**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=[L^{1}] [L^{1}] [L^{1}]=[L^{3}]

=[M^{0} L^{3 }T^{o}]

## 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=Gm_{1 }m_{2}/r_{2}

or G= Fr^{2}/m_{1}m_{2}

or [G]= [M^{1}L^{1}T^{-2}] [L1]2/[M^{1}][M^{1}]

=[M^{-1}L^{3}T^{-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]=[L^{1}T

^{-1}] [U]=[L

^{1}T

^{-1]}[at]=[L

^{1}T

^{-2}][T

^{1]}

=[L^{1} T^{-1} ]

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

**b) S= ut + 1/2 at ^{2}**

**Ans: **Given,

S= ut + 1/2 at^{2}

The Dimension of each term can be written as-

[ S]= [L^{1}] [ut]= [L

^{1}T

^{-1}][T

^{-1}]

=[L^{1}]
[1/2 at^{2}]= [L^{1}T^{-2}] [T^{1}] ( 1/2 is a dimensionless constant)

= [L^{1}]

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

**c) V ^{2}= u^{2 }+ 2as**

**Ans: **

Given,

V^{2}= u^{2} + 2as

The Dimension of each term can be written as-

[ V^{2}]= [L

^{1 }T

^{-1}]

^{2}

= [L^{2}T^{-2}]
[u^{2}]= [L^{1}T^{-1}]^{2}

=[L^{2}T^{-2}]
[2as]= [L^{1}T^{-2}] [T^{1}] ( 2 is a dimensionless constant)

=[L^{2}T^{-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]= [L^{1 }T^{-1}]

and [Bt]= [V]

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

=[L^{1}T^{-1}]/ [L^{1}T^{-2}]

= [ L^{1}T^{-2}]

### Q7. The distance(x) travelled by a particle is related to time “t” as x=at + bt^{2}. Find the dimensions of a and b.

**Ans:** Given,

x=at + bt^{2}

By the principal of homogenity of dimension-

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

or [a]=[L^{1}]/[T^{-1}]

=[ L^{1 }T^{-1}]

& [bt^{2}]=[x]

or [b]= [x]/[t^{2}]

=[L^{1}]/[T^{1}]^{2}

=[L^{1}T^{-2}]

### Q8. If the Pressure, P is given by P=a-x^{2}/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-x^{2}/bt

^{2}]( Since, like quantities can be subtracted)

=[L^{2}]

Now, Pbt= a-x^{2}

or[M^{1}L^{-1}T^{-2}] **X **[b][T^{1}]=[L^{2}]

or [b]= [L^{2}]/[M^{1}L^{-1}T^{-2}] [T^{1}]

=[M^{-1}L^{3}T^{1}]

Thus, [a]= [M^{0}L^{2}T^{0}]
[b]=[M^{-1}L^{3}T^{1}]

### 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∝ m^{a}

∝ l^{b}

∝ g^{c}

Combining,

T∝ m^{a} l^{b} g^{c}

or T= K m^{a} l^{b} g^{c} ———(1)

Where K is a dimensionless constant.

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

[T^{1}]= [M

^{1}]

^{a }[L

^{1}]

^{b }[L

^{1}T

^{-2}]

^{c}

or [M^{0}L^{0}T^{0}]= [M^{a} L^{a+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 m^{0 }L^{1/2} 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∝ m^{a}

∝ v^{b}

∝ r^{c}

Combining,

F∝ m^{a} v^{b} r^{c}

or F= K m^{a} v^{b} r^{c} ————(1) ; where K is a dimensionless constant.

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

[M^{1}L

^{1}T

^{-2}]= [M

^{1}]

^{a}[L

^{1 }T

^{-1}]

^{b }[L

^{1}]

^{c}

or [M^{1} L^{1} T^{-2} = [M^{a }L^{b+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 m^{1} v^{2} r^{-1}

or F= K mv^{2/r}

Experimentally, K= 1

∴ F= mv^{2}/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

^{1}L

^{1}T

^{-2}]

∴ a= 1, b=1 and c= -2

**In SI: **M_{1}= 1kg, L_{1}= 1m, T_{1}= 1s and n_{1}= 10

**In CGS: **M_{2}= 1g, L_{2}= 1cm, T_{2}= 1s and n_{2}= ?

Using, n_{2}=n_{1} [M1/M_{2}]^{a} [L_{1}/L_{2}]^{b} [T_{1}/T_{2}]^{c}

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

=10 **X **10^{3}g/1g **X **10^{2}cm/1cm **X **1

=10 **X **10^{5}

=10^{6}

∴ 10N = 10^{6 }dyne

### Q13. (NCERT 2.3) A calorie is a unit of heat or energy and it equals about 4.2 J where 1J= 1kg m^{2} 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} γ^{2} in terms of the new units.

**Ans:** Here,

1cal= 4.2 J

1 J= 1kg m^{2} s^{-2}

Dimensional formula of energy,

[E]= [M^{1}L

^{2}T

^{-2}]

∴ a= 1, b=2, and c= -2

**In** **SI**: M_{1}=1kg, L_{1}=1m, T_{1}=1s, and n_{1}=4.2

**In New System: **M_{2}= α kg, L_{2}= β m, T_{2}=γ s and n_{2}= ?

Using, n_{2}=n_{1} [M1/M_{2}]^{a} [L_{1}/L_{2}]^{b} [T_{1}/T_{2}]^{c}

=4.2 [1kg/αkg]^{1} [1m/β m]^{2} [1s/γ s]^{-2}

=4.2 **X **1/α **X** 1/β^{2} **X **1/γ^{-2}

∴ 1Cal= 4.2 J= 4.2 α^{-1} β^{-2} γ^{2 }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} Nm^{2} 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]= [F^{1}]/ [L^{2}]

= [F^{1} L^{-2} T^{0}]

b) Energy,

E=1/2 mv^{2}

or E= 1/2 . F/a . v^{2} ——– (∵F= ma ⇒ m=F/a)

or [E]= [F^{1}]/ [L^{1} T^{-2}] **X **[L^{1} T^{-2}]

= [F^{1}]/[L^{1} T^{-2}] **X **[L^{2} T^{-2}]

= [F^{1}L^{1} T^{0}]

### 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]

e.g: Area= length** X **breadth

or [Area]= [L^{1}] **X **[L^{1}]

=[L^{2}]

=[M^{0} L^{2} T^{0}]

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 + at^{2}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]=[M
^{0}L^{0 }T^{0}] or [θ]=[M^{0}L^{0}T^{0}] - The logarithmic functions are dimensionless. i.e. [log e
- Exponential functions are dimensionless. i.e. [e
^{x}]= [M^{0}L^{0}T^{0}] or [x]= [M^{0}L^{0}T^{0}]

### 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—-10
^{3} - Mega—-M—10
^{6} - Giga—–G—-10
^{9} - Tera—–T—–10
^{12}

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 **10^{2 }cm has 3 significant figures

= 4.72 **X **10^{3} mm has 3 significant figures