Dimensional analysis and its applications MCQ Quiz - Objective Question with Answer for Dimensional analysis and its applications - Download Free PDF

Concept:

Each physical quantity has its unique dimensional formula based on its derived units.

Energy Density is the amount of energy stored per unit volume.

Formula for Energy Density: u = E/V

E is energy, having dimensional formula [M L² T⁻²].

V is volume, having dimensional formula [L³].

Energy Density (u): The dimensional formula of energy density is [M L⁻¹ T⁻²].

Calculation:

We are asked to identify which quantity has the dimensional formula [M L⁻¹ T⁻²].

Option 1: Pressure × Area

• Pressure has the dimensional formula [M L⁻¹ T⁻²].
• Area has the dimensional formula [L²].
• Multiplying them gives [M L¹ T⁻²], which is not the required formula.

Option 2: Force × Pressure

• Force has the dimensional formula [M L T⁻²].
• Pressure has the dimensional formula [M L⁻¹ T⁻²].
• Multiplying them gives [M² L⁰ T⁻⁴], which is not the required formula.

Option 3: Power × Time

• Power has the dimensional formula [M L² T⁻³].
• Time has the dimensional formula [T].
• Multiplying them gives [M L² T⁻²], which is not the required formula.

Option 4: Energy Density

• Energy density has the dimensional formula [M L⁻¹ T⁻²], which matches the required formula.

∴ The correct answer is option 4) Energy Density.

CONCEPT:

Physical Quantities and Dimensions

• Physical quantities are properties or characteristics of a system that can be measured or calculated from other measurements.
• Each physical quantity has a dimension that can be represented by a combination of the basic dimensions (mass, length, time, etc.).
• In order for a mathematical operation involving physical quantities to be physically meaningful, the quantities must have compatible dimensions.

EXPLANATION:

• Let's examine the given options:
  • Option 1: (P-Q)/R
    • For this operation to be meaningful, P and Q must have the same dimensions because only quantities with the same dimensions can be subtracted. If P and Q have the same dimensions, (P-Q) will have the same dimension as P (or Q), but it is given P and Q have a different dimension thus it cannot be meaningful quantity.
  • Option 2: PQ-R
    • For this operation to be meaningful, PQ and R must have the same dimensions. PQ represents the product of P and Q, so the dimensions of PQ are the product of the dimensions of P and Q. Unless R has the same dimensions as PQ.
  • Option 3: PQ/R
    • This operation is meaningful if the dimensions of PQ are compatible with the dimensions of R. Specifically, the dimensions of PQ should be equal to the dimensions of R. This means that the dimensions of P multiplied by the dimensions of Q should result in the dimensions of R.
  • Option 4: (PR - Q²)/R
    • For this operation is meaningful, PR and Q² must have the same dimensions. Since Q² represents Q multiplied by itself, its dimensions are the square of the dimensions of Q. PR represents the product of P and R, so its dimensions are the product of the dimensions of P and R. Additionally, (PR - Q²)/R must be dimensionally consistent.

Therefore, the correct answer is option 1: (P-Q)/R.

Explanation:

(A) 

\([\mathrm{Y}]=\frac{\mathrm{F}}{\mathrm{A}\left(\frac{\Delta \ell}{\ell}\right)} \Rightarrow \frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^{2}}=\mathrm{ML}^{-1} \mathrm{~T}^{-2}\rm (II)\)

(B)

Torque \((\vec{\tau})=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}\)

\((\vec{\tau})=\mathrm{L} \times \mathrm{MLT}^{-2}=\mathrm{ML}^{2} \mathrm{~T}^{-2} \text{ (IV)}\)

(C)

Coefficient of viscosity \(\Rightarrow \mathrm{F}=\eta \mathrm{A} \frac{\mathrm{dV}}{\mathrm{dt}}\)

η → Pa·sec

\([\eta]=\frac{\mathrm{MLT}^{-2}}{\mathrm{L}^{2}} \times \mathrm{T}=\mathrm{ML}^{-1} \mathrm{T}^{-1}\)

(D)

Gravitational constant (G)

\(\mathrm{F}=\frac{\mathrm{GM}_{1} \mathrm{M}_{2}}{\mathrm{r}^{2}}\)

\({[\mathrm{G}]=\frac{\mathrm{F} \cdot \mathrm{r}^{2}}{\mathrm{m}_{1} \mathrm{m}_{2}}=\frac{\mathrm{MLT}^{-2} \times \mathrm{L}^{2}}{\mathrm{M}^{2}}=\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{T}^{-2}} \rm (III)\)

Ezxplanation:

\(\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}\)

⇒ \(\left[\mu_{0}\right]=\left[\frac{\mathrm{B} \times \mathrm{r}}{\mathrm{I}}\right]=\left[\frac{\mathrm{MT}^{-2} \mathrm{~A}^{-1} \times \mathrm{L}}{\mathrm{~A}}\right]=\left[\mathrm{MLT}^{-2} \mathrm{~A}^{-2}\right]\)

magnetic field F = qvB

\(\mathrm{B}=\left[\frac{\mathrm{MLT}^{-2}}{\mathrm{AT} \mathrm{~L} / \mathrm{T}}\right]=\left[\mathrm{MT}^{-2} \mathrm{~A}^{-1}\right]\)

[M] = [NTA] = [M] = [ML²] 

\(\tau=\mathrm{c} \theta \Rightarrow \mathrm{c}=\left|\frac{\tau}{\theta}\right|=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]\)

Explanation:

We know that 

B = μ₀ni

\(\left[\frac{\mathrm{B}}{\mu_{0}}\right]=[\mathrm{ni}]=\mathrm{L}^{-1} \mathrm{~A}^{1}\)

CONCEPT:

• SI Unit: It is an abbreviation from the French name Le Systeme International d’Unites.
  • There are 7 fundamental units and two supplementary units.

• The dimensions of all the units shall be homogeneous.
  • Means dimension of velocity shall be added or subtracted to others having the same dimension.
  • Example - velocity is not subtracted from ampere.

CALCULATION:

Given α = βt + λ

⇒ meter = ( β × time ) + meter

⇒ (meter-meter) = ( β × time )

⇒ β = meter/time ⇒ m t^-1 ⇒ m s-1

Explanation:

Stress - Stress is defined as the force that is acted on an object per unit area. It is measured as N/m².

Stress \( = \frac{{Force}}{{Area}}\) ----(1)

Force = mass \(\times\) acceleration

⇒ F = ma

⇒ [F] = [MLT^-2]

and area, [A]= [L²]

Putting these values in equation (1) we have;

Stress = \( \frac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ {{L^2}} \right]}}\)

⇒ Stress = \( \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]\)

Hence, option 2) is the correct answer.

CONCEPT:

• Dimensional Formula: The expressions or formulae which give knowledge on how and which of the fundamental quantities are present in a physical quantity are referred to as the Dimensional Formula of the Physical Quantity.
  • Dimensional formulae also help in deriving units from one system to different others.
  • Suppose, a physical quantity Z which depends on base dimensions  L (Length), M (Mass), and T (Time) with respective powers a, b and c, then its dimensional formula will be represented as [M^bL^aT^c]
  • A dimensional formula is usually closed in a bracket [ ].
  • Dimensional formulae of trigonometric, angle and solid angle aren't defined as these quantities are dimensionless in nature.

EXPLANATION:

• For Option 1:

Q = -KA(ΔTΔt /Δx)

K = (Qx)/(AΔTΔ t),

K is the thermal conductivity. Q is the amount of heat transferred through the material, x is the distance between the two isothermal planes. A is the area of the surface in square meters. t is time.

The dimension of thermal conductivity is

\(\frac{[energy][length]}{[area][temp][second]}=\frac{[ML^2T^{-2}][L]}{[L^2][K][T]}=M^1L^1T^{−3}K^{−1}\)

• For Option 2

Potential energy = Electric potential × Charge of particle

Electric potential =  Potential energy / Charge of a particle

and electric charge = electric current × time

The dimension of electric potential is

\(\frac{[energy]}{[electric Current][time]}=\frac{[ML^2T^{-2}]}{[A][T]}=M^1L^2T^{−3}A^{−1}\)

• For Option 3: 

As per Biot-Savart’s law, the magnetic induction at a point is given by

\(dB=\frac{\mu_0}{4\pi}\left(\frac{I\,dl\,\sin\theta}{r^2}\right)\)

The unit of magnetic induction dB is Wb/m² or N/A m.

Putting the dimensions of dB, I, dl and r, we get,

\( L^0M^1T^{-2}A^{-1}=\left(\textrm{Dimensions of }\mu_0\right)\left(\frac{L^0M^0T^0A^1× L^1M^0T^0}{L^2M^0T^0}\right)\)

\(The\space dimensions\space of\space\mu_0=\frac{L^0M^1T^{-2}A^{-1}× L^2M^0T^0}{L^0M^0T^0A^1× L^1M^0T^0}\)]

\(The\space dimensions \space of \space permeability\space is \space [L^1M^1T^{-2}A^{-2}]\)

• In the option dimension of permeability is not given correctly.
• For Option 4:

We know the formula \(V=V_0 e^{(-\frac{t}{RC})}\)

dimensional formula of RC will be equal to the dimensional formula of t

So the dimension of RC is \(M^0L^0T^{1}\)

Important Points

•  The dimensional formula of permeability of free space (μ0) is M1L1T-2A-2

CONCEPT:

• Dimensions: When a derived quantity is expressed in terms of fundamental quantities, it is written as a product of different powers of the fundamental quantities.
• The powers to which fundamental quantities must be raised in order to express the given physical quantity are called its dimensions.
• A quantity without dimension will usually be a ratio of two quantities with similar dimensions and hence, will cancel out. Thus, they will have no units and known as dimensionless quantity.
• Angle: It is defined as the ratio of the length of arc to the radius, i.e.,

​\(Angle\;\left( \theta \right) = \frac{{length\;of\;arc\;\left( l \right)}}{{radius\;\left( r \right)}}\)

\( \Rightarrow \theta = \frac{{\left[ {{M^0}L{T^0}} \right]}}{{\left[ {{M^0}L{T^0}} \right]}} = 1\)

∴ An angle is a dimensionless quantity

• Strain: It is defined as the ratio of change in length to the original length. i.e.,

\(Strain = \frac{{change\;in\;length\;\left( {{\rm{\Delta }}l} \right)}}{{original\;length\;\left( l \right)}}\)

\( \Rightarrow Strain = \frac{{\left[ {{M^0}L{T^0}} \right]}}{{\left[ {{M^0}L{T^0}} \right]}} = 1\)

∴ Strain is a dimensionless quantity

• Specific gravity: It is the ratio of the density of a substance to the density of a given reference material i.e.,

\(Specific\;gravity\;\left( \rho \right) = \frac{{Density\;of\;the\;object\;\left( {{\rho _{object}}} \right)}}{{Density\;of\;water\;\left( {{\rho _{water}}} \right)}}\)

\( \Rightarrow Specific\;gravity = \frac{{\left[ {M{L^3}{T^0}} \right]}}{{\left[ {M{L^3}{T^0}} \right]}} = 1\)

∴ Specific gravity is a dimensionless quantity

EXPLANATION:

As from the above discussion, we can say that:

• An angle is the ratio of the length of arc to the radius. As both have the dimension of length so the angle is a dimensionless quantity. So option 1 follows.
• Strain: It is defined as the ratio of change in length to the original length. As both have the dimension of length so the strain is a dimensionless quantity. So option 2 follows.
• Specific gravity: It is the ratio of the density of a substance to the density of given reference material. As both have the dimension of density so the specific gravity is a dimensionless quantity. So option 3 follows.

As the given three options are correct.

So the correct option is 4.

The correct answer is option 1) i.e. [T-1]

CONCEPT:​

• The dimensional formula is used to express any physical quantity in terms of fundamental quantities.
• Dimensional Analysis: It is a technique adopted to derive a relationship between various physical quantities using their dimensions and units of measurement. The two basic rules followed in the dimensional analysis are:
  • Only those quantities that have the same dimensions can be added and subtracted from each other.
  • Two physical quantities are equal if they have the same dimensions.

EXPLANATION:

• (1-e^-at) is a constant value and will have no dimensions.
• Thus, the dimensions of v/A will be equal to the dimension of x.
• Dimension of position, x = [M⁰L¹T⁰]
• Dimension of velocity, v =[M⁰L¹T^-1]

\(\Rightarrow x = \frac{v}{A}\)

\(\Rightarrow [M^0L^1T^0] = \frac{[M^0L^1T^{-1}]}{A}\)

\(\Rightarrow A = [T^{-1}]\)

CONCEPT:

Dimensions:

• Dimensions of a physical quantity are the powers to which the fundamental units are raised to obtain one unit of that quantity.

​EXPLANATION:

​Given F = X cos(Pt) + Y sin(Qs), t = time and s = distance

The dimension of F, t and s is given as,

⇒ [F] = [M¹L¹T^-2]

⇒ [t] = [M0L0T1]

⇒ [s] = [M0L1T0]

• Since the trigonometric function is a dimensionless quantity. So, Pt and Qs is also a dimensionless quantity.

So,

⇒ [P][t] = [M0L0T⁰]

⇒ [P][M0L0T1] = [M0L0T0]

⇒ [P] = [M0L0T^-1]     -----(1)

⇒ [Q][s] = [M0L0T0]

⇒ [Q][M0L1T0] = [M0L0T0]

⇒ [Q] = [M0L^-1T0]     -----(2)

• The dimension of the F, X and Y must be equal.

