ
x m2 = n2y2m2
o
x ds
n2 =
y 2 dθ =
r

25
Units & Dimensions
Solid Angle Rule of rounding off Numbers

It is the ratio of the area (dA) on the surface  If the next digit is less than 5: Keep the
of a sphere to the square of the radius (r) of last significant figure as it is and drop all
the sphere. It is measured in steradians (sr), digits to the right.
which is also a dimensionless unit.
Example: Rounding 3.142 to three
significant figures: 3.142 becomes 3.14.
r dA  If the next digit is 5 or greater: Increase
dΩ
the last significant figure by one and drop
all digits to the right.
o
dA Example: Rounding 3.146 to three
dΩ = sr significant figures: 3.146 becomes 3.15.
r2

4 Significant Figures Addition & Substraction:-

Final result should have same no. of decimal

The digits in a number that contribute to its
precision and accuracy are known as significant placed as that of original no. with minimum
figure. no. of decimal places.

Properties of significant figure: 3.1421 9.99

0.241 - 0 · 00 99
 All non-zero digits are significant. Ans:-3.47. Ans:-9.98
0.09 9 · 9801
eg:- 42.3 → 3 S.F. 3.4731 Ans: 9.98

243.4 → 4 S.F.
Multiplication & Division:-
 Zero between two non-zero digits is
significant. The no. of S.F. equals the smallest no. of S.F.
in any of the original no.
eg:- 4.03 → 3 S.F.
51.028
243.4 → 4 S.F. 25.5
× 1.31 Ans:- 66.8 5 Ans:- 66.8
 Leading Zero or zeros placed to left are 66.84668
never significant.
eg:- 0.543 → 3 S.F. 5 Dimensional Analysis
0.006 → 1 S.F.
Dimension of physical quantity are power to
 Trailing zeros or zero placed to the right of
which units of base quantity are raised.
the number after decimal are significant.
eg:- 4.330 → 4 S.F. eg:- [M]a [L]b [T]c [A]a [K]C

343.000 → 6 S.F.
Dimension of some physical quantity
 
In exponential expression the numerical
 Mass → M
portion given the number of S.F.
eg:- 1.32 × 10-2 → 3 S.F.  Length → L

1.32 ×.104 → 3 S.F.  Time → T

26
Physics
 –1
 Velocity → LT Equate the Dimensions:
–2
 Acceleration → LT Substitute the dimensional formulas of each
–2 quantity into the equation and equate the
 Force → MLT
dimensions on both sides.
2 –2
 Energy → ML T
Solve for the Exponents: Solve the resulting
2 –3
 Power → ML T system of equations to find the values of x,
–2 y, and z.
 Force gradiant → MT

-:Applications:- Q.3. If momentum (p), area (A) and time

(t) are to be fundamental quantities,
1. Checking the Correctness of various energy has the dimensioanl formula
formulae:- (a) [p1A–1t–1] (b) [p2A1t1]
eg:- Z = A +B (c) [p1A–1/2t1] (d) [p1A–1/2t–1]
[Z] = [A] = [B] Sol. Fundamental quantites are momentum
(p), area (A) and time (t). therefore
2. Conversion of one system of unit to x y z
E = kp A t
another.
Substituting the dimensions
n1 U1 = n2 U2
[ML2T–2]= [ML T–1]a [M0L2T0]b [M0L0T]c
A B C A B C
eg:- n1 [M1 L1 T1 ] = n2[M2 L2 T2 ] Comparing power
a + 2b = 2
M2 A
L2 B
T2 C

  n1 = n2 a=1
M1 L1 T1 –a + c =–2
Solving these equations
MR*
1
Different physical quantity ka dimension a = 1, b = , c = –1
2
nikalne ke liye force and energy ka dimension 1/2 –1
Hence, E = pA T
yad rakna hai. Avi tension nahi lena aage
ke chapter ke sath yad hota jayga. Note:
 Formula of force to find dimension of
3. Creation of new formule different physical quantity

Using the principle of homogeneity of m1m2 q1 q2

F =G 2
,F =
dimension, we can drive the formula of r 4ϖε0r2
physical quantity: μ0I1I2d
F = Kx, F =
Assume the physical quantity Y depends on 4ϖr
a, b, and c, which have dimensions [a], [b], F = qE
and (c) respectively. Write the equation in F = qvB
the form: F = 6ϖηrv
x y z
Y = k · a · b · c F = Sl

27
Units & Dimensions
 Formula of energy to find dimn of MR*
different physical quantity Resistance = R = ωL = 1
ωC
E = hf H kA∆T
= 2ϖ
3 t l ω=
E = kBT T
2 Stress = γ × Strain
L
Q2 Time = = LC = RC
E = Q = ms∆T R
2C
Q = mL
H = I2Rt Dimensionless Quantities:-
1  Strain
E = LI2
2
 Refractive index
PV = nRT
 Relative density

