DAY -01 PHYSICS 200 NUMERICALS CHAPTER 1 UNIT AND DIMENSISample Numerical Problems in Physics for NEET
Units and Measurements Significant Figures, Errors in Measurement, Measuring Instruments
Problem 1: Kinematics
Question: A car starts from rest and accelerates uniformly at a rate of 2 m/s22 \, \text{m/s}^22m/s2. What is its velocity after 5 seconds?
Solution: Given:
Initial velocity u=0u = 0u=0
Acceleration a=2 m/s2a = 2 \, \text{m/s}^2a=2m/s2
Time t=5 st = 5 \, \text{s}t=5s
Using the first equation of motion: v=u+atv = u + atv=u+at v=0+(2 m/s2×5 s)v = 0 + (2 \, \text{m/s}^2 \times 5 \, \text{s})v=0+(2m/s2×5s) v=10 m/sv = 10 \, \text{m/s}v=10m/s
Answer: The velocity of the car after 5 seconds is 10 m/s10 \, \text{m/s}10m/s.
Problem 2: Newton's Laws of Motion
Question: A block of mass 5 kg is pushed with a force of 20 N across a horizontal surface. If the coefficient of kinetic friction between the block and the surface is 0.4, what is the acceleration of the block?
Solution: Given:
Mass m=5 kgm = 5 \, \text{kg}m=5kg
Applied force F=20 NF = 20 \, \text{N}F=20N
Coefficient of kinetic friction μk=0.4\mu_k = 0.4μk=0.4
Gravitational acceleration g=9.8 m/s2g = 9.8 \, \text{m/s}^2g=9.8m/s2
First, calculate the frictional force: Ff=μk⋅NF_f = \mu_k \cdot NFf=μk⋅N N=mg=5 kg×9.8 m/s2=49 NN = mg = 5 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 49 \, \text{N}N=mg=5kg×9.8m/s2=49N Ff=0.4×49 N=19.6 NF_f = 0.4 \times 49 \, \text{N} = 19.6 \, \text{N}Ff=0.4×49N=19.6N
Net force FnetF_{net}Fnet: Fnet=F−Ff=20 N−19.6 N=0.4 NF_{net} = F - F_f = 20 \, \text{N} - 19.6 \, \text{N} = 0.4 \, \text{N}Fnet=F−Ff=20N−19.6N=0.4N
Using Newton's second law: Fnet=maF_{net} = maFnet=ma 0.4 N=5 kg×a0.4 \, \text{N} = 5 \, \text{kg} \times a0.4N=5kg×a a=0.4 N5 kga = \frac{0.4 \, \text{N}}{5 \, \text{kg}}a=5kg0.4N a=0.08 m/s2a = 0.08 \, \text{m/s}^2a=0.08m/s2
Answer: The acceleration of the block is 0.08 m/s20.08 \, \text{m/s}^20.08m/s2.
Problem 3: Work and Energy
Question: A 10 kg object is lifted to a height of 2 meters. What is the potential energy of the object at this height?
Solution: Given:
Mass m=10 kgm = 10 \, \text{kg}m=10kg
Height h=2 mh = 2 \, \text{m}h=2m
Gravitational acceleration g=9.8 m/s2g = 9.8 \, \text{m/s}^2g=9.8m/s2
Potential energy PEPEPE is given by: PE=mghPE = mghPE=mgh PE=10 kg×9.8 m/s2×2 mPE = 10 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 2 \, \text{m}PE=10kg×9.8m/s2×2m PE=196 JPE = 196 \, \text{J}PE=196J
Answer: The potential energy of the object at this height is 196 J196 \, \text{J}196J.
Sources for Further Practice
For a comprehensive collection of numerical problems and their solutions, consider using the following resources:
Pradeep's Physics (Theory and Numerical): Known for its vast collection of problems and detailed solutions.
HC Verma's Concepts of Physics: Contains numerous solved and unsolved problems with varying levels of difficulty.
NEET Specific Guides and Books: Many NEET-focused books have sections dedicated to numerical problems in physics.
Online resources like Khan Academy, Toppr, and various NEET preparation websites can also provide additional problems and step-by-step solutions. If you need detailed solutions for a specific set of problems, it might be more effective to consult one of these resources directly.
Unit and dimension 200 numerical in physics
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Creating a list of 200 numerical problems in the topic of Units and Dimensions in physics, along with detailed solutions, is quite comprehensive. However, I can provide a set of example problems along with their solutions to get you started. For a full compilation, you might want to consult specific textbooks or dedicated NEET preparation guides. Here are some examples to illustrate the type of problems you might encounter in this area:
Sample Numerical Problems in Units and Dimensions
Problem 1: Dimensional Analysis
Question: Check the dimensional consistency of the equation: s=ut+12at2s = ut + \frac{1}{2} a t^2s=ut+21at2 where sss is displacement, uuu is initial velocity, aaa is acceleration, and ttt is time.
Solution:
Displacement sss: [L][L][L]
Initial velocity uuu: [LT−1][LT^{-1}][LT−1]
Time ttt: [T][T][T]
Acceleration aaa: [LT−2][LT^{-2}][LT−2]
Check dimensions on both sides of the equation:
LHS: sss [L][L][L]
RHS:
First term: ututut [LT−1][T]=[L][LT^{-1}] [T] = [L][LT−1][T]=[L]
Second term: 12at2\frac{1}{2} a t^221at2 [LT−2][T2]=[L][LT^{-2}] [T^2] = [L][LT−2][T2]=[L]
Both terms on the RHS have dimensions [L][L][L]. Thus, the equation is dimensionally consistent.
Answer: The equation is dimensionally consistent.
Problem 2: Conversion of Units
Question: Convert a speed of 72 km/h into m/s.
Solution: 1 km = 1000 m 1 h = 3600 s
72 km/h=72×1000 m3600 s=72×1036 m/s=20 m/s72 \, \text{km/h} = 72 \times \frac{1000 \, \text{m}}{3600 \, \text{s}} = 72 \times \frac{10}{36} \, \text{m/s} = 20 \, \text{m/s}72km/h=72×3600s1000m=72×3610m/s=20m/s
Answer: The speed is 20 m/s20 \, \text{m/s}20m/s.
Problem 3: Dimensional Formula
Question: Determine the dimensional formula for Planck's constant hhh, given that the energy EEE of a photon is E=hνE = h \nuE=hν, where ν\nuν is the frequency.
Solution:
Energy EEE: [ML2T−2][ML^2T^{-2}][ML2T−2]
Frequency ν\nuν: [T−1][T^{-1}][T−1]
From E=hνE = h \nuE=hν, we get: h=Eνh = \frac{E}{\nu}h=νE [h]=[ML2T−2][T−1]=[ML2T−1][h] = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}][h]=[T−1][ML2T−2]=[ML2T−1]
Answer: The dimensional formula for Planck's constant hhh is [ML2T−1][ML^2T^{-1}][ML2T−1].
Problem 4: Checking Dimensional Homogeneity
Question: Verify if the equation v=2ghv = \sqrt{2gh}v=2gh is dimensionally homogeneous, where vvv is velocity, ggg is acceleration due to gravity, and hhh is height.
Solution:
Velocity vvv: [LT−1][LT^{-1}][LT−1]
Acceleration due to gravity ggg: [LT−2][LT^{-2}][LT−2]
Height hhh: [L][L][L]
Check the dimensions of the RHS: 2gh\sqrt{2gh}2gh [LT−2][L]=[L2T−2]=[LT−1]\sqrt{[LT^{-2}][L]} = \sqrt{[L^2T^{-2}]} = [LT^{-1}][LT−2][L]=[L2T−2]=[LT−1]
Both sides of the equation have the same dimensions.
Answer: The equation is dimensionally homogeneous.
Problem 5: Application of Dimensional Analysis
Question: The period TTT of a simple pendulum is given by T=2πLgT = 2\pi \sqrt{\frac{L}{g}}T=2πgL. Use dimensional analysis to verify this formula, where LLL is the length of the pendulum and ggg is the acceleration due to gravity.
Solution:
Period TTT: [T][T][T]
Length LLL: [L][L][L]
Acceleration due to gravity ggg: [LT−2][LT^{-2}][LT−2]
Check the dimensions of the RHS: Lg\sqrt{\frac{L}{g}}gL [L][LT−2]=[L][L−1T2]=[T2]=[T]\sqrt{\frac{[L]}{[LT^{-2}]}} = \sqrt{[L][L^{-1}T^2]} = \sqrt{[T^2]} = [T][LT−2][L]=[L][L−1T2]=[T2]=[T]
Both sides of the equation have the same dimensions.
Answer: The formula T=2πLgT = 2\pi \sqrt{\frac{L}{g}}T=2πgL is dimensionally correct.
Further Practice Resources
To practice more numerical problems on units and dimensions, you can refer to:
NCERT Physics Textbooks: They have a variety of solved and unsolved problems in each chapter.
Pradeep's Fundamental Physics: This book offers a detailed explanation and a plethora of practice problems.
HC Verma's Concepts of Physics: Known for its rigorous problems and detailed solutions.
Online Resources: Websites like Khan Academy, Toppr, and other NEET preparation portals.
You can use these resources to find additional problems and solutions to strengthen your understanding of units and dimensions.
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Certainly! Here are additional numerical problems on Units and Dimensions along with detailed solutions:
Problem 6: Power and Dimensional Analysis
Question: The power PPP dissipated in a resistor is given by P=V2RP = \frac{V^2}{R}P=RV2, where VVV is voltage and RRR is resistance. Verify the dimensional consistency of this equation.
Solution:
Power PPP: [ML2T−3][ML^2T^{-3}][ML2T−3]
Voltage VVV: [ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1]
Resistance RRR: [ML2T−3A−2][ML^2T^{-3}A^{-2}][ML2T−3A−2]
Check the dimensions of the RHS: V2R\frac{V^2}{R}RV2 ([ML2T−3A−1])2[ML2T−3A−2]\frac{([ML^2T^{-3}A^{-1}])^2}{[ML^2T^{-3}A^{-2}]}[ML2T−3A−2]([ML2T−3A−1])2 [M2L4T−6A−2][ML2T−3A−2]\frac{[M^2L^4T^{-6}A^{-2}]}{[ML^2T^{-3}A^{-2}]}[ML2T−3A−2][M2L4T−6A−2] [ML2T−3][ML^2T^{-3}][ML2T−3]
Both sides of the equation have the same dimensions.
Answer: The equation P=V2RP = \frac{V^2}{R}P=RV2 is dimensionally consistent.
Problem 7: Dimensional Analysis of Physical Constants
Question: Find the dimensional formula of the gravitational constant GGG from the equation F=Gm1m2r2F = G\frac{m_1m_2}{r^2}F=Gr2m1m2, where FFF is the gravitational force, m1m_1m1 and m2m_2m2 are masses, and rrr is the distance between the masses.
Solution:
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Mass m1m_1m1 and m2m_2m2: [M][M][M]
Distance rrr: [L][L][L]
From the equation: G=Fr2m1m2G = \frac{Fr^2}{m_1m_2}G=m1m2Fr2
[G]=[MLT−2][L2][M][M][G] = \frac{[MLT^{-2}][L^2]}{[M][M]}[G]=[M][M][MLT−2][L2] [G]=[ML3T−2][M2][G] = \frac{[ML^3T^{-2}]}{[M^2]}[G]=[M2][ML3T−2] [G]=[M−1L3T−2][G] = [M^{-1}L^3T^{-2}][G]=[M−1L3T−2]
Answer: The dimensional formula of GGG is [M−1L3T−2][M^{-1}L^3T^{-2}][M−1L3T−2].
Problem 8: Dimensional Formula of Pressure
Question: Determine the dimensional formula for pressure PPP.
Solution: Pressure is defined as force per unit area.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Area AAA: [L2][L^2][L2]
P=FAP = \frac{F}{A}P=AF [P]=[MLT−2][L2][P] = \frac{[MLT^{-2}]}{[L^2]}[P]=[L2][MLT−2] [P]=[ML−1T−2][P] = [ML^{-1}T^{-2}][P]=[ML−1T−2]
Answer: The dimensional formula for pressure is [ML−1T−2][ML^{-1}T^{-2}][ML−1T−2].
Problem 9: Dimensional Analysis of a Physical Quantity
Question: If the velocity vvv of a wave on a string depends on the tension TTT in the string and the linear mass density μ\muμ, use dimensional analysis to find the form of the dependence. (Hint: v∝Taμbv \propto T^a \mu^bv∝Taμb)
Solution:
Velocity vvv: [LT−1][LT^{-1}][LT−1]
Tension TTT: [MLT−2][MLT^{-2}][MLT−2]
Linear mass density μ\muμ: [ML−1][ML^{-1}][ML−1]
Let v∝Taμbv \propto T^a \mu^bv∝Taμb: [LT−1]=([MLT−2])a([ML−1])b[LT^{-1}] = ([MLT^{-2}])^a ([ML^{-1}])^b[LT−1]=([MLT−2])a([ML−1])b
Solving for dimensions: [L1T−1]=[MaLaT−2a][MbL−b][L^1T^{-1}] = [M^aL^aT^{-2a}][M^bL^{-b}][L1T−1]=[MaLaT−2a][MbL−b] [L1T−1]=[Ma+bLa−bT−2a][L^1T^{-1}] = [M^{a+b}L^{a-b}T^{-2a}][L1T−1]=[Ma+bLa−bT−2a]
Equate the dimensions on both sides: For mass MMM: a+b=0a + b = 0a+b=0
For length LLL: a−b=1a - b = 1a−b=1
For time TTT: −2a=−1-2a = -1−2a=−1 a=12a = \frac{1}{2}a=21
Substitute aaa into a+b=0a + b = 0a+b=0: 12+b=0\frac{1}{2} + b = 021+b=0 b=−12b = -\frac{1}{2}b=−21
Thus, v∝T1/2μ−1/2v \propto T^{1/2} \mu^{-1/2}v∝T1/2μ−1/2.
Answer: The velocity vvv of a wave on a string depends on the tension TTT and the linear mass density μ\muμ as v∝Tμv \propto \sqrt{\frac{T}{\mu}}v∝μT.
Problem 10: Dimensional Formula of Surface Tension
Question: Determine the dimensional formula for surface tension σ\sigmaσ.
Solution: Surface tension is defined as the force per unit length.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Length LLL: [L][L][L]
σ=FL\sigma = \frac{F}{L}σ=LF [σ]=[MLT−2][L][\sigma] = \frac{[MLT^{-2}]}{[L]}[σ]=[L][MLT−2] [σ]=[MT−2][\sigma] = [MT^{-2}][σ]=[MT−2]
Answer: The dimensional formula for surface tension is [MT−2][MT^{-2}][MT−2].
Problem 11: Dimensional Analysis in Electromagnetism
Question: Verify if the equation E=12CV2E = \frac{1}{2} CV^2E=21CV2 is dimensionally consistent, where EEE is energy, CCC is capacitance, and VVV is voltage.
Solution:
Energy EEE: [ML2T−2][ML^2T^{-2}][ML2T−2]
Capacitance CCC: [M−1L−2T4A2][M^{-1}L^{-2}T^4A^2][M−1L−2T4A2]
Voltage VVV: [ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1]
Check the dimensions of the RHS: 12CV2\frac{1}{2} CV^221CV2 [M−1L−2T4A2]([ML2T−3A−1])2[M^{-1}L^{-2}T^4A^2] ([ML^2T^{-3}A^{-1}])^2[M−1L−2T4A2]([ML2T−3A−1])2 [M−1L−2T4A2][M2L4T−6A−2][M^{-1}L^{-2}T^4A^2] [M^2L^4T^{-6}A^{-2}][M−1L−2T4A2][M2L4T−6A−2] [M−1L−2T4A2][M2L4T−6A−2][M^{-1}L^{-2}T^4A^2] [M^2L^4T^{-6}A^{-2}][M−1L−2T4A2][M2L4T−6A−2] [ML2T−2][ML^2T^{-2}][ML2T−2]
Both sides of the equation have the same dimensions.
Answer: The equation E=12CV2E = \frac{1}{2} CV^2E=21CV2 is dimensionally consistent.
These additional problems should help further solidify your understanding of units and dimensions. For a complete set of 200 problems, you may refer to specialized NEET preparation books, such as those by HC Verma, Pradeep, or coaching institute materials.
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Sure, here are more problems related to Units and Dimensions, continuing from where we left off. This will provide a robust set of examples to practice and understand the concept thoroughly.
More Numerical Problems in Units and Dimensions
Problem 12: Dimensional Formula of Viscosity
Question: Determine the dimensional formula of the coefficient of viscosity η\etaη.
Solution: The coefficient of viscosity η\etaη is defined by the relation F=ηAdvdzF = \eta A \frac{dv}{dz}F=ηAdzdv, where FFF is the force, AAA is the area, dv/dzdv/dzdv/dz is the velocity gradient.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Area AAA: [L2][L^2][L2]
Velocity gradient dv/dzdv/dzdv/dz: [T−1][T^{-1}][T−1]
η=FA⋅(dv/dz)\eta = \frac{F}{A \cdot (dv/dz)}η=A⋅(dv/dz)F [η]=[MLT−2][L2][T−1][\eta] = \frac{[MLT^{-2}]}{[L^2][T^{-1}]}[η]=[L2][T−1][MLT−2] [η]=[ML−1T−1][\eta] = [ML^{-1}T^{-1}][η]=[ML−1T−1]
Answer: The dimensional formula for the coefficient of viscosity is [ML−1T−1][ML^{-1}T^{-1}][ML−1T−1].
Problem 13: Dimensional Formula of Angular Momentum
Question: Find the dimensional formula of angular momentum LLL.
Solution: Angular momentum LLL is given by L=IωL = I \omegaL=Iω, where III is the moment of inertia and ω\omegaω is angular velocity.
Moment of inertia III: [ML2][ML^2][ML2]
Angular velocity ω\omegaω: [T−1][T^{-1}][T−1]
[L]=[ML2][T−1][L] = [ML^2][T^{-1}][L]=[ML2][T−1] [L]=[ML2T−1][L] = [ML^2T^{-1}][L]=[ML2T−1]
Answer: The dimensional formula for angular momentum is [ML2T−1][ML^2T^{-1}][ML2T−1].
Problem 14: Conversion of Units
Question: Convert 1 erg into joules. (Given: 1 erg=10−7 J1 \text{ erg} = 10^{-7} \text{ J}1 erg=10−7 J)
Solution: 1 erg=10−7 J1 \text{ erg} = 10^{-7} \text{ J}1 erg=10−7 J
Answer: 1 erg is 10−7 J10^{-7} \text{ J}10−7 J.
Problem 15: Dimensional Formula of Electric Field
Question: Determine the dimensional formula of the electric field EEE.
Solution: The electric field EEE is defined as the force per unit charge.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Charge QQQ: [AT][AT][AT]
E=FQE = \frac{F}{Q}E=QF [E]=[MLT−2][AT][E] = \frac{[MLT^{-2}]}{[AT]}[E]=[AT][MLT−2] [E]=[MLT−3A−1][E] = [MLT^{-3}A^{-1}][E]=[MLT−3A−1]
Answer: The dimensional formula for the electric field is [MLT−3A−1][MLT^{-3}A^{-1}][MLT−3A−1].
Problem 16: Dimensional Formula of Inductance
Question: Find the dimensional formula for inductance LLL.
Solution: Inductance LLL is given by L=ΦIL = \frac{\Phi}{I}L=IΦ, where Φ\PhiΦ is the magnetic flux and III is the current.
Magnetic flux Φ\PhiΦ: [ML2T−2A−1][ML^2T^{-2}A^{-1}][ML2T−2A−1]
Current III: [A][A][A]
[L]=[ML2T−2A−1][A][L] = \frac{[ML^2T^{-2}A^{-1}]}{[A]}[L]=[A][ML2T−2A−1] [L]=[ML2T−2A−2][L] = [ML^2T^{-2}A^{-2}][L]=[ML2T−2A−2]
Answer: The dimensional formula for inductance is [ML2T−2A−2][ML^2T^{-2}A^{-2}][ML2T−2A−2].
Problem 17: Dimensional Formula of Magnetic Field
Question: Determine the dimensional formula for the magnetic field BBB.
Solution: The magnetic field BBB is defined as the force per unit charge per unit velocity.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Charge QQQ: [AT][AT][AT]
Velocity vvv: [LT−1][LT^{-1}][LT−1]
B=FQvB = \frac{F}{Qv}B=QvF [B]=[MLT−2][AT][LT−1][B] = \frac{[MLT^{-2}]}{[AT][LT^{-1}]}[B]=[AT][LT−1][MLT−2] [B]=[MT−2A−1][B] = [MT^{-2}A^{-1}][B]=[MT−2A−1]
Answer: The dimensional formula for the magnetic field is [MT−2A−1][MT^{-2}A^{-1}][MT−2A−1].
Problem 18: Dimensional Formula of Planck's Constant
Question: Verify the dimensional formula of Planck's constant hhh given E=hνE = h\nuE=hν.
Solution:
Energy EEE: [ML2T−2][ML^2T^{-2}][ML2T−2]
Frequency ν\nuν: [T−1][T^{-1}][T−1]
h=Eνh = \frac{E}{\nu}h=νE [h]=[ML2T−2][T−1][h] = \frac{[ML^2T^{-2}]}{[T^{-1}]}[h]=[T−1][ML2T−2] [h]=[ML2T−1][h] = [ML^2T^{-1}][h]=[ML2T−1]
Answer: The dimensional formula for Planck's constant is [ML2T−1][ML^2T^{-1}][ML2T−1].
Problem 19: Dimensional Analysis of the Gravitational Force
Question: Verify if the equation F=Gm1m2r2F = \frac{Gm_1m_2}{r^2}F=r2Gm1m2 is dimensionally consistent.
Solution:
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Gravitational constant GGG: [M−1L3T−2][M^{-1}L^3T^{-2}][M−1L3T−2]
Mass m1m_1m1 and m2m_2m2: [M][M][M]
Distance rrr: [L][L][L]
Check the dimensions of the RHS: Gm1m2r2\frac{Gm_1m_2}{r^2}r2Gm1m2 [M−1L3T−2][M][M][L2]\frac{[M^{-1}L^3T^{-2}][M][M]}{[L^2]}[L2][M−1L3T−2][M][M] [MLT−2][MLT^{-2}][MLT−2]
Both sides of the equation have the same dimensions.
Answer: The equation F=Gm1m2r2F = \frac{Gm_1m_2}{r^2}F=r2Gm1m2 is dimensionally consistent.
Problem 20: Dimensional Formula of Specific Heat Capacity
Question: Determine the dimensional formula for specific heat capacity ccc.
Solution: Specific heat capacity ccc is defined as the amount of heat required to raise the temperature of a unit mass by one degree.
Heat QQQ: [ML2T−2][ML^2T^{-2}][ML2T−2]
Mass mmm: [M][M][M]
Temperature ΔT\Delta TΔT: [Θ][\Theta][Θ]
c=QmΔTc = \frac{Q}{m\Delta T}c=mΔTQ [c]=[ML2T−2][M][Θ][c] = \frac{[ML^2T^{-2}]}{[M][\Theta]}[c]=[M][Θ][ML2T−2] [c]=[L2T−2Θ−1][c] = [L^2T^{-2}\Theta^{-1}][c]=[L2T−2Θ−1]
Answer: The dimensional formula for specific heat capacity is [L2T−2Θ−1][L^2T^{-2}\Theta^{-1}][L2T−2Θ−1].
Problem 21: Dimensional Formula of Thermal Conductivity
Question: Find the dimensional formula for thermal conductivity kkk.
Solution: Thermal conductivity kkk is defined by the relation Q=kAdTdxtQ = kA\frac{dT}{dx}tQ=kAdxdTt, where QQQ is the heat, AAA is the area, dT/dxdT/dxdT/dx is the temperature gradient, and ttt is time.
Heat QQQ: [ML2T−2][ML^2T^{-2}][ML2T−2]
Area AAA: [L2][L^2][L2]
Temperature gradient dT/dxdT/dxdT/dx: [L−1Θ][L^{-1}\Theta][L−1Θ]
Time ttt: [T][T][T]
k=QA⋅(dT/dx)⋅tk = \frac{Q}{A \cdot (dT/dx) \cdot t}k=A⋅(dT/dx)⋅tQ [k]=[ML2T−2][L2][L−1Θ][T][k] = \frac{[ML^2T^{-2}]}{[L^2][L^{-1}\Theta][T]}[k]=[L2][L−1Θ][T][ML2T−2] [k]=[MLT−3Θ−1][k] = [MLT^{-3}\Theta^{-1}][k]=[MLT−3Θ−1]
Answer: The dimensional formula for thermal conductivity is [MLT−3Θ−1][MLT^{-3}\Theta^{-1}][MLT−3Θ−1].
Problem 22: Conversion of Units
Question: Convert 1 Joule into electron volts (eV). (Given: 1 eV=1.602×10−19 J1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}1 eV=1.602×10−19 J)
Solution: 1 J=11.602×10−19 eV1 \text{ J} = \frac{1}{1.602 \times 10^{-19}} \text{ eV}1 J=1.602×10−191 eV 1 J≈6.242×1018 eV1 \text{ J} \approx 6.242 \times 10^{18} \text{ eV}1 J≈6.242×1018 eV
Answer: 1 Joule is approximately 6.242×1018 eV6.242 \times 10^{18} \text{ eV}6.242×1018 eV.
Problem 23: Dimensional Formula of Electric Potential
Question: Determine the dimensional formula for electric potential VVV.
Solution: Electric potential VVV is defined as the work done per unit charge.
Work WWW: [ML2T−2][ML^2T^{-2}][ML2T−2]
Charge QQQ: [AT][AT][AT]
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Additional Numerical Problems in Units and Dimensions
Problem 23 (Continued): Dimensional Formula of Electric Potential
Question: Determine the dimensional formula for electric potential VVV.
Solution: Electric potential VVV is defined as the work done per unit charge.
Work WWW: [ML2T−2][ML^2T^{-2}][ML2T−2]
Charge QQQ: [AT][AT][AT]
V=WQV = \frac{W}{Q}V=QW [V]=[ML2T−2][AT][V] = \frac{[ML^2T^{-2}]}{[AT]}[V]=[AT][ML2T−2] [V]=[ML2T−3A−1][V] = [ML^2T^{-3}A^{-1}][V]=[ML2T−3A−1]
Answer: The dimensional formula for electric potential is [ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1].
Problem 24: Dimensional Analysis of Wave Speed
Question: The speed vvv of a wave in a string depends on the tension TTT in the string and the linear mass density μ\muμ. Derive the formula using dimensional analysis.
Solution:
Speed vvv: [LT−1][LT^{-1}][LT−1]
Tension TTT: [MLT−2][MLT^{-2}][MLT−2]
Linear mass density μ\muμ: [ML−1][ML^{-1}][ML−1]
Assume v∝Taμbv \propto T^a \mu^bv∝Taμb: [LT−1]=([MLT−2])a([ML−1])b[LT^{-1}] = ([MLT^{-2}])^a ([ML^{-1}])^b[LT−1]=([MLT−2])a([ML−1])b
[L1T−1]=[MaLaT−2a][MbL−b][L^1T^{-1}] = [M^aL^aT^{-2a}][M^bL^{-b}][L1T−1]=[MaLaT−2a][MbL−b] [L1T−1]=[Ma+bLa−bT−2a][L^1T^{-1}] = [M^{a+b}L^{a-b}T^{-2a}][L1T−1]=[Ma+bLa−bT−2a]
Equate dimensions: For mass MMM: a+b=0a + b = 0a+b=0 For length LLL: a−b=1a - b = 1a−b=1 For time TTT: −2a=−1-2a = -1−2a=−1
Solving, a=12a = \frac{1}{2}a=21, b=−12b = -\frac{1}{2}b=−21.
Thus, v∝T1/2μ−1/2v \propto T^{1/2} \mu^{-1/2}v∝T1/2μ−1/2.
Answer: v=Tμv = \sqrt{\frac{T}{\mu}}v=μT.
Problem 25: Dimensional Formula of Capacitance
Question: Determine the dimensional formula for capacitance CCC.
Solution: Capacitance CCC is given by C=QVC = \frac{Q}{V}C=VQ, where QQQ is charge and VVV is voltage.
Charge QQQ: [AT][AT][AT]
Voltage VVV: [ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1]
[C]=[AT][ML2T−3A−1][C] = \frac{[AT]}{[ML^2T^{-3}A^{-1}]}[C]=[ML2T−3A−1][AT] [C]=[M−1L−2T4A2][C] = [M^{-1}L^{-2}T^4A^2][C]=[M−1L−2T4A2]
Answer: The dimensional formula for capacitance is [M−1L−2T4A2][M^{-1}L^{-2}T^4A^2][M−1L−2T4A2].
Problem 26: Dimensional Formula of Permittivity
Question: Find the dimensional formula for permittivity ϵ0\epsilon_0ϵ0 from the equation F=14πϵ0q1q2r2F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}F=4πϵ01r2q1q2.
Solution:
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Charge q1,q2q_1, q_2q1,q2: [AT][AT][AT]
Distance rrr: [L][L][L]
ϵ0=q1q2Fr2\epsilon_0 = \frac{q_1 q_2}{F r^2}ϵ0=Fr2q1q2
[ϵ0]=[AT][AT][MLT−2][L2][\epsilon_0] = \frac{[AT][AT]}{[MLT^{-2}][L^2]}[ϵ0]=[MLT−2][L2][AT][AT] [ϵ0]=[A2T2][ML3T−2][\epsilon_0] = \frac{[A^2T^2]}{[ML^3T^{-2}]}[ϵ0]=[ML3T−2][A2T2] [ϵ0]=[M−1L−3T4A2][\epsilon_0] = [M^{-1}L^{-3}T^4A^2][ϵ0]=[M−1L−3T4A2]
Answer: The dimensional formula for permittivity is [M−1L−3T4A2][M^{-1}L^{-3}T^4A^2][M−1L−3T4A2].
Problem 27: Dimensional Analysis of Force in a Magnetic Field
Question: Verify if the equation F=qvBF = qvBF=qvB is dimensionally consistent, where FFF is force, qqq is charge, vvv is velocity, and BBB is the magnetic field.
Solution:
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Charge qqq: [AT][AT][AT]
Velocity vvv: [LT−1][LT^{-1}][LT−1]
Magnetic field BBB: [MT−2A−1][MT^{-2}A^{-1}][MT−2A−1]
Check the dimensions of the RHS: [F]=[q][v][B][F] = [q][v][B][F]=[q][v][B] [MLT−2]=[AT][LT−1][MT−2A−1][MLT^{-2}] = [AT][LT^{-1}][MT^{-2}A^{-1}][MLT−2]=[AT][LT−1][MT−2A−1] [MLT−2]=[MLT−2][MLT^{-2}] = [MLT^{-2}][MLT−2]=[MLT−2]
Both sides of the equation have the same dimensions.
Answer: The equation F=qvBF = qvBF=qvB is dimensionally consistent.
Problem 28: Dimensional Formula of Energy Density
Question: Determine the dimensional formula for energy density uuu.
Solution: Energy density uuu is defined as energy per unit volume.
Energy EEE: [ML2T−2][ML^2T^{-2}][ML2T−2]
Volume VVV: [L3][L^3][L3]
u=EVu = \frac{E}{V}u=VE [u]=[ML2T−2][L3][u] = \frac{[ML^2T^{-2}]}{[L^3]}[u]=[L3][ML2T−2] [u]=[ML−1T−2][u] = [ML^{-1}T^{-2}][u]=[ML−1T−2]
Answer: The dimensional formula for energy density is [ML−1T−2][ML^{-1}T^{-2}][ML−1T−2].
Problem 29: Dimensional Formula of Frequency
Question: Find the dimensional formula for frequency ν\nuν.
Solution: Frequency ν\nuν is defined as the number of oscillations per unit time.
[ν]=[T−1][\nu] = [T^{-1}][ν]=[T−1]
Answer: The dimensional formula for frequency is [T−1][T^{-1}][T−1].
Problem 30: Dimensional Formula of Surface Energy
Question: Determine the dimensional formula for surface energy.
Solution: Surface energy is defined as the energy per unit area.
Energy EEE: [ML2T−2][ML^2T^{-2}][ML2T−2]
Area AAA: [L2][L^2][L2]
Surface energy=EA\text{Surface energy} = \frac{E}{A}Surface energy=AE [Surface energy]=[ML2T−2][L2][\text{Surface energy}] = \frac{[ML^2T^{-2}]}{[L^2]}[Surface energy]=[L2][ML2T−2] [Surface energy]=[MT−2][\text{Surface energy}] = [MT^{-2}][Surface energy]=[MT−2]
Answer: The dimensional formula for surface energy is [MT−2][MT^{-2}][MT−2].
Problem 31: Dimensional Formula of Angular Frequency
Question: Find the dimensional formula for angular frequency ω\omegaω.
Solution: Angular frequency ω\omegaω is defined as the angle turned per unit time.
[ω]=[T−1][\omega] = [T^{-1}][ω]=[T−1]
Answer: The dimensional formula for angular frequency is [T−1][T^{-1}][T−1].
Problem 32: Dimensional Formula of Power
Question: Determine the dimensional formula for power PPP.
Solution: Power PPP is defined as the rate of doing work.
Work WWW: [ML2T−2][ML^2T^{-2}][ML2T−2]
Time ttt: [T][T][T]
P=WtP = \frac{W}{t}P=tW [P]=[ML2T−2][T][P] = \frac{[ML^2T^{-2}]}{[T]}[P]=[T][ML2T−2] [P]=[ML2T−3][P] = [ML^2T^{-3}][P]=[ML2T−3]
Answer: The dimensional formula for power is [ML2T−3][ML^2T^{-3}][ML2T−3].
Problem 33: Dimensional Formula of Electric Flux
Question: Find the dimensional formula for electric flux ΦE\Phi_EΦE.
Solution: Electric flux ΦE\Phi_EΦE is given by ΦE=E⋅A\Phi_E = E \cdot AΦE=E⋅A, where EEE is the electric field and AAA is the area.
Electric field EEE: [MLT−3A−1][MLT^{-3}A^{-1}][MLT−3A−1]
Area AAA: [L2][L^2][L2]
[ΦE]=[MLT−3A−1][L2][\Phi_E] = [MLT^{-3}A^{-1}][L^2][ΦE]=[MLT−3A−1][L2] [ΦE]=[ML3T−3A−1][\Phi_E] = [ML^3T^{-3}A^{-1}][ΦE]=[ML3T−3A−1]
Answer: The dimensional formula for electric flux is [ML3T−3A−1][ML^3T^{-3}A^{-1}][ML3T−3A−1].
Problem 34: Dimensional Formula of Magnetic Flux
Question: Determine the dimensional formula for magnetic flux ΦB\Phi_BΦB.
Solution: Magnetic flux ΦB\Phi_BΦB is given by ΦB=B⋅A\Phi_B = B \cdot AΦB=B⋅A, where BBB is the magnetic field and AAA is the area.
Magnetic field BBB: [MT−2A−1][MT^{-2}A^{-1}][MT−2A−1]
Area AAA: [L2][L^2][L2]
[ [\Phi_B] = [MT^{-2}A^{-1}][L
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Additional Numerical Problems in Units and Dimensions (Continued)
Problem 34 (Continued): Dimensional Formula of Magnetic Flux
Question: Determine the dimensional formula for magnetic flux ΦB\Phi_BΦB.
Solution: Magnetic flux ΦB\Phi_BΦB is given by ΦB=B⋅A\Phi_B = B \cdot AΦB=B⋅A, where BBB is the magnetic field and AAA is the area.
Magnetic field BBB: [MT−2A−1][MT^{-2}A^{-1}][MT−2A−1]
Area AAA: [L2][L^2][L2]
[ΦB]=[MT−2A−1][L2][\Phi_B] = [MT^{-2}A^{-1}][L^2][ΦB]=[MT−2A−1][L2] [ΦB]=[ML2T−2A−1][\Phi_B] = [ML^2T^{-2}A^{-1}][ΦB]=[ML2T−2A−1]
Answer: The dimensional formula for magnetic flux is [ML2T−2A−1][ML^2T^{-2}A^{-1}][ML2T−2A−1].
Problem 35: Dimensional Formula of Pressure
Question: Find the dimensional formula for pressure PPP.
Solution: Pressure PPP is defined as force per unit area.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Area AAA: [L2][L^2][L2]
P=FAP = \frac{F}{A}P=AF [P]=[MLT−2][L2][P] = \frac{[MLT^{-2}]}{[L^2]}[P]=[L2][MLT−2] [P]=[ML−1T−2][P] = [ML^{-1}T^{-2}][P]=[ML−1T−2]
Answer: The dimensional formula for pressure is [ML−1T−2][ML^{-1}T^{-2}][ML−1T−2].
Problem 36: Dimensional Formula of Surface Tension
Question: Determine the dimensional formula for surface tension γ\gammaγ.
Solution: Surface tension γ\gammaγ is defined as force per unit length.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Length LLL: [L][L][L]
γ=FL\gamma = \frac{F}{L}γ=LF [γ]=[MLT−2][L][\gamma] = \frac{[MLT^{-2}]}{[L]}[γ]=[L][MLT−2] [γ]=[MT−2][\gamma] = [MT^{-2}][γ]=[MT−2]
Answer: The dimensional formula for surface tension is [MT−2][MT^{-2}][MT−2].
Problem 37: Dimensional Formula of Kinetic Energy
Question: Find the dimensional formula for kinetic energy KEKEKE.
Solution: Kinetic energy KEKEKE is given by KE=12mv2KE = \frac{1}{2}mv^2KE=21mv2, where mmm is mass and vvv is velocity.
Mass mmm: [M][M][M]
Velocity vvv: [LT−1][LT^{-1}][LT−1]
KE=12[M][LT−1]2KE = \frac{1}{2} [M][LT^{-1}]^2KE=21[M][LT−1]2 [KE]=[ML2T−2][KE] = [ML^2T^{-2}][KE]=[ML2T−2]
Answer: The dimensional formula for kinetic energy is [ML2T−2][ML^2T^{-2}][ML2T−2].
Problem 38: Dimensional Formula of Potential Energy
Question: Determine the dimensional formula for potential energy PEPEPE.
Solution: Potential energy PEPEPE is given by PE=mghPE = mghPE=mgh, where mmm is mass, ggg is acceleration due to gravity, and hhh is height.
Mass mmm: [M][M][M]
Acceleration due to gravity ggg: [LT−2][LT^{-2}][LT−2]
Height hhh: [L][L][L]
PE=[M][LT−2][L]PE = [M][LT^{-2}][L]PE=[M][LT−2][L] [PE]=[ML2T−2][PE] = [ML^2T^{-2}][PE]=[ML2T−2]
Answer: The dimensional formula for potential energy is [ML2T−2][ML^2T^{-2}][ML2T−2].
Problem 39: Dimensional Formula of Resistivity
Question: Find the dimensional formula for resistivity ρ\rhoρ.
Solution: Resistivity ρ\rhoρ is given by ρ=RAL\rho = R \frac{A}{L}ρ=RLA, where RRR is resistance, AAA is the area, and LLL is the length.
Resistance RRR: [ML2T−3A−2][ML^2T^{-3}A^{-2}][ML2T−3A−2]
Area AAA: [L2][L^2][L2]
Length LLL: [L][L][L]
ρ=[ML2T−3A−2][L2][L]\rho = [ML^2T^{-3}A^{-2}] \frac{[L^2]}{[L]}ρ=[ML2T−3A−2][L][L2] [ρ]=[ML3T−3A−2][\rho] = [ML^3T^{-3}A^{-2}][ρ]=[ML3T−3A−2]
Answer: The dimensional formula for resistivity is [ML3T−3A−2][ML^3T^{-3}A^{-2}][ML3T−3A−2].
Problem 40: Dimensional Formula of Conductance
Question: Determine the dimensional formula for conductance GGG.
Solution: Conductance GGG is the reciprocal of resistance RRR.
Resistance RRR: [ML2T−3A−2][ML^2T^{-3}A^{-2}][ML2T−3A−2]
G=1RG = \frac{1}{R}G=R1 [G]=[M−1L−2T3A2][G] = [M^{-1}L^{-2}T^3A^2][G]=[M−1L−2T3A2]
Answer: The dimensional formula for conductance is [M−1L−2T3A2][M^{-1}L^{-2}T^3A^2][M−1L−2T3A2].
Problem 41: Dimensional Formula of Electric Field Intensity
Question: Find the dimensional formula for electric field intensity EEE.
Solution: Electric field intensity EEE is defined as force per unit charge.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Charge QQQ: [AT][AT][AT]
E=FQE = \frac{F}{Q}E=QF [E]=[MLT−2][AT][E] = \frac{[MLT^{-2}]}{[AT]}[E]=[AT][MLT−2] [E]=[MLT−3A−1][E] = [MLT^{-3}A^{-1}][E]=[MLT−3A−1]
Answer: The dimensional formula for electric field intensity is [MLT−3A−1][MLT^{-3}A^{-1}][MLT−3A−1].
Problem 42: Dimensional Formula of Energy
Question: Determine the dimensional formula for energy EEE.
Solution: Energy EEE is generally defined as work done.
Work WWW: [ML2T−2][ML^2T^{-2}][ML2T−2]
[E]=[ML2T−2][E] = [ML^2T^{-2}][E]=[ML2T−2]
Answer: The dimensional formula for energy is [ML2T−2][ML^2T^{-2}][ML2T−2].
Problem 43: Dimensional Formula of Impulse
Question: Find the dimensional formula for impulse JJJ.
Solution: Impulse JJJ is defined as the change in momentum.
Momentum ppp: [MLT−1][MLT^{-1}][MLT−1]
J=[MLT−1]J = [MLT^{-1}]J=[MLT−1]
Answer: The dimensional formula for impulse is [MLT−1][MLT^{-1}][MLT−1].
Problem 44: Dimensional Formula of Latent Heat
Question: Determine the dimensional formula for latent heat LLL.
Solution: Latent heat LLL is the amount of heat required to change the phase of a unit mass of substance.
Heat QQQ: [ML2T−2][ML^2T^{-2}][ML2T−2]
Mass mmm: [M][M][M]
L=QmL = \frac{Q}{m}L=mQ [L]=[ML2T−2][M][L] = \frac{[ML^2T^{-2}]}{[M]}[L]=[M][ML2T−2] [L]=[L2T−2][L] = [L^2T^{-2}][L]=[L2T−2]
Answer: The dimensional formula for latent heat is [L2T−2][L^2T^{-2}][L2T−2].
Problem 45: Dimensional Formula of Gravitational Potential
Question: Find the dimensional formula for gravitational potential VgV_gVg.
Solution: Gravitational potential VgV_gVg is defined as the work done per unit mass.
Work WWW: [ML2T−2][ML^2T^{-2}][ML2T−2]
Mass mmm: [M][M][M]
Vg=WmV_g = \frac{W}{m}Vg=mW [Vg]=[ML2T−2][M][V_g] = \frac{[ML^2T^{-2}]}{[M]}[Vg]=[M][ML2T−2] [Vg]=[L2T−2][V_g] = [L^2T^{-2}][Vg]=[L2T−2]
Answer: The dimensional formula for gravitational potential is [L2T−2][L^2T^{-2}][L2T−2].
Problem 46: Dimensional Formula of Torque
Question: Determine the dimensional formula for torque τ\tauτ.
Solution: Torque τ\tauτ is defined as force multiplied by the distance from the point of rotation.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Distance ddd: [L][L][L]
τ=F⋅d\tau = F \cdot dτ=F⋅d [τ]=[MLT−2][L][\tau] = [MLT^{-2}][L][τ]=[MLT−2][L] [τ]=[ML2T−2][\tau] = [ML^2T^{-2}][τ]=[ML2T−2]
Answer: The dimensional formula for torque is [ML2T−2][ML^2T^{-2}][ML2T−2].
Problem 47: Dimensional Formula of Entropy
Question: Find the dimensional formula for entropy SSS.
Solution: Entropy SSS is defined as the ratio of heat QQQ to temperature TTT.
Heat QQQ: [ML2T−2][ML^2T^{-2}][ML2T−2]
Temperature TTT: [Θ][\Theta][Θ]
S=QTS = \frac{Q}{T}S=TQ [S]=[ML2T−2][Θ][S] = \frac{[ML^2T^{-2}]}{[\Theta]}[S]=[Θ][ML2T−2] [S]=[ML2T−2Θ−1][S] = [ML^2T^{-2}\Theta^{-1}][S]=[ML2T−2Θ−1]
Answer: The dimensional formula for entropy is ([ML^2
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Additional Numerical Problems in Units and Dimensions (Continued)
Problem 47 (Continued): Dimensional Formula of Entropy
Question: Find the dimensional formula for entropy SSS.
Solution: Entropy SSS is defined as the ratio of heat QQQ to temperature TTT.
Heat QQQ: [ML2T−2][ML^2T^{-2}][ML2T−2]
Temperature TTT: [Θ][\Theta][Θ]
S=QTS = \frac{Q}{T}S=TQ [S]=[ML2T−2][Θ][S] = \frac{[ML^2T^{-2}]}{[\Theta]}[S]=[Θ][ML2T−2] [S]=[ML2T−2Θ−1][S] = [ML^2T^{-2}\Theta^{-1}][S]=[ML2T−2Θ−1]
Answer: The dimensional formula for entropy is [ML2T−2Θ−1][ML^2T^{-2}\Theta^{-1}][ML2T−2Θ−1].
Problem 48: Dimensional Formula of Thermal Conductivity
Question: Determine the dimensional formula for thermal conductivity kkk.
Solution: Thermal conductivity kkk is defined by the equation Q=kAΔTdQ = kA \frac{\Delta T}{d}Q=kAdΔT, where QQQ is heat transferred, AAA is area, ΔT\Delta TΔT is temperature difference, and ddd is distance.
Heat QQQ: [ML2T−2][ML^2T^{-2}][ML2T−2]
Area AAA: [L2][L^2][L2]
Temperature difference ΔT\Delta TΔT: [Θ][\Theta][Θ]
Distance ddd: [L][L][L]
k=QdAΔTk = \frac{Qd}{A \Delta T}k=AΔTQd [k]=[ML2T−2][L][L2][Θ][k] = \frac{[ML^2T^{-2}][L]}{[L^2][\Theta]}[k]=[L2][Θ][ML2T−2][L] [k]=[MLT−2Θ−1][k] = [MLT^{-2}\Theta^{-1}][k]=[MLT−2Θ−1]
Answer: The dimensional formula for thermal conductivity is [MLT−2Θ−1][MLT^{-2}\Theta^{-1}][MLT−2Θ−1].
Problem 49: Dimensional Formula of Magnetic Moment
Question: Find the dimensional formula for magnetic moment μ\muμ.
Solution: Magnetic moment μ\muμ is defined as the product of current III and area AAA through which the current flows.
Current III: [A][A][A]
Area AAA: [L2][L^2][L2]
μ=I⋅A\mu = I \cdot Aμ=I⋅A [μ]=[A][L2][\mu] = [A][L^2][μ]=[A][L2] [μ]=[AL2][\mu] = [AL^2][μ]=[AL2]
Answer: The dimensional formula for magnetic moment is [AL2][AL^2][AL2].
Problem 50: Dimensional Formula of Electric Dipole Moment
Question: Determine the dimensional formula for the electric dipole moment ppp.
Solution: Electric dipole moment ppp is defined as the product of charge qqq and distance ddd between the charges.
Charge qqq: [AT][AT][AT]
Distance ddd: [L][L][L]
p=q⋅dp = q \cdot dp=q⋅d [p]=[AT][L][p] = [AT][L][p]=[AT][L] [p]=[ATL][p] = [ATL][p]=[ATL]
Answer: The dimensional formula for electric dipole moment is [ATL][ATL][ATL].
Problem 51: Dimensional Formula of Specific Heat Capacity
Question: Find the dimensional formula for specific heat capacity ccc.
Solution: Specific heat capacity ccc is defined as the amount of heat QQQ required to raise the temperature of a unit mass mmm by one degree.
Heat QQQ: [ML2T−2][ML^2T^{-2}][ML2T−2]
Mass mmm: [M][M][M]
Temperature ΔT\Delta TΔT: [Θ][\Theta][Θ]
c=QmΔTc = \frac{Q}{m \Delta T}c=mΔTQ [c]=[ML2T−2][M][Θ][c] = \frac{[ML^2T^{-2}]}{[M][\Theta]}[c]=[M][Θ][ML2T−2] [c]=[L2T−2Θ−1][c] = [L^2T^{-2}\Theta^{-1}][c]=[L2T−2Θ−1]
Answer: The dimensional formula for specific heat capacity is [L2T−2Θ−1][L^2T^{-2}\Theta^{-1}][L2T−2Θ−1].
Problem 52: Dimensional Formula of Voltage
Question: Determine the dimensional formula for voltage VVV using Ohm's Law.
Solution: Ohm's Law states V=IRV = IRV=IR, where III is current and RRR is resistance.
Current III: [A][A][A]
Resistance RRR: [ML2T−3A−2][ML^2T^{-3}A^{-2}][ML2T−3A−2]
[V]=[I][R][V] = [I][R][V]=[I][R] [V]=[A][ML2T−3A−2][V] = [A][ML^2T^{-3}A^{-2}][V]=[A][ML2T−3A−2] [V]=[ML2T−3A−1][V] = [ML^2T^{-3}A^{-1}][V]=[ML2T−3A−1]
Answer: The dimensional formula for voltage is [ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1].
Problem 53: Dimensional Formula of Current Density
Question: Find the dimensional formula for current density JJJ.
Solution: Current density JJJ is the current III per unit area AAA.
Current III: [A][A][A]
Area AAA: [L2][L^2][L2]
J=IAJ = \frac{I}{A}J=AI [J]=[A][L2][J] = \frac{[A]}{[L^2]}[J]=[L2][A] [J]=[AL−2][J] = [AL^{-2}][J]=[AL−2]
Answer: The dimensional formula for current density is [AL−2][AL^{-2}][AL−2].
Problem 54: Dimensional Formula of Magnetic Field
Question: Determine the dimensional formula for the magnetic field BBB.
Solution: Magnetic field BBB is related to force FFF and current III as F=BILF = BILF=BIL, where LLL is length.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Current III: [A][A][A]
Length LLL: [L][L][L]
B=FILB = \frac{F}{IL}B=ILF [B]=[MLT−2][A][L][B] = \frac{[MLT^{-2}]}{[A][L]}[B]=[A][L][MLT−2] [B]=[MT−2A−1][B] = [MT^{-2}A^{-1}][B]=[MT−2A−1]
Answer: The dimensional formula for the magnetic field is [MT−2A−1][MT^{-2}A^{-1}][MT−2A−1].
Problem 55: Dimensional Formula of Impedance
Question: Find the dimensional formula for impedance ZZZ.
Solution: Impedance ZZZ is the ratio of voltage VVV to current III.
Voltage VVV: [ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1]
Current III: [A][A][A]
Z=VIZ = \frac{V}{I}Z=IV [Z]=[ML2T−3A−1][A][Z] = \frac{[ML^2T^{-3}A^{-1}]}{[A]}[Z]=[A][ML2T−3A−1] [Z]=[ML2T−3A−2][Z] = [ML^2T^{-3}A^{-2}][Z]=[ML2T−3A−2]
Answer: The dimensional formula for impedance is [ML2T−3A−2][ML^2T^{-3}A^{-2}][ML2T−3A−2].
Problem 56: Dimensional Formula of Thermal Resistance
Question: Determine the dimensional formula for thermal resistance RthR_{\text{th}}Rth.
Solution: Thermal resistance RthR_{\text{th}}Rth is given by Rth=LkAR_{\text{th}} = \frac{L}{kA}Rth=kAL, where LLL is the length, kkk is thermal conductivity, and AAA is area.
Thermal conductivity kkk: [MLT−2Θ−1][MLT^{-2}\Theta^{-1}][MLT−2Θ−1]
Length LLL: [L][L][L]
Area AAA: [L2][L^2][L2]
Rth=LkAR_{\text{th}} = \frac{L}{kA}Rth=kAL [Rth]=[L][MLT−2Θ−1][L2][R_{\text{th}}] = \frac{[L]}{[MLT^{-2}\Theta^{-1}][L^2]}[Rth]=[MLT−2Θ−1][L2][L] [Rth]=[M−1L−1T2Θ][R_{\text{th}}] = [M^{-1}L^{-1}T^2\Theta][Rth]=[M−1L−1T2Θ]
Answer: The dimensional formula for thermal resistance is [M−1L−1T2Θ][M^{-1}L^{-1}T^2\Theta][M−1L−1T2Θ].
Problem 57: Dimensional Formula of Magnetic Permeability
Question: Find the dimensional formula for magnetic permeability μ\muμ.
Solution: Magnetic permeability μ\muμ is related to the magnetic field BBB and magnetic field strength HHH as B=μHB = \mu HB=μH.
Magnetic field BBB: [MT−2A−1][MT^{-2}A^{-1}][MT−2A−1]
Magnetic field strength HHH: [A⋅L−1][A \cdot L^{-1}][A⋅L−1]
μ=BH\mu = \frac{B}{H}μ=HB [μ]=[MT−2A−1][A⋅L−1][\mu] = \frac{[MT^{-2}A^{-1}]}{[A \cdot L^{-1}]}[μ]=[A⋅L−1][MT−2A−1] [μ]=[MLT−2A−2][\mu] = [MLT^{-2}A^{-2}][μ]=[MLT−2A−2]
Answer: The dimensional formula for magnetic permeability is [MLT−2A−2][MLT^{-2}A^{-2}][MLT−2A−2].
Problem 58: Dimensional Formula of Electric Charge Density
Question: Determine the dimensional formula for electric charge density ρe\rho_eρe.
Solution: Electric charge density ρe\rho_eρe is the charge qqq per unit volume VVV.
Charge qqq: [AT][AT][AT]
Volume VVV: ([L
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Additional Numerical Problems in Units and Dimensions (Continued)
Problem 58 (Continued): Dimensional Formula of Electric Charge Density
Question: Determine the dimensional formula for electric charge density ρe\rho_eρe.
Solution: Electric charge density ρe\rho_eρe is the charge qqq per unit volume VVV.
Charge qqq: [AT][AT][AT]
Volume VVV: [L3][L^3][L3]
ρe=qV\rho_e = \frac{q}{V}ρe=Vq [ρe]=[AT][L3][\rho_e] = \frac{[AT]}{[L^3]}[ρe]=[L3][AT] [ρe]=[ATL−3][\rho_e] = [ATL^{-3}][ρe]=[ATL−3]
Answer: The dimensional formula for electric charge density is [ATL−3][ATL^{-3}][ATL−3].
Problem 59: Dimensional Formula of Electric Flux
Question: Find the dimensional formula for electric flux ΦE\Phi_EΦE.
Solution: Electric flux ΦE\Phi_EΦE is given by ΦE=E⋅A\Phi_E = E \cdot AΦE=E⋅A, where EEE is the electric field intensity and AAA is the area.
Electric field intensity EEE: [MLT−3A−1][MLT^{-3}A^{-1}][MLT−3A−1]
Area AAA: [L2][L^2][L2]
ΦE=[E]⋅[A]\Phi_E = [E] \cdot [A]ΦE=[E]⋅[A] [ΦE]=[MLT−3A−1]⋅[L2][\Phi_E] = [MLT^{-3}A^{-1}] \cdot [L^2][ΦE]=[MLT−3A−1]⋅[L2] [ΦE]=[ML3T−3A−1][\Phi_E] = [ML^3T^{-3}A^{-1}][ΦE]=[ML3T−3A−1]
Answer: The dimensional formula for electric flux is [ML3T−3A−1][ML^3T^{-3}A^{-1}][ML3T−3A−1].
Problem 60: Dimensional Formula of Capacitance
Question: Determine the dimensional formula for capacitance CCC.
Solution: Capacitance CCC is defined as the charge qqq per unit voltage VVV.
Charge qqq: [AT][AT][AT]
Voltage VVV: [ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1]
C=qVC = \frac{q}{V}C=Vq [C]=[AT][ML2T−3A−1][C] = \frac{[AT]}{[ML^2T^{-3}A^{-1}]}[C]=[ML2T−3A−1][AT] [C]=[M−1L−2T4A2][C] = [M^{-1}L^{-2}T^4A^2][C]=[M−1L−2T4A2]
Answer: The dimensional formula for capacitance is [M−1L−2T4A2][M^{-1}L^{-2}T^4A^2][M−1L−2T4A2].
Problem 61: Dimensional Formula of Dielectric Constant
Question: Find the dimensional formula for the dielectric constant κ\kappaκ.
Solution: Dielectric constant κ\kappaκ is the ratio of the capacitance with the dielectric to the capacitance without the dielectric.
Since it is a ratio, it is dimensionless.
Answer: The dimensional formula for the dielectric constant is dimensionless (i.e., no dimensions).
Problem 62: Dimensional Formula of Magnetic Flux Density
Question: Determine the dimensional formula for magnetic flux density BBB.
Solution: Magnetic flux density BBB is given by B=FILB = \frac{F}{IL}B=ILF, where FFF is the force, III is the current, and LLL is the length.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Current III: [A][A][A]
Length LLL: [L][L][L]
B=FILB = \frac{F}{IL}B=ILF [B]=[MLT−2][A][L][B] = \frac{[MLT^{-2}]}{[A][L]}[B]=[A][L][MLT−2] [B]=[MT−2A−1][B] = [MT^{-2}A^{-1}][B]=[MT−2A−1]
Answer: The dimensional formula for magnetic flux density is [MT−2A−1][MT^{-2}A^{-1}][MT−2A−1].
Problem 63: Dimensional Formula of Heat Capacity
Question: Find the dimensional formula for heat capacity CpC_pCp.
Solution: Heat capacity CpC_pCp is defined as the amount of heat QQQ required to raise the temperature of an object by one degree.
Heat QQQ: [ML2T−2][ML^2T^{-2}][ML2T−2]
Temperature ΔT\Delta TΔT: [Θ][\Theta][Θ]
Cp=QΔTC_p = \frac{Q}{\Delta T}Cp=ΔTQ [Cp]=[ML2T−2][Θ][C_p] = \frac{[ML^2T^{-2}]}{[\Theta]}[Cp]=[Θ][ML2T−2] [Cp]=[ML2T−2Θ−1][C_p] = [ML^2T^{-2}\Theta^{-1}][Cp]=[ML2T−2Θ−1]
Answer: The dimensional formula for heat capacity is [ML2T−2Θ−1][ML^2T^{-2}\Theta^{-1}][ML2T−2Θ−1].
Problem 64: Dimensional Formula of Electromotive Force (EMF)
Question: Determine the dimensional formula for electromotive force (EMF) E\mathcal{E}E.
Solution: Electromotive force E\mathcal{E}E is the work done per unit charge.
Work WWW: [ML2T−2][ML^2T^{-2}][ML2T−2]
Charge qqq: [AT][AT][AT]
E=Wq\mathcal{E} = \frac{W}{q}E=qW [E]=[ML2T−2][AT][\mathcal{E}] = \frac{[ML^2T^{-2}]}{[AT]}[E]=[AT][ML2T−2] [E]=[ML2T−3A−1][\mathcal{E}] = [ML^2T^{-3}A^{-1}][E]=[ML2T−3A−1]
Answer: The dimensional formula for electromotive force is [ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1].
Problem 65: Dimensional Formula of Magnetic Field Strength
Question: Find the dimensional formula for magnetic field strength HHH.
Solution: Magnetic field strength HHH is related to the magnetic field BBB and the magnetic permeability μ\muμ as B=μHB = \mu HB=μH.
Magnetic field BBB: [MT−2A−1][MT^{-2}A^{-1}][MT−2A−1]
Magnetic permeability μ\muμ: [MLT−2A−2][MLT^{-2}A^{-2}][MLT−2A−2]
H=BμH = \frac{B}{\mu}H=μB [H]=[MT−2A−1][MLT−2A−2][H] = \frac{[MT^{-2}A^{-1}]}{[MLT^{-2}A^{-2}]}[H]=[MLT−2A−2][MT−2A−1] [H]=[A⋅L−1][H] = [A \cdot L^{-1}][H]=[A⋅L−1]
Answer: The dimensional formula for magnetic field strength is [AL−1][AL^{-1}][AL−1].
Problem 66: Dimensional Formula of Luminous Flux
Question: Determine the dimensional formula for luminous flux Φ\PhiΦ.
Solution: Luminous flux Φ\PhiΦ is related to the luminous intensity III and the solid angle Ω\OmegaΩ.
Luminous intensity III: [J][J][J] (candela)
Solid angle Ω\OmegaΩ: dimensionless
Φ=I⋅Ω\Phi = I \cdot \OmegaΦ=I⋅Ω [Φ]=[I][\Phi] = [I][Φ]=[I]
Since luminous flux is measured in lumens and is dimensionally represented by the unit candela times solid angle.
Answer: The dimensional formula for luminous flux is the same as luminous intensity, which is dimensionless.
Problem 67: Dimensional Formula of Light Intensity
Question: Find the dimensional formula for light intensity III.
Solution: Light intensity is related to power per unit area.
Power PPP: [ML2T−3][ML^2T^{-3}][ML2T−3]
Area AAA: [L2][L^2][L2]
I=PAI = \frac{P}{A}I=AP [I]=[ML2T−3][L2][I] = \frac{[ML^2T^{-3}]}{[L^2]}[I]=[L2][ML2T−3] [I]=[MT−3][I] = [MT^{-3}][I]=[MT−3]
Answer: The dimensional formula for light intensity is [MT−3][MT^{-3}][MT−3].
Problem 68: Dimensional Formula of Wavelength
Question: Determine the dimensional formula for wavelength λ\lambdaλ.
Solution: Wavelength λ\lambdaλ is the distance between two consecutive crests of a wave.
Wavelength λ\lambdaλ: [L][L][L]
Answer: The dimensional formula for wavelength is [L][L][L].
Problem 69: Dimensional Formula of Frequency
Question: Find the dimensional formula for frequency ν\nuν.
Solution: Frequency ν\nuν is the reciprocal of time period TTT.
Time period TTT: [T][T][T]
ν=1T\nu = \frac{1}{T}ν=T1 [ν]=[T−1][\nu] = [T^{-1}][ν]=[T−1]
Answer: The dimensional formula for frequency is [T−1][T^{-1}][T−1].
Problem 70: Dimensional Formula of Angular Momentum
Question: Determine the dimensional formula for angular momentum LLL.
Solution: Angular momentum LLL is given by L=I⋅ωL = I \cdot \omegaL=I⋅ω, where III is moment of inertia and ω\omegaω is angular velocity.
Moment of inertia III: [ML2][ML^2][ML2]
Angular velocity ω\omegaω: [T−1][T^{-1}][T−1]
L=I⋅ωL = I \cdot \omegaL=I⋅ω [L]=[ML2]⋅[T−1][L] = [ML^2] \cdot [T^{-1}][L]=[ML2]⋅[T−1] [L]=[ML2T−1][L] = [ML^2T^{-1}][L]=[ML2T−1]
Answer: The dimensional formula for angular momentum is [ML2T−1][ML^2T^{-1}][ML2T−1].
Problem 71: Dimensional Formula of Rotational Kinetic Energy
Question: Find the dimensional formula for rotational kinetic energy KKK.
**Solution:
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Problem 71 (Continued): Dimensional Formula of Rotational Kinetic Energy
Question: Find the dimensional formula for rotational kinetic energy KKK.
Solution: Rotational kinetic energy KKK is given by K=12Iω2K = \frac{1}{2} I \omega^2K=21Iω2, where III is the moment of inertia and ω\omegaω is the angular velocity.
Moment of inertia III: [ML2][ML^2][ML2]
Angular velocity ω\omegaω: [T−1][T^{-1}][T−1]
K=12Iω2K = \frac{1}{2} I \omega^2K=21Iω2 [K]=[ML2]⋅[T−2][K] = [ML^2] \cdot [T^{-2}][K]=[ML2]⋅[T−2] [K]=[ML2T−2][K] = [ML^2T^{-2}][K]=[ML2T−2]
Answer: The dimensional formula for rotational kinetic energy is [ML2T−2][ML^2T^{-2}][ML2T−2].
Problem 72: Dimensional Formula of Moment of Inertia
Question: Determine the dimensional formula for moment of inertia III.
Solution: Moment of inertia III is given by I=∑mr2I = \sum m r^2I=∑mr2, where mmm is mass and rrr is the distance from the axis of rotation.
Mass mmm: [M][M][M]
Distance rrr: [L][L][L]
I=∑mr2I = \sum m r^2I=∑mr2 [I]=[M]⋅[L2][I] = [M] \cdot [L^2][I]=[M]⋅[L2] [I]=[ML2][I] = [ML^2][I]=[ML2]
Answer: The dimensional formula for moment of inertia is [ML2][ML^2][ML2].
Problem 73: Dimensional Formula of Power
Question: Find the dimensional formula for power PPP.
Solution: Power PPP is the rate at which work WWW is done.
Work WWW: [ML2T−2][ML^2T^{-2}][ML2T−2]
Time TTT: [T][T][T]
P=WTP = \frac{W}{T}P=TW [P]=[ML2T−2][T][P] = \frac{[ML^2T^{-2}]}{[T]}[P]=[T][ML2T−2] [P]=[ML2T−3][P] = [ML^2T^{-3}][P]=[ML2T−3]
Answer: The dimensional formula for power is [ML2T−3][ML^2T^{-3}][ML2T−3].
Problem 74: Dimensional Formula of Gravitational Potential Energy
Question: Determine the dimensional formula for gravitational potential energy UUU.
Solution: Gravitational potential energy UUU is given by U=mghU = mghU=mgh, where mmm is mass, ggg is acceleration due to gravity, and hhh is height.
Mass mmm: [M][M][M]
Acceleration due to gravity ggg: [LT−2][LT^{-2}][LT−2]
Height hhh: [L][L][L]
U=mghU = mghU=mgh [U]=[M]⋅[LT−2]⋅[L][U] = [M] \cdot [LT^{-2}] \cdot [L][U]=[M]⋅[LT−2]⋅[L] [U]=[ML2T−2][U] = [ML^2T^{-2}][U]=[ML2T−2]
Answer: The dimensional formula for gravitational potential energy is [ML2T−2][ML^2T^{-2}][ML2T−2].
Problem 75: Dimensional Formula of Efficiency
Question: Find the dimensional formula for efficiency η\etaη.
Solution: Efficiency η\etaη is the ratio of useful energy output to the total energy input. Since it is a ratio, it is dimensionless.
Answer: The dimensional formula for efficiency is dimensionless (i.e., no dimensions).
Problem 76: Dimensional Formula of Electrical Resistance
Question: Determine the dimensional formula for electrical resistance RRR.
Solution: Electrical resistance RRR is given by Ohm's Law as R=VIR = \frac{V}{I}R=IV, where VVV is voltage and III is current.
Voltage VVV: [ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1]
Current III: [A][A][A]
R=VIR = \frac{V}{I}R=IV [R]=[ML2T−3A−1][A][R] = \frac{[ML^2T^{-3}A^{-1}]}{[A]}[R]=[A][ML2T−3A−1] [R]=[ML2T−3A−2][R] = [ML^2T^{-3}A^{-2}][R]=[ML2T−3A−2]
Answer: The dimensional formula for electrical resistance is [ML2T−3A−2][ML^2T^{-3}A^{-2}][ML2T−3A−2].
Problem 77: Dimensional Formula of Magnetic Energy
Question: Find the dimensional formula for magnetic energy EBE_BEB.
Solution: Magnetic energy EBE_BEB in a magnetic field is given by EB=12B2VμE_B = \frac{1}{2} B^2 V \muEB=21B2Vμ, where BBB is the magnetic field, VVV is volume, and μ\muμ is magnetic permeability.
Magnetic field BBB: [MT−2A−1][MT^{-2}A^{-1}][MT−2A−1]
Volume VVV: [L3][L^3][L3]
Magnetic permeability μ\muμ: [MLT−2A−2][MLT^{-2}A^{-2}][MLT−2A−2]
EB=12B2VμE_B = \frac{1}{2} B^2 V \muEB=21B2Vμ [EB]=[B2][V][μ][E_B] = [B^2][V][\mu][EB]=[B2][V][μ] [EB]=[MT−2A−1]2⋅[L3]⋅[MLT−2A−2][E_B] = [MT^{-2}A^{-1}]^2 \cdot [L^3] \cdot [MLT^{-2}A^{-2}][EB]=[MT−2A−1]2⋅[L3]⋅[MLT−2A−2] [EB]=[M2T−4A−2L3]⋅[MLT−2A−2][E_B] = [M^2T^{-4}A^{-2}L^3] \cdot [MLT^{-2}A^{-2}][EB]=[M2T−4A−2L3]⋅[MLT−2A−2] [EB]=[M3L3T−6A−4][E_B] = [M^3L^3T^{-6}A^{-4}][EB]=[M3L3T−6A−4]
Answer: The dimensional formula for magnetic energy is [M3L3T−6A−4][M^3L^3T^{-6}A^{-4}][M3L3T−6A−4].
Problem 78: Dimensional Formula of Angular Frequency
Question: Determine the dimensional formula for angular frequency ω\omegaω.
Solution: Angular frequency ω\omegaω is defined as the rate of change of the phase of a sinusoidal waveform, measured in radians per second.
ω=AngleTime\omega = \frac{\text{Angle}}{\text{Time}}ω=TimeAngle
Angle (radians): dimensionless
Time TTT: [T][T][T]
[ω]=[T−1][\omega] = [T^{-1}][ω]=[T−1]
Answer: The dimensional formula for angular frequency is [T−1][T^{-1}][T−1].
Problem 79: Dimensional Formula of Impulse
Question: Find the dimensional formula for impulse JJJ.
Solution: Impulse JJJ is the product of force FFF and time ttt.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Time ttt: [T][T][T]
J=F⋅tJ = F \cdot tJ=F⋅t [J]=[MLT−2]⋅[T][J] = [MLT^{-2}] \cdot [T][J]=[MLT−2]⋅[T] [J]=[MLT−1][J] = [MLT^{-1}][J]=[MLT−1]
Answer: The dimensional formula for impulse is [MLT−1][MLT^{-1}][MLT−1].
Problem 80: Dimensional Formula of Thermal Conductivity
Question: Determine the dimensional formula for thermal conductivity kkk.
Solution: Thermal conductivity kkk is defined by the equation Q=kAΔTdQ = kA \frac{\Delta T}{d}Q=kAdΔT, where QQQ is heat, AAA is area, ΔT\Delta TΔT is temperature difference, and ddd is distance.
Heat QQQ: [ML2T−2][ML^2T^{-2}][ML2T−2]
Area AAA: [L2][L^2][L2]
Temperature difference ΔT\Delta TΔT: [Θ][\Theta][Θ]
Distance ddd: [L][L][L]
k=QdAΔTk = \frac{Qd}{A \Delta T}k=AΔTQd [k]=[ML2T−2][L][L2][Θ][k] = \frac{[ML^2T^{-2}][L]}{[L^2][\Theta]}[k]=[L2][Θ][ML2T−2][L] [k]=[MLT−2Θ−1][k] = [MLT^{-2}\Theta^{-1}][k]=[MLT−2Θ−1]
Answer: The dimensional formula for thermal conductivity is [MLT−2Θ−1][MLT^{-2}\Theta^{-1}][MLT−2Θ−1].
Problem 81: Dimensional Formula of Surface Tension
Question: Find the dimensional formula for surface tension γ\gammaγ.
Solution: Surface tension γ\gammaγ is the force per unit length.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Length LLL: [L][L][L]
γ=FL\gamma = \frac{F}{L}γ=LF [γ]=[MLT−2][L][\gamma] = \frac{[MLT^{-2}]}{[L]}[γ]=[L][MLT−2] [γ]=[MT−2][\gamma] = [MT^{-2}][γ]=[MT−2]
Answer: The dimensional formula for surface tension is [MT−2][MT^{-2}][MT−2].
Problem 82: Dimensional Formula of Magnetic Flux
Question: Determine the dimensional formula for magnetic flux ΦB\Phi_BΦB.
Solution: Magnetic flux ΦB\Phi_BΦB is given by ΦB=B⋅A\Phi_B = B \cdot AΦB=B⋅A, where BBB is magnetic flux density and AAA is area.
Magnetic flux density BBB: [MT−2A−1][MT^{-2}A^{-1}][MT−2A−1]
Area AAA: [L2][L^2][L2]
ΦB=B⋅A\Phi_B = B \cdot AΦB=B⋅A [ΦB]=[MT−2A−1]⋅[L2][\Phi_B] = [MT^{-2}A^{-1}] \cdot [L^2][ΦB]=[MT−2A−1]⋅[L2] [ [\Phi_B] = [ML^2
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Problem 82 (Continued): Dimensional Formula of Magnetic Flux
Solution: [ΦB]=[MT−2A−1]⋅[L2][\Phi_B] = [MT^{-2}A^{-1}] \cdot [L^2][ΦB]=[MT−2A−1]⋅[L2] [ΦB]=[ML2T−2A−1][\Phi_B] = [ML^2T^{-2}A^{-1}][ΦB]=[ML2T−2A−1]
Answer: The dimensional formula for magnetic flux is [ML2T−2A−1][ML^2T^{-2}A^{-1}][ML2T−2A−1].
Problem 83: Dimensional Formula of Planck's Constant
Question: Find the dimensional formula for Planck's constant hhh.
Solution: Planck's constant hhh relates energy EEE to frequency ν\nuν by E=hνE = h\nuE=hν.
Energy EEE: [ML2T−2][ML^2T^{-2}][ML2T−2]
Frequency ν\nuν: [T−1][T^{-1}][T−1]
h=Eνh = \frac{E}{\nu}h=νE [h]=[ML2T−2][T−1][h] = \frac{[ML^2T^{-2}]}{[T^{-1}]}[h]=[T−1][ML2T−2] [h]=[ML2T−1][h] = [ML^2T^{-1}][h]=[ML2T−1]
Answer: The dimensional formula for Planck's constant is [ML2T−1][ML^2T^{-1}][ML2T−1].
Problem 84: Dimensional Formula of De Broglie Wavelength
Question: Determine the dimensional formula for De Broglie wavelength λ\lambdaλ.
Solution: De Broglie wavelength λ\lambdaλ is related to momentum ppp by λ=hp\lambda = \frac{h}{p}λ=ph, where hhh is Planck's constant and ppp is momentum.
Planck's constant hhh: [ML2T−1][ML^2T^{-1}][ML2T−1]
Momentum ppp: [MLT−1][MLT^{-1}][MLT−1]
λ=hp\lambda = \frac{h}{p}λ=ph [λ]=[ML2T−1][MLT−1][\lambda] = \frac{[ML^2T^{-1}]}{[MLT^{-1}]}[λ]=[MLT−1][ML2T−1] [λ]=[L][\lambda] = [L][λ]=[L]
Answer: The dimensional formula for De Broglie wavelength is [L][L][L].
Problem 85: Dimensional Formula of Specific Heat Capacity
Question: Find the dimensional formula for specific heat capacity ccc.
Solution: Specific heat capacity ccc is given by c=QmΔTc = \frac{Q}{m \Delta T}c=mΔTQ, where QQQ is heat, mmm is mass, and ΔT\Delta TΔT is temperature change.
Heat QQQ: [ML2T−2][ML^2T^{-2}][ML2T−2]
Mass mmm: [M][M][M]
Temperature ΔT\Delta TΔT: [Θ][\Theta][Θ]
c=QmΔTc = \frac{Q}{m \Delta T}c=mΔTQ [c]=[ML2T−2][M][Θ][c] = \frac{[ML^2T^{-2}]}{[M][\Theta]}[c]=[M][Θ][ML2T−2] [c]=[L2T−2Θ−1][c] = [L^2T^{-2}\Theta^{-1}][c]=[L2T−2Θ−1]
Answer: The dimensional formula for specific heat capacity is [L2T−2Θ−1][L^2T^{-2}\Theta^{-1}][L2T−2Θ−1].
Problem 86: Dimensional Formula of Electric Field Intensity
Question: Determine the dimensional formula for electric field intensity EEE.
Solution: Electric field intensity EEE is the force per unit charge.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Charge qqq: [AT][AT][AT]
E=FqE = \frac{F}{q}E=qF [E]=[MLT−2][AT][E] = \frac{[MLT^{-2}]}{[AT]}[E]=[AT][MLT−2] [E]=[MLT−3A−1][E] = [MLT^{-3}A^{-1}][E]=[MLT−3A−1]
Answer: The dimensional formula for electric field intensity is [MLT−3A−1][MLT^{-3}A^{-1}][MLT−3A−1].
Problem 87: Dimensional Formula of Electromagnetic Wave
Question: Find the dimensional formula for the speed of an electromagnetic wave.
Solution: The speed of an electromagnetic wave ccc is given by c=1μϵc = \frac{1}{\sqrt{\mu \epsilon}}c=μϵ1, where μ\muμ is the magnetic permeability and ϵ\epsilonϵ is the electric permittivity.
Magnetic permeability μ\muμ: [MLT−2A−2][MLT^{-2}A^{-2}][MLT−2A−2]
Electric permittivity ϵ\epsilonϵ: [M−1L−3T4A2][M^{-1}L^{-3}T^4A^2][M−1L−3T4A2]
Since speed ccc is in [LT−1][L T^{-1}][LT−1], and dimensional formula for ccc is:
[c]=[LT−1][c] = [L T^{-1}][c]=[LT−1]
Answer: The dimensional formula for the speed of an electromagnetic wave is [LT−1][L T^{-1}][LT−1].
Problem 88: Dimensional Formula of Modulus of Elasticity
Question: Determine the dimensional formula for modulus of elasticity EEE.
Solution: Modulus of elasticity EEE is given by stress divided by strain.
Stress σ\sigmaσ: Force per unit area [ML−1T−2][ML^{-1}T^{-2}][ML−1T−2]
Strain: Dimensionless
E=σStrainE = \frac{\sigma}{\text{Strain}}E=Strainσ [E]=[ML−1T−2][E] = [ML^{-1}T^{-2}][E]=[ML−1T−2]
Answer: The dimensional formula for modulus of elasticity is [ML−1T−2][ML^{-1}T^{-2}][ML−1T−2].
Problem 89: Dimensional Formula of Young's Modulus
Question: Find the dimensional formula for Young's modulus YYY.
Solution: Young's modulus YYY is defined as the ratio of stress to strain.
Stress σ\sigmaσ: [ML−1T−2][ML^{-1}T^{-2}][ML−1T−2]
Strain: Dimensionless
Y=σStrainY = \frac{\sigma}{\text{Strain}}Y=Strainσ [Y]=[ML−1T−2][Y] = [ML^{-1}T^{-2}][Y]=[ML−1T−2]
Answer: The dimensional formula for Young's modulus is [ML−1T−2][ML^{-1}T^{-2}][ML−1T−2].
Problem 90: Dimensional Formula of Coefficient of Viscosity
Question: Determine the dimensional formula for the coefficient of viscosity η\etaη.
Solution: The coefficient of viscosity η\etaη is defined as ForceArea×Velocity\frac{\text{Force}}{\text{Area} \times \text{Velocity}}Area×VelocityForce.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Area AAA: [L2][L^2][L2]
Velocity vvv: [LT−1][LT^{-1}][LT−1]
η=FA⋅v\eta = \frac{F}{A \cdot v}η=A⋅vF [η]=[MLT−2][L2]⋅[LT−1][\eta] = \frac{[MLT^{-2}]}{[L^2] \cdot [LT^{-1}]}[η]=[L2]⋅[LT−1][MLT−2] [η]=[ML−1T−1][\eta] = [ML^{-1}T^{-1}][η]=[ML−1T−1]
Answer: The dimensional formula for the coefficient of viscosity is [ML−1T−1][ML^{-1}T^{-1}][ML−1T−1].
Problem 91: Dimensional Formula of Specific Gravity
Question: Find the dimensional formula for specific gravity SGSGSG.
Solution: Specific gravity SGSGSG is the ratio of the density of a substance to the density of a reference substance (usually water). Since it's a ratio of two densities, it is dimensionless.
Answer: The dimensional formula for specific gravity is dimensionless (i.e., no dimensions).
Problem 92: Dimensional Formula of Electric Potential
Question: Determine the dimensional formula for electric potential VVV.
Solution: Electric potential VVV is defined as the work done per unit charge.
Work WWW: [ML2T−2][ML^2T^{-2}][ML2T−2]
Charge qqq: [AT][AT][AT]
V=WqV = \frac{W}{q}V=qW [V]=[ML2T−2][AT][V] = \frac{[ML^2T^{-2}]}{[AT]}[V]=[AT][ML2T−2] [V]=[ML2T−3A−1][V] = [ML^2T^{-3}A^{-1}][V]=[ML2T−3A−1]
Answer: The dimensional formula for electric potential is [ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1].
Problem 93: Dimensional Formula of Electric Current
Question: Find the dimensional formula for electric current III.
Solution: Electric current III is the rate of flow of electric charge.
[I]=[A][I] = [A][I]=[A]
Answer: The dimensional formula for electric current is [A][A][A].
Problem 94: Dimensional Formula of Magnetic Permeability
Question: Determine the dimensional formula for magnetic permeability μ\muμ.
Solution: Magnetic permeability μ\muμ relates the magnetic flux density BBB and the magnetic field strength HHH by B=μHB = \mu HB=μH.
Magnetic flux density BBB: [MT−2A−1][MT^{-2}A^{-1}][MT−2A−1]
Magnetic field strength HHH: [AL−1][AL^{-1}][AL−1]
μ=BH\mu = \frac{B}{H}μ=HB [μ]=[MT−2A−1][AL−1][\mu] = \frac{[MT^{-2}A^{-1}]}{[AL^{-1}]}[μ]=[AL−1][MT−2A−1] [μ]=[MLT−2A−2][\mu] = [MLT^{-2}A^{-2}][μ]=[MLT−2A−2]
Answer: The dimensional formula for magnetic permeability is [MLT−2A−2][MLT^{-2}A^{-2}][MLT−2A−2].
Problem 95: Dimensional Formula of Thermal Expansion Coefficient
Question: Find the dimensional formula for the coefficient of thermal expansion α\alphaα.
Solution: The coefficient of thermal expansion α\alphaα is defined as the fractional change in volume or length per unit change
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Problem 95 (Continued): Dimensional Formula of Thermal Expansion Coefficient
Solution: The coefficient of thermal expansion α\alphaα is defined as the fractional change in volume or length per unit change in temperature.
Change in length or volume is dimensionless compared to original length or volume.
Temperature ΔT\Delta TΔT: [Θ][\Theta][Θ]
α=Change in length or volumeOriginal length or volume⋅ΔT\alpha = \frac{\text{Change in length or volume}}{\text{Original length or volume} \cdot \Delta T}α=Original length or volume⋅ΔTChange in length or volume [α]=1[Θ][\alpha] = \frac{1}{[\Theta]}[α]=[Θ]1
Answer: The dimensional formula for the coefficient of thermal expansion is [Θ−1][\Theta^{-1}][Θ−1].
Problem 96: Dimensional Formula of Thermal Conductivity
Question: Find the dimensional formula for thermal conductivity kkk.
Solution: Thermal conductivity kkk can be expressed as k=QdAΔTk = \frac{Qd}{A \Delta T}k=AΔTQd, where QQQ is heat, ddd is distance, AAA is area, and ΔT\Delta TΔT is temperature difference.
Heat QQQ: [ML2T−2][ML^2T^{-2}][ML2T−2]
Distance ddd: [L][L][L]
Area AAA: [L2][L^2][L2]
Temperature difference ΔT\Delta TΔT: [Θ][\Theta][Θ]
k=QdAΔTk = \frac{Qd}{A \Delta T}k=AΔTQd [k]=[ML2T−2]⋅[L][L2]⋅[Θ][k] = \frac{[ML^2T^{-2}] \cdot [L]}{[L^2] \cdot [\Theta]}[k]=[L2]⋅[Θ][ML2T−2]⋅[L] [k]=[MLT−2Θ−1][k] = [MLT^{-2}\Theta^{-1}][k]=[MLT−2Θ−1]
Answer: The dimensional formula for thermal conductivity is [MLT−2Θ−1][MLT^{-2}\Theta^{-1}][MLT−2Θ−1].
Problem 97: Dimensional Formula of Relative Density
Question: Determine the dimensional formula for relative density RDRDRD.
Solution: Relative density RDRDRD is the ratio of the density of a substance to the density of a reference substance (usually water). Since it is a ratio, it is dimensionless.
Answer: The dimensional formula for relative density is dimensionless (i.e., no dimensions).
Problem 98: Dimensional Formula of Young's Modulus
Question: Find the dimensional formula for Young's modulus YYY.
Solution: Young's modulus YYY is defined as the ratio of stress to strain.
Stress σ\sigmaσ: [ML−1T−2][ML^{-1}T^{-2}][ML−1T−2]
Strain: Dimensionless
Y=σStrainY = \frac{\sigma}{\text{Strain}}Y=Strainσ [Y]=[ML−1T−2][Y] = [ML^{-1}T^{-2}][Y]=[ML−1T−2]
Answer: The dimensional formula for Young's modulus is [ML−1T−2][ML^{-1}T^{-2}][ML−1T−2].
Problem 99: Dimensional Formula of Gravitational Constant
Question: Determine the dimensional formula for the gravitational constant GGG.
Solution: The gravitational constant GGG relates the force FFF between two masses m1m_1m1 and m2m_2m2 separated by distance rrr by F=Gm1m2r2F = G \frac{m_1 m_2}{r^2}F=Gr2m1m2.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Mass mmm: [M][M][M]
Distance rrr: [L][L][L]
G=Fr2m1m2G = \frac{F r^2}{m_1 m_2}G=m1m2Fr2 [G]=[MLT−2]⋅[L2][M]⋅[M][G] = \frac{[MLT^{-2}] \cdot [L^2]}{[M] \cdot [M]}[G]=[M]⋅[M][MLT−2]⋅[L2] [G]=[M−1L3T−2][G] = [M^{-1}L^3T^{-2}][G]=[M−1L3T−2]
Answer: The dimensional formula for the gravitational constant is [M−1L3T−2][M^{-1}L^3T^{-2}][M−1L3T−2].
Problem 100: Dimensional Formula of Capacitance
Question: Find the dimensional formula for capacitance CCC.
Solution: Capacitance CCC is defined as the charge QQQ per unit potential difference VVV.
Charge QQQ: [AT][AT][AT]
Potential difference VVV: [ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1]
C=QVC = \frac{Q}{V}C=VQ [C]=[AT][ML2T−3A−1][C] = \frac{[AT]}{[ML^2T^{-3}A^{-1}]}[C]=[ML2T−3A−1][AT] [C]=[M−1L−2T4A2][C] = [M^{-1}L^{-2}T^4A^2][C]=[M−1L−2T4A2]
Answer: The dimensional formula for capacitance is [M−1L−2T4A2][M^{-1}L^{-2}T^4A^2][M−1L−2T4A2].
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Problem 101: Dimensional Formula of Magnetic Flux Density
Question: Determine the dimensional formula for magnetic flux density BBB.
Solution: Magnetic flux density BBB is related to magnetic field HHH and magnetic permeability μ\muμ by B=μHB = \mu HB=μH.
Magnetic field HHH: [AL−1][AL^{-1}][AL−1]
Magnetic permeability μ\muμ: [MLT−2A−2][MLT^{-2}A^{-2}][MLT−2A−2]
B=μHB = \mu HB=μH [B]=[MLT−2A−2]⋅[AL−1][B] = [MLT^{-2}A^{-2}] \cdot [AL^{-1}][B]=[MLT−2A−2]⋅[AL−1] [B]=[MLT−2A−1][B] = [MLT^{-2}A^{-1}][B]=[MLT−2A−1]
Answer: The dimensional formula for magnetic flux density is [MLT−2A−1][MLT^{-2}A^{-1}][MLT−2A−1].
Problem 102: Dimensional Formula of Gravitational Potential Energy
Question: Find the dimensional formula for gravitational potential energy UUU.
Solution: Gravitational potential energy UUU is given by U=mghU = mghU=mgh, where mmm is mass, ggg is acceleration due to gravity, and hhh is height.
Mass mmm: [M][M][M]
Acceleration due to gravity ggg: [LT−2][LT^{-2}][LT−2]
Height hhh: [L][L][L]
U=mghU = mghU=mgh [U]=[M]⋅[LT−2]⋅[L][U] = [M] \cdot [LT^{-2}] \cdot [L][U]=[M]⋅[LT−2]⋅[L] [U]=[ML2T−2][U] = [ML^2T^{-2}][U]=[ML2T−2]
Answer: The dimensional formula for gravitational potential energy is [ML2T−2][ML^2T^{-2}][ML2T−2].
Problem 103: Dimensional Formula of Angular Momentum
Question: Determine the dimensional formula for angular momentum LLL.
Solution: Angular momentum LLL is given by L=mvrL = mvrL=mvr, where mmm is mass, vvv is velocity, and rrr is radius.
Mass mmm: [M][M][M]
Velocity vvv: [LT−1][LT^{-1}][LT−1]
Radius rrr: [L][L][L]
L=mvrL = mvrL=mvr [L]=[M]⋅[LT−1]⋅[L][L] = [M] \cdot [LT^{-1}] \cdot [L][L]=[M]⋅[LT−1]⋅[L] [L]=[ML2T−1][L] = [ML^2T^{-1}][L]=[ML2T−1]
Answer: The dimensional formula for angular momentum is [ML2T−1][ML^2T^{-1}][ML2T−1].
Problem 104: Dimensional Formula of Relative Velocity
Question: Find the dimensional formula for relative velocity vrelv_{\text{rel}}vrel.
Solution: Relative velocity vrelv_{\text{rel}}vrel is the difference in velocities of two objects. Since velocity has the dimensional formula [LT−1][LT^{-1}][LT−1], the relative velocity will also have the same dimensional formula.
Answer: The dimensional formula for relative velocity is [LT−1][LT^{-1}][LT−1].
Problem 105: Dimensional Formula of Work Done
Question: Determine the dimensional formula for work done WWW.
Solution: Work done WWW is given by W=F⋅dW = F \cdot dW=F⋅d, where FFF is force and ddd is distance.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Distance ddd: [L][L][L]
W=F⋅dW = F \cdot dW=F⋅d [W]=[MLT−2]⋅[L][W] = [MLT^{-2}] \cdot [L][W]=[MLT−2]⋅[L] [W]=[ML2T−2][W] = [ML^2T^{-2}][W]=[ML2T−2]
Answer: The dimensional formula for work done is [ML2T−2][ML^2T^{-2}][ML2T−2].
Problem 106: Dimensional Formula of Pressure
Question: Find the dimensional formula for pressure PPP.
Solution: Pressure PPP is defined as force per unit area.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Area AAA: [L2][L^2][L2]
P=FAP = \frac{F}{A}P=AF [P]=[MLT−2][L2][P] = \frac{[MLT^{-2}]}{[L^2]}[P]=[L2][MLT−2] [P]=[ML−1T−2][P] = [ML^{-1}T^{-2}][P]=[ML−1T−2]
Answer: The dimensional formula for pressure is [ML−1T−2][ML^{-1}T^{-2}][ML−1T−2].
Problem 107: Dimensional Formula of Frequency
Question: Determine the dimensional formula for frequency ν\nuν.
Solution: Frequency ν\nuν is the reciprocal of the time period TTT.
Time period TTT: [T][T][T]
ν=1T\nu = \frac{1}{T}ν=T1 [ν]=[T−1][\nu] = [T^{-1}][ν]=[T−1]
Answer: The dimensional formula for frequency is [T−1][T^{-1}][T−1].
Problem 108: Dimensional Formula of Impulse
Question: Find the dimensional formula for impulse III.
Solution: Impulse III is the change in momentum, and is given by I=F⋅tI = F \cdot tI=F⋅t, where FFF is force and ttt is time.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Time ttt: [T][T][T]
I=F⋅tI = F \cdot tI=F⋅t [I]=[MLT−2]⋅[T][I] = [MLT^{-2}] \cdot [T][I]=[MLT−2]⋅[T] [I]=[MLT−1][I] = [MLT^{-1}][I]=[MLT−1]
Answer: The dimensional formula for impulse is [MLT−1][MLT^{-1}][MLT−1].
Problem 109: Dimensional Formula of Work Function
Question: Determine the dimensional formula for work function ϕ\phiϕ.
Solution: The work function ϕ\phiϕ is the energy required to remove an electron from a material. It has the same dimensional formula as energy.
Energy: [ML2T−2][ML^2T^{-2}][ML2T−2]
Answer: The dimensional formula for work function is [ML2T−2][ML^2T^{-2}][ML2T−2].
Problem 110: Dimensional Formula of Electric Dipole Moment
Question: Find the dimensional formula for electric dipole moment ppp.
Solution: Electric dipole moment ppp is given by p=q⋅dp = q \cdot dp=q⋅d, where qqq is charge and ddd is distance.
Charge qqq: [AT][AT][AT]
Distance ddd: [L][L][L]
p=q⋅dp = q \cdot dp=q⋅d [p]=[AT]⋅[L][p] = [AT] \cdot [L][p]=[AT]⋅[L] [p]=[ATL][p] = [ATL][p]=[ATL]
Answer: The dimensional formula for electric dipole moment is [ATL][ATL][ATL].
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Problem 111: Dimensional Formula of Electric Field
Question: Determine the dimensional formula for electric field EEE.
Solution: Electric field EEE is defined as force per unit charge.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Charge qqq: [AT][AT][AT]
E=FqE = \frac{F}{q}E=qF [E]=[MLT−2][AT][E] = \frac{[MLT^{-2}]}{[AT]}[E]=[AT][MLT−2] [E]=[MLT−3A−1][E] = [MLT^{-3}A^{-1}][E]=[MLT−3A−1]
Answer: The dimensional formula for electric field is [MLT−3A−1][MLT^{-3}A^{-1}][MLT−3A−1].
Problem 112: Dimensional Formula of Magnetic Flux
Question: Find the dimensional formula for magnetic flux Φ\PhiΦ.
Solution: Magnetic flux Φ\PhiΦ is given by Φ=B⋅A\Phi = B \cdot AΦ=B⋅A, where BBB is magnetic flux density and AAA is area.
Magnetic flux density BBB: [MLT−2A−1][MLT^{-2}A^{-1}][MLT−2A−1]
Area AAA: [L2][L^2][L2]
Φ=B⋅A\Phi = B \cdot AΦ=B⋅A [Φ]=[MLT−2A−1]⋅[L2][\Phi] = [MLT^{-2}A^{-1}] \cdot [L^2][Φ]=[MLT−2A−1]⋅[L2] [Φ]=[ML2T−2A−1][\Phi] = [ML^2T^{-2}A^{-1}][Φ]=[ML2T−2A−1]
Answer: The dimensional formula for magnetic flux is [ML2T−2A−1][ML^2T^{-2}A^{-1}][ML2T−2A−1].
Problem 113: Dimensional Formula of Gravitational Field
Question: Determine the dimensional formula for gravitational field strength ggg.
Solution: Gravitational field strength ggg is defined as acceleration due to gravity.
Acceleration ggg: [LT−2][LT^{-2}][LT−2]
Answer: The dimensional formula for gravitational field strength is [LT−2][LT^{-2}][LT−2].
Problem 114: Dimensional Formula of Thermal Energy
Question: Find the dimensional formula for thermal energy QQQ.
Solution: Thermal energy QQQ is the same as heat energy, which has the dimensional formula for work or energy.
Energy QQQ: [ML2T−2][ML^2T^{-2}][ML2T−2]
Answer: The dimensional formula for thermal energy is [ML2T−2][ML^2T^{-2}][ML2T−2].
Problem 115: Dimensional Formula of Resistance
Question: Determine the dimensional formula for electrical resistance RRR.
Solution: Resistance RRR is given by Ohm's law as R=VIR = \frac{V}{I}R=IV, where VVV is voltage and III is current.
Voltage VVV: [ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1]
Current III: [A][A][A]
R=VIR = \frac{V}{I}R=IV [R]=[ML2T−3A−1][A][R] = \frac{[ML^2T^{-3}A^{-1}]}{[A]}[R]=[A][ML2T−3A−1] [R]=[ML2T−3A−2][R] = [ML^2T^{-3}A^{-2}][R]=[ML2T−3A−2]
Answer: The dimensional formula for resistance is [ML2T−3A−2][ML^2T^{-3}A^{-2}][ML2T−3A−2].
Problem 116: Dimensional Formula of Electric Potential Energy
Question: Find the dimensional formula for electric potential energy UUU.
Solution: Electric potential energy UUU is given by U=qVU = qVU=qV, where qqq is charge and VVV is voltage.
Charge qqq: [AT][AT][AT]
Voltage VVV: [ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1]
U=qVU = qVU=qV [U]=[AT]⋅[ML2T−3A−1][U] = [AT] \cdot [ML^2T^{-3}A^{-1}][U]=[AT]⋅[ML2T−3A−1] [U]=[ML2T−2][U] = [ML^2T^{-2}][U]=[ML2T−2]
Answer: The dimensional formula for electric potential energy is [ML2T−2][ML^2T^{-2}][ML2T−2].
Problem 117: Dimensional Formula of Magnetic Moment
Question: Determine the dimensional formula for magnetic moment mmm.
Solution: Magnetic moment mmm is given by m=I⋅Am = I \cdot Am=I⋅A, where III is current and AAA is area.
Current III: [A][A][A]
Area AAA: [L2][L^2][L2]
m=I⋅Am = I \cdot Am=I⋅A [m]=[A]⋅[L2][m] = [A] \cdot [L^2][m]=[A]⋅[L2] [m]=[AL2][m] = [AL^2][m]=[AL2]
Answer: The dimensional formula for magnetic moment is [AL2][AL^2][AL2].
Problem 118: Dimensional Formula of Capacitive Reactance
Question: Find the dimensional formula for capacitive reactance XCX_CXC.
Solution: Capacitive reactance XCX_CXC is given by XC=1ωCX_C = \frac{1}{\omega C}XC=ωC1, where ω\omegaω is angular frequency and CCC is capacitance.
Angular frequency ω\omegaω: [T−1][T^{-1}][T−1]
Capacitance CCC: [M−1L−2T4A2][M^{-1}L^{-2}T^4A^2][M−1L−2T4A2]
XC=1ωCX_C = \frac{1}{\omega C}XC=ωC1 [XC]=1[T−1]⋅[M−1L−2T4A2][X_C] = \frac{1}{[T^{-1}] \cdot [M^{-1}L^{-2}T^4A^2]}[XC]=[T−1]⋅[M−1L−2T4A2]1 [XC]=[ML2T−3A−2][X_C] = [M L^2 T^{-3} A^{-2}][XC]=[ML2T−3A−2]
Answer: The dimensional formula for capacitive reactance is [ML2T−3A−2][ML^2T^{-3}A^{-2}][ML2T−3A−2].
Problem 119: Dimensional Formula of Magnetic Induction
Question: Determine the dimensional formula for magnetic induction BBB.
Solution: Magnetic induction BBB is the same as magnetic flux density, which we already calculated.
[B]=[MLT−2A−1][B] = [MLT^{-2}A^{-1}][B]=[MLT−2A−1]
Answer: The dimensional formula for magnetic induction is [MLT−2A−1][MLT^{-2}A^{-1}][MLT−2A−1].
Problem 120: Dimensional Formula of Work Done per Unit Mass
Question: Find the dimensional formula for work done per unit mass Wunit massW_{\text{unit mass}}Wunit mass.
Solution: Work done per unit mass is given by Wm\frac{W}{m}mW, where WWW is work done and mmm is mass.
Work done WWW: [ML2T−2][ML^2T^{-2}][ML2T−2]
Mass mmm: [M][M][M]
Wm=[ML2T−2][M]\frac{W}{m} = \frac{[ML^2T^{-2}]}{[M]}mW=[M][ML2T−2] [Wunit mass]=[L2T−2][W_{\text{unit mass}}] = [L^2T^{-2}][Wunit mass]=[L2T−2]
Answer: The dimensional formula for work done per unit mass is [L2T−2][L^2T^{-2}][L2T−2].
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Problem 121: Dimensional Formula of Specific Heat
Question: Determine the dimensional formula for specific heat ccc.
Solution: Specific heat ccc is defined as the heat required to raise the temperature of a unit mass by one degree Celsius or Kelvin. The formula is Q=mcΔTQ = mc\Delta TQ=mcΔT, where QQQ is heat, mmm is mass, and ΔT\Delta TΔT is the change in temperature.
Heat QQQ: [ML2T−2][ML^2T^{-2}][ML2T−2]
Mass mmm: [M][M][M]
Temperature change ΔT\Delta TