Dimensional formulae and dimensional equations MCQ Quiz - Objective Question with Answer for Dimensional formulae and dimensional equations - Download Free PDF

Calculation:
Given: P = a³ × b² × c^−1/2 × d⁻¹

Taking logarithm on both sides:

ln P = 3 ln a + 2 ln b − (1/2) ln c − ln d

Now, taking error on both sides:

|ΔP / P| = 3 × |Δa / a| + 2 × |Δb / b| + (1/2) × |Δc / c| + |Δd / d|

⇒ Percentage error in P

= 3(1%) + 2(3%) + (1/2)(4%) + 2%

= (3 + 6 + 2 + 2)%

= 13%

Concept:

Dimensional Analysis:

• Dimensional analysis is a method used to check the consistency of physical equations by analyzing the dimensions of the quantities involved.
• In the equation of motion of a damped oscillator, we have terms involving mass (m), damping coefficient (b), and spring constant (k), which are related to position (x) and time (t).
• The dimensions of a quantity can be determined using the fundamental units of mass (M), length (L), and time (T). In this case, we are interested in finding the dimensional formula for the term b / √(km), where b is the damping coefficient, k is the spring constant, and m is the mass of the block.

Calculation:

Given the equation of motion of the damped oscillator:

m d²x/dt² + b dx/dt + kx = 0

We need to determine the dimensions of the term b / √(km).

• The dimensions of mass (m) are [M].
• The dimensions of spring constant (k) are [M][T]⁻².
• The dimensions of damping coefficient (b) are [M][T]⁻¹.

Now, the dimensional formula of b / √(km) will be:

[b] / √([k][m]) = [M][T]⁻¹ / √([M][T]⁻² × [M])

On simplifying:

Dimensions of b / √(km) = [M][T]⁻¹ / [M][T]⁻¹ = [T]⁰[M]⁰[T]⁰

Therefore, the correct dimensional formula for b / √(km) is [M]⁰[L]⁰[T]⁰.

Concept:

Dimensional Analysis:

• Dimensional analysis is a method used to check the consistency of physical equations by analyzing the dimensions of the quantities involved.
• In the equation of motion of a damped oscillator, we have terms involving mass (m), damping coefficient (b), and spring constant (k), which are related to position (x) and time (t).
• The dimensions of a quantity can be determined using the fundamental units of mass (M), length (L), and time (T). In this case, we are interested in finding the dimensional formula for the term b / √(km), where b is the damping coefficient, k is the spring constant, and m is the mass of the block.

Calculation:

Given the equation of motion of the damped oscillator:

m d²x/dt² + b dx/dt + kx = 0

We need to determine the dimensions of the term b / √(km).

• The dimensions of mass (m) are [M].
• The dimensions of spring constant (k) are [M][T]⁻².
• The dimensions of damping coefficient (b) are [M][T]⁻¹.

Now, the dimensional formula of b / √(km) will be:

[b] / √([k][m]) = [M][T]⁻¹ / √([M][T]⁻² × [M])

On simplifying:

Dimensions of b / √(km) = [M][T]⁻¹ / [M][T]⁻¹ = [T]⁰[M]⁰[T]⁰

Therefore, the correct dimensional formula for b / √(km) is [M]⁰[L]⁰[T]⁰.

Explanation:

Torque and Energy

Torque = force × distance = M L² T⁻²

Energy = work = force × distance = M L² T⁻²

Surface Tension and Impulse

Surface tension = force / length = M T⁻²

Impulse = force × time = M L T⁻¹

Angular Momentum and Planck’s Constant

Angular momentum = M L² T⁻¹

Planck's constant = M L² T⁻¹

Pressure and Young’s Modulus

Pressure = force / area = M L⁻¹ T⁻²

Young’s modulus = stress / strain = M L⁻¹ T⁻²

∴ Correct answer is Surface tension and impulse.

Explanation:

Angular impulse = [M L² T^–1]

Latent Heat = [M⁰ L² T^–2]

Electrical resistivity = [M L³ T^–3 A^–2]

Electromotive force = [M L² T^–3 A^–1] 

∴ Correct match is (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

CONCEPT:

• Principle of homogeneity of dimensions: According to this principle, a physical equation will be dimensionally correct if the dimensions of all the terms occurring on both sides of the equation are the same.
  • This principle is based on the fact that only the physical quantities of the same kind can be added, subtracted, or compared.
  • Thus, the velocity can be added to velocity but not to force.

EXPLANATION:

Given that:

F = ax2 + bx + c

• From the principle of dimensional homogeneity, the left-hand side of the equation dimensionally equal to the right-hand side of the equation.

The dimensional formula of Force (F) = [M L T^-2]

The dimensional formula of Displacement (x) = [L]

LHS = RHS

[M L T-2] = [b] ×  [L]

[b] = M L⁰ T^-2

Hence option 4 is correct.

CONCEPT:

• The dimensional formula is defined as the expression of the physical quantity in terms of mass, length, time, and ampere.

EXPLANATION:

• Density: It is defined as mass per unit volume.

i.e. \(Density = \frac{{Mass}}{{Volume}}\)

Now,

Dimensional formula of mass = [M]

Dimensional formula of volume = [L³]

\(Density = \frac{{\left[ M \right]}}{{\left[ {{L^3}} \right]}}\)

∴ Density = ML^-3 T⁰

∴ Dimensional formula of density is [ML^-3 T⁰].

• Some Basic Dimensional formula:

CONCEPT:

Work done:

• The dot product of force and displacement is called work done.
• SI unit of Work is Joules.
• Work done is given by,

Work done (W) = F.s cos θ _____(1)

Where,

F is force

 s is displacement

θ is the angle between F and s.

EXPLANATION:

As we know that Work done (W) = F.s cos θ _____(1)

∴ Force = mass × acceleration

The dimensional formula of force (F) = [MLT-2]

The dimensional formula of displacement (s) = [L]  

Put the values of force and displacement in equation (1) neglecting angle,

Work done (W) =  F.s 

Work done (W) = MLT-2.L

Work done (W) = ML2T-2 

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

CONCEPT:

• Dimensional Formula: The dimensions of a physical quantity represents its nature. 
• ​Dimensional Formula can help us in finding the relationship between different physical quantities.
  • For example, Dimensional Formula for work done and Kinetic Energy is the same. So, we can say they are related.
• For dimensionless quantities, dimensional analysis can't be possible.
  • For example, Relative density is a dimensionless quantity which is the ratio of two densities. we cannot find the dimension of relative density
• The dimension of fundamental quantities are known and dimensional formulas for other quantities are derived from fundamental units.
• Example:
  • Finding Dimension of Density:
  1. The expression for density is Mass / Volume
  2. Mass is a fundamental unit with dimension M.
  3. Volume is a cube of length. So, Dimensional Formula for Length L3
  4. Dimensional Formula for density M / L3 = ML -3

EXPLANATION:

From the definition of acceleration,

\(A = \frac{{dv}}{{dt}} = \frac{{{d^2}x}}{{d{t^2}}}\)

\(\smallint dv = \smallint A\;dt\)

V = A × T

\(\frac{x}{T} = A\;T\)

⇒ x = A T²

To write in the form of Dimensions

⇒ x = [F⁰AT²]

The correct answer is option 3) i.e. M1 I-1 L2 T-2

CONCEPT:

• Magnetic flux: Magnetic flux is the measure of the number of magnetic field lines passing through a given surface area.

The magnetic flux ϕ is given by:

ϕ = BAcosθ 

Where B is the magnetic field strength, A is the area vector, and θ is the angle made by the area vector with respect to the magnetic field line.

EXPLANATION:

We know, ϕ = BAcosθ 

For dimensional analysis, ϕ = BA

The dimensional formula of A (area) = L²       ----(1)

From lenz's law we know that F = q(v × B) ⇒ B = \(\frac{F}{qv}\)

Force, F = mass × acceleration ⇒ M¹ L¹T^-2      ----(2)

Charge, q = Current × time ⇒ I¹ T¹      ----(3)

Velocity, v ⇒ L¹ T^-1      ----(4)

Substituting (1), (2), (3), and (4) in ϕ = BA

[ϕ] = \(\frac{F}{qv}\)A = \(\frac{M^1 L^1T^{-2} }{(I^1T^1)(L^1T^{-1})} L^2 \) = M1 I-1 L2 T-2

Concept:

Surface tension: 

• Surface tension is the property by virtue of which liquid tires to minimize its free surface area.
• In spherical shape the surface area is minimum and for this reason, the raindrops are spherical.
• Surface tension is measured as the force acting per unit length of an imaginary line drawn on the liquid surface.
  \(Surface\;tension = \frac{{Force}}{{length}}\)

• Temperature: If temperature increases then the surface tension of a liquid decreases.
• Soluble Impurities: In the case of less soluble impurities, the surface tension decreases. But, for highly soluble impurities in the liquid the surface tension increases.

