# Dimensional Analysis Question and Answer|Problems on Dimensional Analysis

• 4
Shares

Dimensional Analysis Question and Answer: Friends, if you are searching for Dimensional Analysis Question and Answer then you are on the right on place, today here I am going to provide you Dimensional Analysis Question and Answer. All these Problems on Dimensional Analysis which I am going to provide are very very important for your board as well as for your Entrance Exam.

## Q1. What is the Dimension of Physical Quantities?

Ans: The Dimension of a Physical quantity are the powers to which the fundamental or base quantities are raised to represent that quantity.

## Q2. What is Dimensional Formula?

Ans: The expression which shows how and which of the fundamental or base quantities represent the Dimensions of a physical quantity is called the dimensional formula of the given physical quantity.

E.g. Dimension of Volume= Length X Breadth X Height=[L1] [L1] [L1]=[L3]

=[M0 L3 To]

## Q3. What is Dimensional Equation?

Ans: When a physical quantity is equated to its dimensional formula, the equation obtained is called the dimensional equation of the quantity.

## Q3. Find the Dimensions of Universal gravitational constant?

Ans: According to Newton’s Law of gravitation,

F=Gm1 m2/r2

or G= Fr2/m1m2

or [G]= [M1L1T-2] [L1]2/[M1][M1]

=[M-1L3T-2]

## Q4. What do you mean by Principal of Homogeneity of Dimensions?

Ans: The principal of homogeneity states that the dimensional formula of each term on the two sides of a correct equation must be same.

If the dimensions of all the terms of an equation are not the same, the equation is wrong.

## Q.5 Check the Correctness of the following equation:

a) V= u + at

Ans: Given, V=u + at

The Dimension of each term can be written as-

[V]=[L1T-1] [U]=[L1T-1]

[at]=[L1T-2][T1]

=[L1 T-1 ]

Since all the terms have the same dimensions, the given equation is dimensionally correct.

b) S= ut + 1/2 at2

Ans: Given,

S= ut + 1/2 at2

The Dimension of each term can be written as-

[ S]= [L1] [ut]= [L1T-1][T-1]

=[L1] [1/2  at2]= [L1T-2] [T1] ( 1/2 is a dimensionless constant)

= [L1]

Since all the terms have the same dimensions, the given equation is dimensionally correct.

c) V2= u2 + 2as

Ans:

Given,

V2= u2 +  2as

The Dimension of each term can be written as-

[ V2]= [L1 T-1]2

= [L2T-2] [u2]= [L1T-1]2

=[L2T-2] [2as]= [L1T-2] [T1] ( 2 is a dimensionless constant)

=[L2T-2]

Since all the terms have the same dimensions, the given equation is dimensionally correct.

## Q6. The Velocity(v) of a particle is related to time “t” as V=A + Bt. Find it’s dimensions of “A” and “B”

Ans: Given,

V= A + Bt

By the principle of homogenity of dimensions-

[A] = [V]

= [L1 T-1]

and [Bt]= [V]

or [B]=[V]/[t]

=[L1T-1]/ [L1T-2]

= [ L1T-2]

### Q7. The distance(x) travelled by a particle is related to time “t” as x=at + bt2. Find the dimensions of a and  b.

Ans: Given,

x=at + bt2

By the principal of homogenity of dimension-

[at]= [x]

or [a]=[x]/[t]

or [a]=[L1]/[T-1]

=[ L1 T-1]

& [bt2]=[x]

or [b]= [x]/[t2]

=[L1]/[T1]2

=[L1T-2]

### Q8. If the Pressure, P is given by P=a-x2/bt; where “x” is distance, “t” is time and “a” and “b” are constant. Find the dimensional formulae of “a” and “b”.

Ans: Given,

P=a-x2/bt

[a]=[x2]( Since, like quantities can be subtracted)

=[L2]

Now, Pbt= a-x2

or[M1L-1T-2] X [b][T1]=[L2]

or [b]= [L2]/[M1L-1T-2] [T1]

=[M-1L3T1]

Thus, [a]= [M0L2T0] [b]=[M-1L3T1]

### Q9. What are the uses of Dimensional analysis?

Ans: The uses of dimensional analysis are given below:

• To check the correctness of a given relation.
• To derive relationship among various physical quantities.
• To convert the value of a physical quantity from one system of units to another.

### Q 10. The time period of a simple pendulum(l) and  acceleration due to gravity(g) at a place. Find the expression for time period by the method of dimension.

Ans: Let, T∝ ma

∝ lb

∝ gc

Combining,

T∝ ma lb gc

or T= K ma lb gc ———(1)

Where K is a dimensionless constant.

Writing the dimensions of all the physical quantities in equatition (1) we get,

[T1]= [M1]a [L1]b [L1T-2]c

or [M0L0T0]= [Ma La+b T-2c]

Equating the powers of M, L and T, on the both sides we get,

a=0

b+c= 0 ———-(2)

-2c= 1 or c=-1/2

Equation (2) ⇒

b-1/2= 0

or b= 1/2

Equation (1) ⇒

T= K m0 L1/2 g-1/2

or T= K√l X 1/√g

or T= K√l/√g

Experimentally,

K= 2π

∴ T= 2π√l/√g

### Q 11. The centripetal force(F) acting on a body depends on mass of the body(m), velocity(v) of the body and radius of the circle. Derive the expression for centripetal force by the method of dimension.

Ans: Let, F∝ ma

∝ vb

∝ rc

Combining,

F∝ ma vb rc

or F= K ma vb rc ————(1) ; where K is a dimensionless constant.

Writing the dimensions of all the physical quantities in equation(1) we get,

[M1 L1 T-2]= [M1]a [L1 T-1]b [L1]c

or [M1 L1 T-2 = [Ma Lb+c T-b]

Equating the power of M, L and T on both sides we get,

a= 1

b+c= 1 —————(2)

-b= -2

or b= 2

Putting b= 2 in (2) we get,

2+c= 1

⇒c= 1-2

⇒c= -1

Equation (1) ⇒

F= K m1 v2 r-1

or F= K mv2/r

Experimentally, K= 1

∴ F= mv2/r

### Q12. Name Some Physical Quantities which same dimensional formula:

• Impulse and momentum.
• Work, moment of force, energy, energy, torque.
• Force constant, surface energy and surface tension.
• Stress, pressure and modulus of elasticity.
• Angular velocity, velocity gradient and frequency.
• Angular momentum, Planck’s constant and rotational impulse.
• Force and thrust.
• Latent heat, gravitational potential.
• Power and luminous flux
• Thermal capacity, entropy, universal gas constant and Boltzmann’s constant.

### Q13. Convert 10 Newton into dyne.

Ans: Dimensional formula of force

[F]= [M1L1T-2]

∴ a= 1, b=1 and c= -2

In SI: M1= 1kg,    L1= 1m,   T1= 1s  and  n1= 10

In CGS: M2= 1g,   L2= 1cm,     T2= 1s  and  n2= ?

Using, n2=n1 [M1/M2]a [L1/L2]b [T1/T2]c

=10 [1kg/1g]1 [1m/1cm]1 [1s/1s]-2

=10 X 103g/1g X 102cm/1cm X 1

=10 X 105

=106

∴ 10N = 106 dyne

### Q13. (NCERT 2.3) A calorie is a unit of heat or energy and it equals about 4.2 J where 1J= 1kg m2 s-2. Suppose we employ a system of units in  which the unit of mass is γ s. Show that a calorie has a magnitude 4.2 α -1 β-2 γ2 in terms of the new units.

Ans: Here,

1cal= 4.2 J

1 J= 1kg m2 s-2

Dimensional formula of energy,

[E]= [M1 L2 T-2]

∴ a= 1,   b=2,   and c= -2

In SI: M1=1kg,  L1=1m,  T1=1s,  and n1=4.2

In New System: M2= α kg,   L2= β m,  T2=γ s  and   n2= ?

Using, n2=n1 [M1/M2]a [L1/L2]b [T1/T2]c

=4.2 [1kg/αkg]1 [1m/β m]2 [1s/γ s]-2

=4.2 X 1/α X 1/β2 X 1/γ-2

∴ 1Cal= 4.2 J= 4.2 α-1 β-2 γ2 in terms of new units.

### Q 13. What are the limitations of dimensional analysis?

Ans: The limitations of dimensional analysis are :

• Dimensionless constants cannot be obtained by this method.
• Dimension method cannot provide exact expression for a physical quantity.
• It does not distinguish between physical quantities having same dimension.
• A dimensionally correct equation may not be always correct.

### Q 14. Define dimensional variables.

Ans: The physical quantities which have dimensions and posses variable values are called dimensional variables.

e.g: Area, Volume, Velocity, Acceleration etc.

### Q 15. Define dimensionless variables.

Ans: The quantities which have no dimensions but posses variable values are called dimensionless variables.

e.g: Angle, Strain, Sinθ, Cosθ etc.

### Q 16. Define dimensional constant.

Ans: The quantities which have dimension and posses constant values are called dimensional constants.

e.g: G= 6.67 X 10-11 Nm2 Kg-2,  Plank’s Constant(h)= 6.63 X 10-34 Js

### Q 17. Define Dimensionless Constant.

Ans: The quantities which have no dimensions but posses constant values are called dimensionless constant.

e.g: π= 3.14, e=2.718, 1, 2, 3 etc.

### Q 18. Taking Force (F), Length(L) and time(T) as fundamental quantities. Find the dimensions of-

a) Pressure

b) Energy

Ans: a) Pressure,

P= Force/ Area

⇒ [P]= [F1]/ [L2]

= [F1 L-2 T0]

b) Energy,

E=1/2 mv2

or E= 1/2 . F/a . v2 ——– (∵F= ma ⇒ m=F/a)

or [E]= [F1]/ [L1 T-2] X [L1 T-2]

= [F1]/[L1 T-2] X [L2 T-2]

= [F1L1 T0]

### Q 19. What are the seven dimensions of physical world?

Ans: The seven dimensions of the physical world are :

• Length—[L]
• Mass—[M]
• Time—[T]
• Electric Current—-[A]
• Thermodynamic temperature—[K]
• Amount of substance—[mol]
• Luminious intensity—[cd]

or [Area]= [L1] X [L1]

=[L2]

=[M0 L2 T0]

Thus the dimensions of area are 0,2 and 0 in mass, length and time respectively.

Important Points:

• We can add or subtract similar physical quantities.
• If the dimensions of all the terms of an equation are not same, the equation is wrong.
• For an equation to be correct, it should be dimensionally correct.
• A dimensionally correct may not always be correct. e.g- the equation S=ut + at2 is dimensionally correct but it is not a correct equation.
• [L] + [L]= [L]
• [L] – [L]= [L]
• All the trigonometric ratios are dimensionless.                                                                                                 i.e.[sinθ]=[M0 L0 T0] or  [θ]=[M0L0T0]
• The logarithmic functions are dimensionless.                                                                                                            i.e. [log e
• Exponential functions are dimensionless.                                                                                                                   i.e. [ex]= [M0L0T0]    or  [x]= [M0L0T0]

### Q 20. What are the types of physical quantities?

Ans: There are two types of physical quantities-

1) Fundamental or Base quantities: The physical quantities which are independent of one another and in      terms of which all other quantities can be expressed are called fundamental or base quantities.

e.g: Length, Mass, Time etc.

2) Derives Quantities: The physical quantities which can be expressed in terms of fundamental or base quantities are called derived quantities.

e.g: Area, Volume, Speed, Work, etc.

### Q 21. What is Unit?

Ans: The quantity which is used as the refernce standard for the measurement of a physical quantity is known as the unit.

There are two types of unit:

a) Fundamental or base Units: The units in terms of which all possible physical quantities can be expressed are known as fundamental or base units.

e.g: The units of length, mass, time, electric current etc.

b) Derived Units: The units which can be expressed in terms of fundamental or base units are called derived units.

e.g: The units of area, volume, density, velocity, acceleration etc.

System of units:

System  ————    Length  ————–  Mass  ————– Time

a) CGS   —- —–  Centimetre(cm) ———-gram(g) ———–Second(s)

b) FPS —————- Foot(ft.) ————–pound(lb) ———-Second(s)

c) MKS ————– metre(m) ————-kilogram(kg) ——–Second(s)

### Q 22. What are advantages of SI Units?

Ans: The advantages of SI units are gven below:

• S.I is coherent sysytem of units, i.e. all derived units can be obtained by simple multiplication or division of fundamental units without introducing any numerical factor.
• S.I is a rational sysytem of units, i.e. it uses only one unit for a given physical quantity.
• S.I is a metric system, i.e the multiples and submultiples of SI units can be expressed as powers of 10.

Metric Prefixes:

a) Multiples

•  Kilo—–K—-103
• Mega—-M—106
• Giga—–G—-109
• Tera—–T—–1012

b) Submultiples

• Milli—–m—–10-3
• micro—-μ—–10-6
• nano—-n——10-9
• pico—–p——10-12

### Q 23. Defination of Significant figures.

Ans: The reliable digits plus the first uncertain digit in a measurement are known as significant digits or significant figures. e.g- If we say the period of oscillations of a simple pendulum is 1.62 s, then the digits 1 and 6 are reliable but the digit 2 is uncertain.

Rules for determining the number of significant figures:

• All non-zero digits are significant. e.g- (a) 1452 has 4 significant figures ,  (b) 1.452 has 4 significant figures.
• All zeroes occuring between two non-zero digits are significant. e.g- (a) 1005 has 4 Significant figures and (b) 10.05 has 4 significant figures.
• If the no. is less than 1, the zeroes on the right of dicimal points but to the left of the first non-zero digits are not significant. e.g- In 0.0063008, the first three zeroes are not significant. It has 5 significant figures.
• The terminal zeroes in a no. without a decimal point are not significant. e.g- In 3600, the zeroes are not significant. It has 2 significant figures.
• The terminal zeroes in a no. with a decimal point are significant . e.g- (a) 3.500 has 4 significant figures  and  (b) 0.06900 has 4 significant figures

Note: 1)Change of system of units does not change the no. of significant figures.

e.g- l= 4.803 cm has 4 significant figures.

=0.04803 m has 4 significant figures

=48.03 mm has 4 significant figures

2)The powers of 10 are irrevalent to the determination of the no. of significant figures.

e.g- l= 4.72 m has 3 significant figures

= 4.72 X 102 cm has 3 significant figures

= 4.72 X 103 mm has 3 significant figures