⇒ [F] = [X] = [Y] = [M1L1T-2]     -----(3)

By equation 1 and equation 3,

\(\Rightarrow \frac{\left [ P \right ]}{\left [ X \right ]}=\frac{\left [ M^{0}L^{0}T^{-1} \right ]}{\left [ M^{1}L^{1}T^{-2} \right ]}\)

\(\Rightarrow \frac{\left [ P \right ]}{\left [ X \right ]}=\left [ M^{-1}L^{-1}T^{1} \right ]\)

By equation 1 and equation 3,

\(\Rightarrow \frac{\left [ Q \right ]}{\left [ Y \right ]}=\frac{\left [ M^{0}L^{-1}T^{0} \right ]}{\left [ M^{1}L^{1}T^{-2} \right ]}\)

\(\Rightarrow \frac{\left [ P \right ]}{\left [ X \right ]}=\left [ M^{-1}L^{-2}T^{2} \right ]\)

• Hence, option 1 is correct.

CONCEPT:

• Dimensional analysis: The study of relationships between physical quantities with the help of their dimensions and units of measurements is called dimensional analysis.
• Advantages of dimensional analysis:
  • Consistency of a dimensional equation.
  • Derivation of the relation between physical quantities in physical phenomena.
  • To change units from one system to another.
• Limitations of dimensional analysis:
  • By this method, the value of a dimensionless constant cannot be calculated.
  • By this method, the equation containing trigonometrical, exponential, and logarithmic terms cannot be analysed.
  • In a physical quantity depends on more than three factors, and then relation among them cannot be established.

EXPLANATION:

• In dimensional analysis, the advantages are:
  • Conversion of units from one system to another. So option 1 is correct.
  • Checking the dimensional balance of the equation. So option 2 is correct.
  • Relation among the various physical quantities. So option 3 is correct.
• In dimensional analysis limitation is
  • The value of a dimensionless constant cannot be calculated. So option 4 is a limitation, not an advantage. So option 4 is correct.

CONCEPT:

• Work done: The dot product of force and displacement is called work done.

Work done (W) = F.s cos θ 

Where F is force, s is displacement and θ is the angle between F and s.

Dimensional formula of work (W) = [ML2T-2]

• Energy: The capacity to do work is called energy.

Energy (E) = Work done (W)

Dimensional formula of Energy (E) = dimensional formula of work done (W) = [ML2T-2]

• Power: It is defined as the rate of doing work.

 \(\therefore P = \frac{W}{t}\)

Where, P = power, W = work done and t = time.

The dimensional formula of power P is [ML2T-3].

• Pressure: The pressure is defined as force per unit area.

\(P = \frac{F}{A}\)

As force = mass × acceleration

∴ Dimensional formula of force (F) = [MLT-2]

Hence the dimensional formula of pressure is

\(P = \frac{F}{A} = \frac{{ML{T^{ - 2}}}}{{{L^2}}} = M{L^{ - 1}}{T^{ - 2}}\)

EXPLANATION:

From the above explanation it is clear that

• Work done has same dimensional formula as that of energy. So option 4 is correct.

CONCEPT:

• From the study of dimensional formulae of various quantities, we can easily divide them into four types:

EXPLANATION:

• From the above, it is clear that the velocity of light in a vacuum is a dimensional constant. Therefore option 4 is correct.
• All other quantities changes with change in position, like acceleration due to gravity changes when we go up and down from the surface of the earth.
• The value of surface tension changes when the temperature changes.

Concept:

• Magnetic Field: It is a vector field that is produced by moving charge or magnetized material.
  • It is denoted by\(\vec B\).
  • Its SI unit is Tesla.
  • Smaller unit is Gauss, 1 Tesla = 10,000 Gauss.

Mathematically \(\vec B = \frac{{\vec F}}{{q\;\left| {\vec v\sin \theta } \right|\;}}\) 

• The dimension of the Magnetic Field is [M T^-2 A^-1]

CALCULATION:

\(\vec F = q\;\vec B\;\left| {\vec v\sin \theta } \right|\)

[F] = [M L T^-2]

[Q] = [A T]

[V] = [L T^-1]

Sin θ = Dimensionless

\(\left[ B \right] = \frac{{\left[ F \right]}}{{\left[ q \right]\;\left[ v \right]}} = \frac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ {A{T}} \right]\;\left[ {L{T^{ - 1}}} \right]}} = \left[ {M\;{L^0}\;{T^{ - 2}}\;{A^{ - 1}}} \right] = \left[ {M{T^{ - 2}}{A^{ - 1}}} \right]\)

• Some Basic Dimensions are: