Dimensional Analysis Question and Answer|Problems on Dimensional Analysis

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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