6 Dimensional Formula  Plane Angle

 Solid Angle
 Pressure = stress = Young’s modules
= ML-1 T-2  Poissons ratio
2
 Work = Energy = Torque = M L T-2  Exponential function
2 -3
 Power P = M L T
 Trigonometry function
–1 3 –2
 Gravitational constant G = M L T
 Relative permittivity
-2
 Force constant = Spring constant = M T
-1
 Pure number
 Coefficient of viscosity = ML T-1
2 -2  Efficiency
 Latent heat L = L T
P 2 -3 -1
 Length gratiant
 Electric potential = = M L T A
I
 Coef. of friction
μ0 2 -3 -2
 Resistance = = M L T A
ε0 MR*
Pressure = Stress = Young modulus
-1
 Capacitance = M L-2 T4 A2
= Bulk modulus
-1
 Permittivity ε0 = M L-3 T4 A2 1
= strain × stress = modulus of regitity
 Angular momentum = planck’s constant 2
= M1 L2 T-1 B2 1 nRT
= = ε0E2 = energy density =
Time Period:- 2μ0 2 V
dimensionally addition, substraction ko
L M R
Tα α α
g k g equal le ke solve karte hai.
Kisi be dimensionless function ya quantity
L
= RC = LC ko one likh sakte hai.
R

28
Physics
Q.4. If velocity V = Ax + Bt + C find dimension β
of A, B and C. Q.6. Acceleration a = αt + find dimension
t– δ
MR*
of α, β and δ.
Sol. V = Ax = Bt = C
V –1 β MR* Ka feel
A = = T Sol. a = αt = t = δ
x
B = V = LT–2 ⇒δ=t
t a
–3
⇒α= = LT
C = V = LT–1 t β
⇒ a=
t
Q.5. Force F = αe–βt then find dimension of α
and β. β = αt = LT–1

Sol. F = α βt = 1
Q.7. Fill in the blanks with correct statement,
–2
α = MLT β = T–1
according to given statement

Dimension (1) (2) (c) A physical quantity (d) A physical quantity

.............. .............. have dimension does not have dimension
Unit (a) A physical (b) A physical (3) (4)
quantity have quantity does .............. ..............
unit not have unit
MR*
Ans. (1) May have dimension/may be dimensionless
(2) Must be dimensionless/does not have dimension
(3) Must have unit
(4) May or may not have unit.

Q.8. Fill in the blanks with correct statement, according to given statement

Physically (1) (2) (c) Equation is (d) Equation is

correctness .............. .............. physically wrong physically correct
Dimensional (a) Equation (b) Equation (3) (4)
correctness is dimensional is dimensional .............. ..............
wrong correct
MR*

Ans. (1) Must be physically wrong          (2) May or may not physically correct
(3) May or may be dimensionally correct   (4) Must be dimensionally correct.
a
Snth = u + (2n – 1)
2
(Snth → dimensionally correct because it is displacement in one sec.)

29
Units & Dimensions
Q.9. If force, acceleration and time taken (a) 1, 1, 1 (b) 1, –1, –1
as fundamental physical quantity then (c) –1, –1, 1 (d) –1, –1, –1
find dimension of energy?
MR*
2 –1 2
(a) F A T (b) FAT Velocity me mass hai nahi to η, δ and
r ko arrange kar velocity lena hai hence
(c) F–1AT–2 (d) FA–1T
mass cancell, radius me bhi mass nahi
–3 –1 –3
MR* hai, δ = ML and η = ML T
E(ML2T–2) → Mass ka dimension force hi δ and η divide karne se mass kat jayga
dega ek mass energy me hai to F1 hona to ek ka power positive ek ka negative
chahiye. hona chahiya.

Now L ka square hai ek length force dega Q.12. If energy (E), velocity (V) and time
ek acceleration hence A1 hona chahiye. (T) are chosen as the fundamental
quantities the dimensional formula of
Q.10. Planks constant (h), speed of light (c), surface tension will be
gravitational constant (G) taken as 2 2
(a) EV T (b) EV2T–2
fundamental quantity then dimension
(c) EV–2T–2 (d) E–2V2T2
of length in terms of them.
MR*
hG hc –2
MR* → Surface tension (MT ) Ramlal
(a) (b)
c 3/ 2
G yaha length nahi to length katne ka
socho. Sirf (c) me length kat ho raha
hG Gc hai.
(c) (d)
c5 / 2 h 3/ 2

Limitation of dimensional analysis:

MR*
0 0 X Y Z
(1) It is not use to derive dimensionless
M T l = h c G physicall quantity and constant.
We need dimension of length, then mass (2) This can not decide weather the give
should be cancell out by arranging h, quantity is vector or a scalar.
c and G. c me to mass hai nahi; h → (3) It can not be use to derive an equation
2 –1
ML T and G = M–1L3T–2 to h and G involving more than three physicall
ko multiply karne se mass kat jayga. quantity.
Hence option (b) and (d) wrong ho (4) It can not derive dimensionless function
x
gaya. Now option (a) and (c) dono me having sinθ, cosθ, e etc.
root hai to root laga ke sirf length ka (5) Can not use if one quantity depends
dimension likho phir c se divide kar ke on two other quantity having same
1
ek length (L ) sirf rakho. dimension.
(6) It can not derive equation which cantain
Q.11. Dimension of critical velocity V of liquid
+ve and –ve terms.
flowing through the tube are expressed
as ηxδyrz, where η is coefficient of 7 Instruments
viscosity, δ is density of liquid and r is
radius of the tube then the value of x, y  Least Count (L.C) :-

and z are given by.  Smallest value that

30
Physics
mm Vernier Screw n-1
  1 VSD = n MSD
Scale Scale Gauge

  n-1 1MSD
 LC = 1 MSD – MSD = n
1mm 0.1mm 0.01mm n

Vernier calipers:-  Total = 1 MSR + coinciding ×LC

Reading VSR
 L. C. = 1MSD - 1VSD

 If nVsD Coincides with (n-1) MSD then:- MSD → Main Scale division
(n-1) MSD = nVSD V.S.D → Vernier scale devision

Vernier Screw
Jaws scale clamp
for measuring
outer
dimensions

Main scale Stem for measuring

depths

0 1 2 4 6 7 8 9 10

Jaws
for measuring
inner
dimensions

Screw gauge:-
Measured length Used instrument
Stud Pitch Scale Circular Scale
Screw 1.56 cm vernier calliper
Ratchet
6.8 cm metre scale
8.96 mm screw gauge
LSR = Linear 9.812 cm screw gaye
Thimble Scale

frame
Reading
8.3 mm vernier calliper
CSR = Circular
Accuracy: It is the measure of how close the
Scale
Sleeve Reading

measured value is to the true value. Closness

MSR
 Pitch = of measured and true value.
no. of rotation
Precision: It tells us to what resolution or
Pitch
limit the quantity is measured.
 L.C. =
Total no. of division on Circular Scale
Zero error:
 Total = 1 LSR +CSR × LC
Reading Types:
 Positive Zero Error: When the zero of the
MSR → Main Scale division
vernier scale is to the right of the zero of
CSR → Circular scale reading
the main scale.

31
Units & Dimensions
 Negative Zero Error: When the zero of the So, Z = –VC × 0 = – 0 mm
vernier scale is to the left of the zero of \ L = MSR + xVC – Z
the main scale.
= 55 + (8 × 0.1) – (–0) = 55.8 mm
Calculation:
Q.14 In a vernier caliper, when the jaws are
 Positive Zero Error: Number of divisions touching each other the zero mark on
of the vernier scale (VS) that coincide the vernier scale is left behind slightly
with the main scale (MS) multiplied by of the zero mark on the main scale. At
the least count (LC). that moment the 5th division mark on
 Negative Zero Error: (Total divisions on vernier caliper matches perfectly with
vernier scale - Number of divisions of VS a division on the main scale. When the
that coincide with MS) multiplied by LC. zero marks in both the scale coincides,
10 vernier scale divisions matches
Correction:
with 9 main scale divisions (in mm). A
 Correct Reading for Positive Zero Error: rod of length l is measured with this
 Correct Reading=Observed Reading− vernier caliper. It shows 55 in main
Positive Zero Error scale reading and in the vernier scale
8th mark matches with a main scale
 Correct Reading for Negative Zero Error:
division. What is the value of L?
Correct Reading = Observed Reading −
Negative Zero Error Sol. The verniner constat,

9
Q.13. I n a vernier caliper, when the jaws are VC = 1 – mm = 0.1 mm
10
touching each other the zero marks
in both the scale coincides, 10 vernier Here the zero mark in verniner scale is
scale divisions matches with 9 main left behind the zero mark in main scale.
scale divisions (in mm). A rod of length So, a negative zero error exists.
L is measured with this vernier caliper.
So, Z = –VC × 5 = – 0.5 mm
if It shows 55 in main scale reading and
in the vernier scale 8th mark matches \ L = MSR + xVC – Z

with a main scale division. What is the = 55 + (8 × 0.1) – (–0.5) = 56.3 mm

value of L?
Q.15. If true value of length is 6.57 m then
Sol. The verniner constat, which of the following reading is most
9 accurate and most precise.
VC = 1 – mm = 0.1 mm
10 (a) 6.52 m (b) 6.61 m
Here the zero mark in vernier scale
(c) 6.513 m (d) 6.68 m
matches with the zero mark in main
scale. So, no zero error exists. Sol. Most accurate (b), most precise (c)

32
Physics
 8 Error in Measurement

Difference between true value

& measured value of a quantity

Systematic Errors Random Errors

Errors which tend to occur Irregular and at random
only in one direction, either in magnitude & direction
positive or negative

Instrumental Experimental Personal

Due to inbuilt defect Limitation in Due to individual bias,
of measuring experimental Lack of proper setting
instrument technique of apparatus

Absolute Error:-  Unit and dimension less

Δa = | ai – amean |  It tell about accuracy of measurment

a1 + a2 .... +an
amean =
 Random error can be decreased by
n
increasing no of obserbation.
 Always positive
 Unit and dimension same as physical nx = costn, n = no. of obserbation,
quantity
x = Random error
 Least count error can be taken as
absolute error n
 It can not tell about accuracy of
measurment

Mean Absolute error

Δa1 + Δa2 .... + Δan

Δamean = x
n

Relative Error:-  In 5 reading random error is 3% and

Δ amean systematic error is 4%. If we increased no.

amean of observation to 30 then random error

Δamean 1/2% and systematic error remains 4%.

Percentage Error = × 100
amean

33
Units & Dimensions
Combination of errors

Formula Absolute error Relative error Percentage error

Operations
Z ΔZ ΔZ/Z 100 × ΔZ/Z
Sum A + B ΔA + ΔB ΔA + ΔB ΔA + ΔB
× 100
A+B A+B
Difference A – B ΔA + ΔB ΔA + ΔB ΔA + ΔB
× 100
A–B A–B
Multiplication A × B AΔB + BΔA ΔA ΔB ΔA ΔB
+ + × 100
A B A B
Division A BΔA + AΔB ΔA ΔB ΔA ΔB
+ + × 100
B B2 A B A B
Power An nAn–1 ΔA ΔA ΔA
n n × 100
A A
Root A1/n 1 1 ΔA 1 ΔA
A((1/n) – 1)ΔA × 100
n n A n A

General Rule:- Q.16. In an experiment the angles are

p q
A B required to be measured using an
Z =
,
r
C instrument. 29 divisions of the main
Then max. fracn relative error in Z will be:-
scale exactly coincide with the 30
ΔZ ΔA ΔB ΔC divisions of the vernier scale. If the
= P + q +r
Z A B C
smallest division of the main scale is
MR* half-a-degree (= 0.5°) then the least
 Addition/Substration me pahle absolute
count of the instrument is:
error nikalenge phir relative.
 Power/multiplication/division me pahle
(a) one minute (b) half minute
relative error nikalenge phir absolute. (c) one degree (d) half degree
Example: y = 3A2 ← Power hai to direct
relative error likho, constant ko remove Sol. MSD = 0.5°
karo, power ko aage multiply kar do. 30 VSD = 29 MSD
∆y ∆A Least Count
= 2×
y A
= Length of 1 main scale division / No.
Example:
of divisions of Vernier scale
4
y = 2A B
3 = 0.5° / 30
C
∆y ∆A 1 ∆B ∆C = 0.5 × 60min / 30
= 4× + × + 3×
y A 2 B C = 1 min

34
Physics
Q.17. 
A vernier callipers has 1 mm marks 4
Sol. A = 4πr2 V = πr3
on the main scale. It has 20 equal 3
divisions on the Vernier scale which ΔA Δr
=2 ...(i)
match with 16 main scale divisions. A r
For this Vernier callipers, the least ΔV Δr
=3 ...(ii)
count V r

(a) 0.02 mm (b) 0.05 mm ΔV ΔA 3

(ii)/(i) =
V A 2
(c) 0.1 mm (d) 0.2 mm
ΔV 3 ΔA 3
Sol. MSD = 1 mm × 100 = × 100 = ×3
V 2 A 2
20 VSD = 16 MSD = 4.5%
⇒ VSD = 16/20 MSD ℓ
Q.19. 
If T = 2π then find percentage
g
= 0.8 MSD = 0.8 mm
error in measurement of acceleration
Least Count = MSD–VSD due to gravity.
= 1–0.8 = 0.2 mm Sol. Ignore constant
ℓ ℓ
Q.18. If the error in the measurement of area 2
T = ,g=
g T2
of sphere is 3% then find percentage
Δg Δℓ ΔT
error in measurement of volume of = +2
sphere g ℓ T

MR*
[kqy tk;saxs lHkh jkLrs]
rw #dkoVksa ls yM+ rks lghA
lc gksxk gkfly]
rw viuh ftn~n ij vM+ rks lghAA

35
Units & Dimensions
 36
Y

Fa
LI1I2

2
Planck's

HL
K = Q Q
tAT

y
ML2T-2
S.H =
MK

Physics