Dimensions of Physical Quantities and Dimensional Analysis

13 questions

Question 1Question →

In the dimensional equation for the period of oscillation of a simple pendulum, T=kgalbT = k g^a l^b, where TT is the period, gg is the acceleration due to gravity, ll is the length of the pendulum, and kk is a dimensionless constant, what is the numerical value of the exponent aa?

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Answer: -0.5

Answer

The numerical value of the exponent aa is -0.5.
Applying dimensional analysis to T=kgalbT = k g^a l^b, the dimension of the left-hand side is T1\text{T}^1. The right-hand side has dimensions (L T−2)a(L)b=La+bT−2a(\text{L T}^{-2})^a (\text{L})^b = \text{L}^{a+b} \text{T}^{-2a}. Equating the exponents of time T\text{T} yields 1=−2a1 = -2a, which gives a=−0.5a = -0.5.

Step-by-Step Solution

1
Express the dimensions of all physical quantities involved in fundamental base dimensions (M, L, T).
The dimension of period TT is [T][\text{T}], length ll is [L][\text{L}], and gravitational acceleration gg is [L T−2][\text{L T}^{-2}].
Dimensional analysis requires substituting each quantity with its fundamental dimensions.
2
Formulate the dimensional homogeneity equation.
M0L0T1=(L T−2)a(L)b=La+bT−2a\text{M}^0 \text{L}^0 \text{T}^1 = (\text{L T}^{-2})^a (\text{L})^b = \text{L}^{a+b} \text{T}^{-2a}.
The principle of dimensional homogeneity states that the exponents of base dimensions on both sides of a physically correct equation must be equal.
3
Equate exponents of time T\text{T} and solve for aa.
1=−2a  ⟹  a=−12=−0.51 = -2a \implies a = -\frac{1}{2} = -0.5.
Comparing the powers of T\text{T} gives a linear equation in aa.

Key Concept

Principle of Dimensional Homogeneity
Estimated Time:1m 15s
Question 2Question →

The gravitational force FF between two point masses m1m_1 and m2m_2 separated by a distance rr is expressed by the equation F=Gm1m2r2F = \frac{G m_1 m_2}{r^2}, where GG represents the universal gravitational constant. What is the dimensional formula for GG?

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Answer: M−1L3T−2M^{-1} L^3 T^{-2}

Answer

The dimensional formula for the universal gravitational constant GG is M−1L3T−2M^{-1} L^3 T^{-2}.
Isolating GG gives G=Fr2m1m2G = \frac{F r^2}{m_1 m_2}. Substituting the fundamental dimensions for force (MLT−2M L T^{-2}), distance (LL), and mass (MM) yields (MLT−2)(L2)M2=M−1L3T−2\frac{(M L T^{-2})(L^2)}{M^2} = M^{-1} L^3 T^{-2}.

Step-by-Step Solution

1
Make GG the subject of the formula in Newton's law of gravitation.
G=Fr2m1m2G = \frac{F r^2}{m_1 m_2}
To derive the dimensions of GG, isolate it in terms of force, distance, and mass.
2
Substitute fundamental dimensions for force, distance, and mass.
[G] = \frac{[F][r]^2}{[m_1][m_2]} = \frac{(M L T^{-2})(L^2)}{M \cdot M}
Force has dimensions MLT−2M L T^{-2}, distance has dimension LL, and mass has dimension MM.
3
Simplify the powers of fundamental dimensions MM, LL, and TT.
[G] = M^{1-2} L^{1+2} T^{-2} = M^{-1} L^3 T^{-2}
Applying exponent rules simplifies the combined base dimensions.

Key Concept

Dimensions of Physical Constants
Estimated Time:1m 15s
Question 3Question →

The viscous force FF acting on a small sphere of radius rr moving with velocity vv through a liquid is given by Stokes' law, F=6πηrvF = 6\pi \eta r v, where η\eta is the coefficient of viscosity. Which of the following expressions represents the base dimensions of η\eta?

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Answer: ML−1T−1M L^{-1} T^{-1}

Answer

ML−1T−1M L^{-1} T^{-1}
Rearranging Stokes' law gives η=F6πrv\eta = \frac{F}{6\pi r v}. Substituting base dimensions [F]=MLT−2[F] = M L T^{-2}, [r]=L[r] = L, and [v]=LT−1[v] = L T^{-1} yields [η]=MLT−2L2T−1=ML−1T−1[\eta] = \frac{M L T^{-2}}{L^2 T^{-1}} = M L^{-1} T^{-1}.

Step-by-Step Solution

1
Express Stokes' law in terms of the coefficient of viscosity
η=F6πrv\eta = \frac{F}{6\pi r v}
Isolating η\eta allows substitution of fundamental dimensions.
2
Substitute fundamental dimensions for force, radius, and velocity
[η]=MLT−2L⋅LT−1[\eta] = \frac{M L T^{-2}}{L \cdot L T^{-1}}
The constant 6π6\pi is dimensionless, while [F]=MLT−2[F] = M L T^{-2}, [r]=L[r] = L, and [v]=LT−1[v] = L T^{-1}.
3
Simplify the powers of base quantities MM, LL, and TT
[η]=ML1−2T−2−(−1)=ML−1T−1[\eta] = M L^{1 - 2} T^{-2 - (-1)} = M L^{-1} T^{-1}
Applying algebraic rules of exponents simplifies the expression.

Key Concept

Dimensions of Physical Quantities and Dimensional Analysis
Question 4Question →

Power is defined as the rate at which work is done or energy is transferred. Which of the following expressions represents the correct dimensional formula for power?

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Answer: ML2T−3M L^2 T^{-3}

Answer

ML2T−3M L^2 T^{-3}
Power is defined as work done per unit time (P=WtP = \frac{W}{t}). The dimensional formula for work is [W]=ML2T−2[W] = M L^2 T^{-2}, and for time is [t]=T[t] = T. Dividing work by time gives [P]=ML2T−2T=ML2T−3[P] = \frac{M L^2 T^{-2}}{T} = M L^2 T^{-3}.

Step-by-Step Solution

1
Write the fundamental definition of power in terms of work and time.
Power=WorkTime\text{Power} = \frac{\text{Work}}{\text{Time}}
By definition, power measures the rate of doing work.
2
Determine the dimensions of work.
[Work]=[Force]×[Distance]=(MLT−2)×L=ML2T−2[\text{Work}] = [\text{Force}] \times [\text{Distance}] = (M L T^{-2}) \times L = M L^2 T^{-2}
Force has dimensions of mass times acceleration (MLT−2M L T^{-2}), and multiplying by distance (LL) yields energy or work dimensions.
3
Divide the dimensions of work by the dimension of time (TT).
[Power]=ML2T−2T=ML2T−3[\text{Power}] = \frac{M L^2 T^{-2}}{T} = M L^2 T^{-3}
Dividing by time increases the negative exponent of time from −2-2 to −3-3.

Key Concept

Dimensional Formula for Power
Question 5Question →

Which of the following expressions represents the correct dimensions of pressure?

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Answer: ML−1T−2M L^{-1} T^{-2}

Answer

ML−1T−2M L^{-1} T^{-2}
Pressure is defined as Force divided by Area. Substituting the fundamental dimensions gives Force as MLT−2M L T^{-2} and Area as L2L^2. Dividing Force by Area results in MLT−2L2=ML−1T−2\frac{M L T^{-2}}{L^2} = M L^{-1} T^{-2}. Therefore, the expression ML−1T−2M L^{-1} T^{-2} is correct.

Step-by-Step Solution

1
State the physical definition formula for pressure
Pressure P=ForceAreaP = \frac{\text{Force}}{\text{Area}}
Pressure is defined as force applied perpendicular to a surface per unit area.
2
Substitute fundamental dimensions for force and area
[P]=[MLT−2][L2][P] = \frac{[M L T^{-2}]}{[L^2]}
Force has dimensions MLT−2M L T^{-2} and area has dimensions L2L^2.
3
Simplify the index of length LL
[P]=ML1−2T−2=ML−1T−2[P] = M L^{1 - 2} T^{-2} = M L^{-1} T^{-2}
Applying the rules of indices for length gives L1⋅L−2=L−1L^1 \cdot L^{-2} = L^{-1}.

Key Concept

Dimensions of Pressure
Question 6Question →

In atomic physics, the energy EE of a photon is related to its frequency ff by the formula E=hfE = h f, where hh is Planck's constant. What are the fundamental dimensions of Planck's constant hh?

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Answer: ML2T−1M L^2 T^{-1}

Answer

The fundamental dimensions of Planck's constant are ML2T−1M L^2 T^{-1}.
Planck's constant hh is given by h=E/fh = E / f. Since energy EE has fundamental dimensions of ML2T−2M L^2 T^{-2} and frequency ff has dimensions of T−1T^{-1}, dividing energy by frequency yields ML2T−2T−1=ML2T−1\frac{M L^2 T^{-2}}{T^{-1}} = M L^2 T^{-1}.

Step-by-Step Solution

1
Express Planck's constant in terms of energy and frequency
h=Efh = \frac{E}{f}
Rearranging the equation E=hfE = h f isolates hh.
2
Determine the fundamental dimensions of energy (EE) and frequency (ff)
[E]=ML2T−2[E] = M L^2 T^{-2} and [f]=T−1[f] = T^{-1}
Energy has dimensions of work (F×d=MLT−2⋅L=ML2T−2F \times d = M L T^{-2} \cdot L = M L^2 T^{-2}) and frequency is inverse time (T−1T^{-1}).
3
Divide the dimensions of energy by the dimensions of frequency
[h]=ML2T−2T−1=ML2T−2−(−1)=ML2T−1[h] = \frac{M L^2 T^{-2}}{T^{-1}} = M L^2 T^{-2 - (-1)} = M L^2 T^{-1}
Applying the rules of indices simplifies the exponent of time.

Key Concept

Dimensional Analysis of Physical Constants
Estimated Time:1m 0s
Question 7Question →

The speed of sound vv in a gas depends on the gas pressure PP and density ρ\rho according to the relation v=kPaρbv = k P^a \rho^b, where kk is a dimensionless constant. What is the numerical value of the exponent aa?

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Answer: 0.5

Answer

The numerical value of the exponent aa is 0.50.5.
Applying dimensional homogeneity to the relation v=kPaρbv = k P^a \rho^b yields [M0L1T−1]=[ML−1T−2]a[ML−3]b[M^0 L^1 T^{-1}] = [M L^{-1} T^{-2}]^a [M L^{-3}]^b. Equating the powers of time TT gives −2a=−1-2a = -1, which simplifies to a=0.5a = 0.5.

Step-by-Step Solution

1
Determine the base dimensions of all physical quantities in the relationship
[v]=M0LT−1[v] = M^0 L T^{-1}, [P]=ML−1T−2[P] = M L^{-1} T^{-2}, and [ρ]=ML−3[\rho] = M L^{-3}
Physical quantities must be expressed in fundamental dimensions (M,L,TM, L, T) to apply dimensional analysis.
2
Formulate the dimensional balance equation
M0L1T−1=Ma+bL−a−3bT−2aM^0 L^1 T^{-1} = M^{a+b} L^{-a-3b} T^{-2a}
By the principle of dimensional homogeneity, the total dimensions on the left side must equal those on the right side.
3
Equate exponents of TT to solve for aa
−2a=−1  ⟹  a=0.5-2a = -1 \implies a = 0.5
Comparing powers of time TT directly isolates the variable aa.

Key Concept

Dimensional Homogeneity and Derivation of Exponents
Question 8Question →

In electrostatics, Coulomb's law states that the force FF between two point charges q1q_1 and q2q_2 separated by a distance rr in a vacuum is given by F=q1q24πε0r2F = \frac{q_1 q_2}{4\pi \varepsilon_0 r^2}, where ε0\varepsilon_0 is the permittivity of free space. What are the fundamental dimensions of ε0\varepsilon_0 expressed in terms of mass (MM), length (LL), time (TT), and electric current (II)?

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Answer: M−1L−3T4I2M^{-1} L^{-3} T^4 I^2

Answer

The dimensions of permittivity of free space ε0\varepsilon_0 are M−1L−3T4I2M^{-1} L^{-3} T^4 I^2.
The correct answer is derived by isolating ε0=q1q24πFr2\varepsilon_0 = \frac{q_1 q_2}{4\pi F r^2}. Substituting [q]=IT[q] = I T, [F]=MLT−2[F] = M L T^{-2}, and [r]=L[r] = L gives [ε0]=I2T2ML3T−2=M−1L−3T4I2[\varepsilon_0] = \frac{I^2 T^2}{M L^3 T^{-2}} = M^{-1} L^{-3} T^4 I^2.

Step-by-Step Solution

1
Rearrange Coulomb's Law to isolate permittivity of free space ε0\varepsilon_0
ε0=q1q24πFr2\varepsilon_0 = \frac{q_1 q_2}{4\pi F r^2}
Isolating ε0\varepsilon_0 allows us to substitute the dimensions of each constituent physical quantity.
2
Determine the dimensions of charge qq, force FF, distance rr, and the constant 4π4\pi
[q]=IT[q] = I T, [F]=MLT−2[F] = M L T^{-2}, [r2]=L2[r^2] = L^2, and [4π]=1[4\pi] = 1 (dimensionless)
Electric current is a base unit (II), so electric charge is current multiplied by time (ITI T). Force is mass times acceleration (MLT−2M L T^{-2}).
3
Substitute the fundamental dimensions into the rearranged equation and simplify exponent powers
[ε0]=(IT)(IT)(MLT−2)(L2)=I2T2ML3T−2=M−1L−3T4I2[\varepsilon_0] = \frac{(I T)(I T)}{(M L T^{-2})(L^2)} = \frac{I^2 T^2}{M L^3 T^{-2}} = M^{-1} L^{-3} T^4 I^2
Applying exponent laws: T2/T−2=T2−(−2)=T4T^2 / T^{-2} = T^{2 - (-2)} = T^4, 1/M=M−11 / M = M^{-1}, and 1/L3=L−31 / L^3 = L^{-3}.

Key Concept

Dimensional analysis of physical constants in electromagnetism
Question 9Question →

The volume flow rate QQ of a viscous liquid through a pipe depends on the radius rr of the pipe, the coefficient of viscosity η\eta, and the pressure gradient ΔPL\frac{\Delta P}{L} according to the dimensional equation Q=krxηy(ΔPL)zQ = k r^x \eta^y \left(\frac{\Delta P}{L}\right)^z, where kk is a dimensionless constant. What is the value of the exponent xx?

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Answer: 4

Answer

The value of the exponent xx is 4.
Applying the principle of dimensional homogeneity, the dimensions of volume flow rate [Q]=L3T−1[Q] = L^3 T^{-1} are equated to [r]x[η]y[ΔPL]z=Lx(ML−1T−1)y(ML−2T−2)z[r]^x [\eta]^y \left[\frac{\Delta P}{L}\right]^z = L^x (M L^{-1} T^{-1})^y (M L^{-2} T^{-2})^z. Equating powers yields y+z=0y + z = 0 for mass, −y−2z=−1-y - 2z = -1 for time, and x−y−2z=3x - y - 2z = 3 for length. Solving these simultaneous equations gives z=1z = 1, y=−1y = -1, and x=4x = 4.

Step-by-Step Solution

1
Identify the base dimensions of each physical quantity in the given equation.
Flow rate [Q]=M0L3T−1[Q] = M^0 L^3 T^{-1}, radius [r]=L[r] = L, viscosity [η]=ML−1T−1[\eta] = M L^{-1} T^{-1}, and pressure gradient [ΔPL]=ML−2T−2\left[\frac{\Delta P}{L}\right] = M L^{-2} T^{-2}.
Expressing each quantity in terms of fundamental dimensions (MM, LL, TT) is necessary for dimensional analysis.
2
Substitute dimensions into the power-law equation and collect powers of base dimensions.
M0L3T−1=My+zLx−y−2zT−y−2zM^0 L^3 T^{-1} = M^{y+z} L^{x-y-2z} T^{-y-2z}.
The principle of dimensional homogeneity requires both sides of a physically valid equation to have identical dimensions.
3
Set up and solve linear equations for the exponents xx, yy, and zz.
Solving y+z=0y + z = 0, −y−2z=−1-y - 2z = -1, and x−y−2z=3x - y - 2z = 3 yields z=1z = 1, y=−1y = -1, and x=4x = 4.
Equating powers of MM, TT, and LL allows step-by-step determination of each unknown exponent.

Key Concept

Dimensions of Physical Quantities and Dimensional Analysis
Question 10Question →

The rate of heat transfer through a uniform metallic rod of cross-sectional area AA and length dd is given by Qt=kAΔTd\frac{Q}{t} = \frac{k A \Delta T}{d}, where QQ is heat energy, tt is time, ΔT\Delta T is temperature difference, and kk is thermal conductivity. What is the dimensional formula for kk in terms of mass (MM), length (LL), time (TT), and temperature (Θ\Theta)?

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Answer: MLT−3Θ−1M L T^{-3} \Theta^{-1}

Answer

The dimensional formula for thermal conductivity is MLT−3Θ−1M L T^{-3} \Theta^{-1}.
Expressing thermal conductivity explicitly gives k=QdAΔTtk = \frac{Q d}{A \Delta T t}. Substituting fundamental dimensions yields [k]=(ML2T−2)(L)(L2)(Θ)(T)=MLT−3Θ−1[k] = \frac{(M L^2 T^{-2})(L)}{(L^2)(\Theta)(T)} = M L T^{-3} \Theta^{-1}. Thus, the dimensional formula MLT−3Θ−1M L T^{-3} \Theta^{-1} is correct.

Step-by-Step Solution

1
Make thermal conductivity kk the subject of the formula.
k=QdAΔTtk = \frac{Q d}{A \Delta T t}
Isolating kk enables direct substitution of base dimensional quantities.
2
Substitute fundamental dimensions for each physical quantity.
[k]=[ML2T−2][L][L2][Θ][T][k] = \frac{[M L^2 T^{-2}][L]}{[L^2][\Theta][T]}
Heat energy QQ has dimensions [ML2T−2][M L^2 T^{-2}], distance dd is [L][L], area AA is [L2][L^2], temperature change ΔT\Delta T is [Θ][\Theta], and time tt is [T][T].
3
Combine exponents for each fundamental dimension.
[k]=M1L2+1−2T−2−1Θ−1=MLT−3Θ−1[k] = M^{1} L^{2+1-2} T^{-2-1} \Theta^{-1} = M L T^{-3} \Theta^{-1}
Applying exponent laws simplifies the expression to base units.

Key Concept

Dimensional Analysis of Physical Constants
Question 11Question →

The fundamental frequency ff of a stretched vibrating string depends on the tension force FF, the length of the string ll, and its linear mass density μ\mu (mass per unit length) according to the relation f=kFalbμcf = k F^a l^b \mu^c, where kk is a dimensionless constant. Which of the following sets of exponents (a,b,c)(a, b, c) correctly satisfies dimensional homogeneity?

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Answer: (12,−1,−12)(\frac{1}{2}, -1, -\frac{1}{2})

Answer

The correct set of exponents is (12,−1,−12)(\frac{1}{2}, -1, -\frac{1}{2}).
Equating the base dimensions on both sides gives T−1=Ma+cLa+b−cT−2aT^{-1} = M^{a+c} L^{a+b-c} T^{-2a}. Solving for the powers gives a=12a = \frac{1}{2}, b=−1b = -1, and c=−12c = -\frac{1}{2}, matching the option specifying (12,−1,−12)(\frac{1}{2}, -1, -\frac{1}{2}).

Step-by-Step Solution

1
Express the dimensions of all physical quantities involved in fundamental dimensions MM, LL, and TT.
[f]=T−1[f] = T^{-1}, [F]=MLT−2[F] = M L T^{-2}, [l]=L[l] = L, and [μ]=ML−1[\mu] = M L^{-1}.
Linear mass density μ\mu is mass per unit length (mass/length\text{mass}/\text{length}).
2
Substitute dimensions into the given formula f=kFalbμcf = k F^a l^b \mu^c (ignoring the dimensionless constant kk).
T−1=(MLT−2)a(L)b(ML−1)c=Ma+cLa+b−cT−2aT^{-1} = (M L T^{-2})^a (L)^b (M L^{-1})^c = M^{a+c} L^{a+b-c} T^{-2a}.
Dimensional homogeneity requires both sides of the equation to have identical dimensional powers.
3
Equate exponents of MM, LL, and TT from both sides to form algebraic equations.
For TT: −2a=−1  ⟹  a=12-2a = -1 \implies a = \frac{1}{2}. For MM: a+c=0  ⟹  c=−a=−12a + c = 0 \implies c = -a = -\frac{1}{2}. For LL: a+b−c=0  ⟹  12+b−(−12)=0  ⟹  b+1=0  ⟹  b=−1a + b - c = 0 \implies \frac{1}{2} + b - (-\frac{1}{2}) = 0 \implies b + 1 = 0 \implies b = -1.
Solving the system yields the unique set of exponents (a,b,c)=(12,−1,−12)(a, b, c) = (\frac{1}{2}, -1, -\frac{1}{2}).

Key Concept

Dimensional Analysis and Method of Dimensions
Question 12Question →

The dynamic pressure PP exerted by a moving fluid depends on its density ρ\rho and flow velocity vv according to the relationship P=kρavbP = k \rho^a v^b, where kk is a dimensionless constant. What is the value of the exponent bb?

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Answer: 2

Answer

The value of the exponent bb is 2.
Equating the exponent of time (T) on both sides of the dimensional equation M L−1T−2=MaL−3a+bT−b\text{M L}^{-1} \text{T}^{-2} = \text{M}^a \text{L}^{-3a + b} \text{T}^{-b} yields −2=−b-2 = -b, giving b=2b = 2.

Step-by-Step Solution

1
Determine the base dimensions of all physical quantities in the equation.
Pressure [P]=M L−1T−2[P] = \text{M L}^{-1} \text{T}^{-2}, Density [ρ]=M L−3[\rho] = \text{M L}^{-3}, and Velocity [v]=L T−1[v] = \text{L T}^{-1}.
Dimensional analysis requires converting derived physical quantities into fundamental dimensions of Mass (M), Length (L), and Time (T).
2
Apply dimensional homogeneity by substituting the dimensions into P=kρavbP = k \rho^a v^b.
\text{M L}^{-1} \text{T}^{-2} = \text{M}^a \text{L}^{-3a + b} \text{T}^{-b}
The principle of dimensional homogeneity states that exponents of M, L, and T must match on both sides of a physically correct equation.
3
Equate the corresponding exponents for time (T) and solve for bb.
-2 = -b \implies b = 2
Matching the powers of T directly yields the numerical value of exponent bb.

Key Concept

Principle of Dimensional Homogeneity and Dimensional Analysis
Question 13Question →

Newton's law of universal gravitation states that the gravitational force FF between two point masses m1m_1 and m2m_2 separated by a distance rr is given by F=Gm1m2r2F = \frac{G m_1 m_2}{r^2}, where GG is the universal gravitational constant. What is the dimensional formula of GG?

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Answer: M−1L3T−2M^{-1} L^3 T^{-2}

Answer

The dimensional formula of the universal gravitational constant GG is M−1L3T−2M^{-1} L^3 T^{-2}.
Rearranging F=Gm1m2r2F = \frac{G m_1 m_2}{r^2} gives G=Fr2m1m2G = \frac{F r^2}{m_1 m_2}. Substituting the dimensions of Force (MLT−2M L T^{-2}), distance squared (L2L^2), and mass squared (M2M^2) gives (MLT−2)(L2)M2=M−1L3T−2\frac{(M L T^{-2})(L^2)}{M^2} = M^{-1} L^3 T^{-2}.

Step-by-Step Solution

1
Rearrange the gravitational force equation to solve for GG.
G=Fr2m1m2G = \frac{F r^2}{m_1 m_2}
Isolating the physical constant allows us to substitute the dimensions of each constituent quantity.
2
Substitute the fundamental dimensions for force, distance, and mass.
[F]=MLT−2[F] = M L T^{-2}, [r]=L[r] = L, [m1]=[m2]=M[m_1] = [m_2] = M
Force is mass times acceleration (M⋅LT−2M \cdot L T^{-2}), distance is length (LL), and mass is [M][M].
3
Substitute these fundamental dimensions into the expression for GG and simplify the exponents.
[G]=(MLT−2)(L2)M2=M1−2L1+2T−2=M−1L3T−2[G] = \frac{(M L T^{-2}) (L^2)}{M^2} = M^{1-2} L^{1+2} T^{-2} = M^{-1} L^3 T^{-2}
Applying standard exponent rules yields the final dimensional formula.

Key Concept

Deriving Dimensions of Physical Constants
Dimensions of Physical Quantities and Dimensional Analysis Practice Questions — JAMB UTME | Examkin