Explanation:

we know that,

\(Surface\;tension = \frac{{Force}}{{length}}\)

Dimensional formula for surface tension is -

\(Surface\;tension = \frac{{Force}}{{length}}=\frac{[ML{T}^{-2}]}{[L]}=[M{T}^{-2}]\)

CONCEPT:

• Energy - Energy is defined as the capacity to do some work and it is equal to E = mc² 
• Here we have, m as the mass and c as the velocity of light.
• Gravitational constant - The Gravitational constant  is denoted by " G " and it comes from,

⇒\(F = \frac {Gm_1 m_2}{r^2}\)

⇒\(G = \frac {F r^2}{m_1 m_2}\)

here, F is the force, m₁, m₂ is the two masses and r is the distance.

CALCULATIONS:

Energy, E = mc2 

• Dimensional formula of energy

[E] = [M¹(LT^-1)²]

⇒[E] = [M¹ L² T^-2]      ------(1)

and \(G = \frac {F r^2}{m_1 m_2}\)

• Dimensional formula of gravitational constant

[G] = [M^-1 L³ T^-2]      ------(2)

Divide equation (1) and (2), we have

\(\frac{{[E]}}{{[G]}} = \frac{{[{M^1}{L^2}{T^{ - 2}}]}}{{[{M^{ - 1}}{L^3}{T^{ - 2}}]}}\)

⇒ \(\left[ {\frac{E}{G}} \right]\) = [M2 L-1 T0]

Hence, correct option is (2)

Concept:

• Measurement of any physical quantity involves comparison with a certain basic, arbitrarily chosen, internationally accepted reference standard called unit, and a Dimension is a mathematical tool used for studying the nature of physical quantities. 
• The basic concept of dimensions is that we can add or subtract only those quantities which have the same dimensions.
• And the dimensional formula is defined as the expression of the physical quantity in terms of mass, length, and time.

Inductance:

• The tendency of any conductor which opposes the change in the electric current flowing through it is called as inductance of that conductor.
• The SI unit of inductance is Henry.
• The Dimension of Inductance is [M¹ L² T^-2 A^-2]

The inductance of a coil of wire is given by,

\(L = \frac{{{\mu _0}{N^2}A}}{L}\)

Where N is the number of turns

A is the cross-sectional area

L is the length of the solenoid.

μ ₀ is the permeability of the free space

Resistance:

• The property by which an electric conductor opposes the flow of current through it is called as resistance of the conductor.
• It is denoted by R.
• The SI unit of resistance is Ohm (Ω).

\(Resistance\;\left( R \right) = \rho \frac{l}{A}\)

Where ρ is the resistivity of a conductor, l is the length of conductor and A is the cross-sectional area.

Dimension for resistance (R) = [M¹ L² T^-3 A^-2]

Calculation:

We know that,

Dimension of L = [M¹ L² T^-2 A^-2]

Dimension of R = [M¹ L² T^-3 A^-2]

Then the Ratio of both –

\(\frac{L}{R} = \frac{{\left[ {{M^1}{L^2}{T^{ - 2}}{A^{ - 2}}} \right]}}{{\left[ {{M^1}{L^2}{T^{ - 3}}{A^{ - 2}}} \right]}} = \left[ {{M^0}{L^0}{T^1}{A^0}} \right]\)

Hence the correct option in terms of M, L & T is [M⁰ L⁰ T]

The correct answer is option 4) i.e. MLT-1

CONCEPT:

• The dimensional formula is used to express any physical quantity in terms of fundamental quantities - mass, length, and time.​​

• Impulse (J): The change in momentum of an object when the object is acted upon by a force for a certain amount of time is called impulse.

Impulse is expressed mathematically as : \(Δ p=FΔ t\)

Where Δp is change in momentum, F is force, and Δt is the time taken

EXPLANATION:

Impulse is given by:

\(Δ p=FΔ t\)

Impulse (J) \(= (m a)\:.t\)      ----(1)

The dimensional formula for Force (ma) = M[LT^-2]      ----(2)   ( \(\because\) a = velocity/time = \(\frac{dispalcement/time}{time}\) = \(\frac{L/T}{T}\) = LT-2 )

The dimensional formula for time (t) = T      ----(3)

Substituting (2) and (3) in (1),

Dimension of Impulse = M[LT-2] × T =  MLT^-1

The correct answer is option 2):(Acceleration)

Concept:

Acceleration:

• Acceleration is the rate of change of the velocity of an object with respect to time.
• Velocity is given by rate of change of distance with respect to Time
• Velocity = \(Distance \over Time\)
• The dimension of Velocity is LT-1 
• Acceleration  = \(Velocity \over Time\)
• The dimension of accelartation is LT-2 

Additional Information

• The dimensional formula of power P is [ML²T^-3]. 
• Momentum is dimensionally represented as [M¹ L¹ T^-1].
• Density is dimensionally represented as [M¹ L^-3 T⁰].