Dimensions of Physical Quantities and Dimensional Analysis

13 soru

Soru 1Soru

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?

Cevabı ve açıklamayı göster

Cevap: -0.5

Cevap

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 T2)a(L)b=La+bT2a(\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.

Adım Adım Çözüm

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 T2][\text{L T}^{-2}].
Dimensional analysis requires substituting each quantity with its fundamental dimensions.
2
Formulate the dimensional homogeneity equation.
M0L0T1=(L T2)a(L)b=La+bT2a\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.

Anahtar Kavram

Principle of Dimensional Homogeneity
Tahmini Süre:1m 15s
Soru 2Soru

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?

Cevabı ve açıklamayı göster

Cevap: M1L3T2M^{-1} L^3 T^{-2}

Cevap

The dimensional formula for the universal gravitational constant GG is M1L3T2M^{-1} L^3 T^{-2}.
Isolating GG gives G=Fr2m1m2G = \frac{F r^2}{m_1 m_2}. Substituting the fundamental dimensions for force (MLT2M L T^{-2}), distance (LL), and mass (MM) yields (MLT2)(L2)M2=M1L3T2\frac{(M L T^{-2})(L^2)}{M^2} = M^{-1} L^3 T^{-2}.

Adım Adım Çözüm

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 MLT2M 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.

Anahtar Kavram

Dimensions of Physical Constants
Tahmini Süre:1m 15s
Soru 3Soru

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?

Cevabı ve açıklamayı göster

Cevap: ML1T1M L^{-1} T^{-1}

Cevap

ML1T1M L^{-1} T^{-1}
Rearranging Stokes' law gives η=F6πrv\eta = \frac{F}{6\pi r v}. Substituting base dimensions [F]=MLT2[F] = M L T^{-2}, [r]=L[r] = L, and [v]=LT1[v] = L T^{-1} yields [η]=MLT2L2T1=ML1T1[\eta] = \frac{M L T^{-2}}{L^2 T^{-1}} = M L^{-1} T^{-1}.

Adım Adım Çözüm

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]=MLT2LLT1[\eta] = \frac{M L T^{-2}}{L \cdot L T^{-1}}
The constant 6π6\pi is dimensionless, while [F]=MLT2[F] = M L T^{-2}, [r]=L[r] = L, and [v]=LT1[v] = L T^{-1}.
3
Simplify the powers of base quantities MM, LL, and TT
[η]=ML12T2(1)=ML1T1[\eta] = M L^{1 - 2} T^{-2 - (-1)} = M L^{-1} T^{-1}
Applying algebraic rules of exponents simplifies the expression.

Anahtar Kavram

Dimensions of Physical Quantities and Dimensional Analysis
Soru 4Soru

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?

Cevabı ve açıklamayı göster

Cevap: ML2T3M L^2 T^{-3}

Cevap

ML2T3M 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]=ML2T2[W] = M L^2 T^{-2}, and for time is [t]=T[t] = T. Dividing work by time gives [P]=ML2T2T=ML2T3[P] = \frac{M L^2 T^{-2}}{T} = M L^2 T^{-3}.

Adım Adım Çözüm

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]=(MLT2)×L=ML2T2[\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 (MLT2M 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]=ML2T2T=ML2T3[\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.

Anahtar Kavram

Dimensional Formula for Power
Soru 5Soru

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

Cevabı ve açıklamayı göster

Cevap: ML1T2M L^{-1} T^{-2}

Cevap

ML1T2M L^{-1} T^{-2}
Pressure is defined as Force divided by Area. Substituting the fundamental dimensions gives Force as MLT2M L T^{-2} and Area as L2L^2. Dividing Force by Area results in MLT2L2=ML1T2\frac{M L T^{-2}}{L^2} = M L^{-1} T^{-2}. Therefore, the expression ML1T2M L^{-1} T^{-2} is correct.

Adım Adım Çözüm

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]=[MLT2][L2][P] = \frac{[M L T^{-2}]}{[L^2]}
Force has dimensions MLT2M L T^{-2} and area has dimensions L2L^2.
3
Simplify the index of length LL
[P]=ML12T2=ML1T2[P] = M L^{1 - 2} T^{-2} = M L^{-1} T^{-2}
Applying the rules of indices for length gives L1L2=L1L^1 \cdot L^{-2} = L^{-1}.

Anahtar Kavram

Dimensions of Pressure
Soru 6Soru

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?

Cevabı ve açıklamayı göster

Cevap: ML2T1M L^2 T^{-1}

Cevap

The fundamental dimensions of Planck's constant are ML2T1M L^2 T^{-1}.
Planck's constant hh is given by h=E/fh = E / f. Since energy EE has fundamental dimensions of ML2T2M L^2 T^{-2} and frequency ff has dimensions of T1T^{-1}, dividing energy by frequency yields ML2T2T1=ML2T1\frac{M L^2 T^{-2}}{T^{-1}} = M L^2 T^{-1}.

Adım Adım Çözüm

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]=ML2T2[E] = M L^2 T^{-2} and [f]=T1[f] = T^{-1}
Energy has dimensions of work (F×d=MLT2L=ML2T2F \times d = M L T^{-2} \cdot L = M L^2 T^{-2}) and frequency is inverse time (T1T^{-1}).
3
Divide the dimensions of energy by the dimensions of frequency
[h]=ML2T2T1=ML2T2(1)=ML2T1[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.

Anahtar Kavram

Dimensional Analysis of Physical Constants
Tahmini Süre:1m 0s
Soru 7Soru

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?

Cevabı ve açıklamayı göster

Cevap: 0.5

Cevap

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 [M0L1T1]=[ML1T2]a[ML3]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.

Adım Adım Çözüm

1
Determine the base dimensions of all physical quantities in the relationship
[v]=M0LT1[v] = M^0 L T^{-1}, [P]=ML1T2[P] = M L^{-1} T^{-2}, and [ρ]=ML3[\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
M0L1T1=Ma+bLa3bT2aM^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.

Anahtar Kavram

Dimensional Homogeneity and Derivation of Exponents
Soru 8Soru

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)?

Cevabı ve açıklamayı göster

Cevap: M1L3T4I2M^{-1} L^{-3} T^4 I^2

Cevap

The dimensions of permittivity of free space ε0\varepsilon_0 are M1L3T4I2M^{-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]=MLT2[F] = M L T^{-2}, and [r]=L[r] = L gives [ε0]=I2T2ML3T2=M1L3T4I2[\varepsilon_0] = \frac{I^2 T^2}{M L^3 T^{-2}} = M^{-1} L^{-3} T^4 I^2.

Adım Adım Çözüm

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]=MLT2[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 (MLT2M L T^{-2}).
3
Substitute the fundamental dimensions into the rearranged equation and simplify exponent powers
[ε0]=(IT)(IT)(MLT2)(L2)=I2T2ML3T2=M1L3T4I2[\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/T2=T2(2)=T4T^2 / T^{-2} = T^{2 - (-2)} = T^4, 1/M=M11 / M = M^{-1}, and 1/L3=L31 / L^3 = L^{-3}.

Anahtar Kavram

Dimensional analysis of physical constants in electromagnetism
Soru 9Soru

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?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The value of the exponent xx is 4.
Applying the principle of dimensional homogeneity, the dimensions of volume flow rate [Q]=L3T1[Q] = L^3 T^{-1} are equated to [r]x[η]y[ΔPL]z=Lx(ML1T1)y(ML2T2)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, y2z=1-y - 2z = -1 for time, and xy2z=3x - y - 2z = 3 for length. Solving these simultaneous equations gives z=1z = 1, y=1y = -1, and x=4x = 4.

Adım Adım Çözüm

1
Identify the base dimensions of each physical quantity in the given equation.
Flow rate [Q]=M0L3T1[Q] = M^0 L^3 T^{-1}, radius [r]=L[r] = L, viscosity [η]=ML1T1[\eta] = M L^{-1} T^{-1}, and pressure gradient [ΔPL]=ML2T2\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.
M0L3T1=My+zLxy2zTy2zM^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, y2z=1-y - 2z = -1, and xy2z=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.

Anahtar Kavram

Dimensions of Physical Quantities and Dimensional Analysis
Soru 10Soru

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)?

Cevabı ve açıklamayı göster

Cevap: MLT3Θ1M L T^{-3} \Theta^{-1}

Cevap

The dimensional formula for thermal conductivity is MLT3Θ1M L T^{-3} \Theta^{-1}.
Expressing thermal conductivity explicitly gives k=QdAΔTtk = \frac{Q d}{A \Delta T t}. Substituting fundamental dimensions yields [k]=(ML2T2)(L)(L2)(Θ)(T)=MLT3Θ1[k] = \frac{(M L^2 T^{-2})(L)}{(L^2)(\Theta)(T)} = M L T^{-3} \Theta^{-1}. Thus, the dimensional formula MLT3Θ1M L T^{-3} \Theta^{-1} is correct.

Adım Adım Çözüm

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]=[ML2T2][L][L2][Θ][T][k] = \frac{[M L^2 T^{-2}][L]}{[L^2][\Theta][T]}
Heat energy QQ has dimensions [ML2T2][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+12T21Θ1=MLT3Θ1[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.

Anahtar Kavram

Dimensional Analysis of Physical Constants
Soru 11Soru

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?

Cevabı ve açıklamayı göster

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

Cevap

The correct set of exponents is (12,1,12)(\frac{1}{2}, -1, -\frac{1}{2}).
Equating the base dimensions on both sides gives T1=Ma+cLa+bcT2aT^{-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}).

Adım Adım Çözüm

1
Express the dimensions of all physical quantities involved in fundamental dimensions MM, LL, and TT.
[f]=T1[f] = T^{-1}, [F]=MLT2[F] = M L T^{-2}, [l]=L[l] = L, and [μ]=ML1[\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).
T1=(MLT2)a(L)b(ML1)c=Ma+cLa+bcT2aT^{-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+bc=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}).

Anahtar Kavram

Dimensional Analysis and Method of Dimensions
Soru 12Soru

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?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The value of the exponent bb is 2.
Equating the exponent of time (T) on both sides of the dimensional equation M L1T2=MaL3a+bTb\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.

Adım Adım Çözüm

1
Determine the base dimensions of all physical quantities in the equation.
Pressure [P]=M L1T2[P] = \text{M L}^{-1} \text{T}^{-2}, Density [ρ]=M L3[\rho] = \text{M L}^{-3}, and Velocity [v]=L T1[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.

Anahtar Kavram

Principle of Dimensional Homogeneity and Dimensional Analysis
Soru 13Soru

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?

Cevabı ve açıklamayı göster

Cevap: M1L3T2M^{-1} L^3 T^{-2}

Cevap

The dimensional formula of the universal gravitational constant GG is M1L3T2M^{-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 (MLT2M L T^{-2}), distance squared (L2L^2), and mass squared (M2M^2) gives (MLT2)(L2)M2=M1L3T2\frac{(M L T^{-2})(L^2)}{M^2} = M^{-1} L^3 T^{-2}.

Adım Adım Çözüm

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]=MLT2[F] = M L T^{-2}, [r]=L[r] = L, [m1]=[m2]=M[m_1] = [m_2] = M
Force is mass times acceleration (MLT2M \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]=(MLT2)(L2)M2=M12L1+2T2=M1L3T2[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.

Anahtar Kavram

Deriving Dimensions of Physical Constants
Dimensions of Physical Quantities and Dimensional Analysis Alıştırma Soruları — JAMB UTME | Examkin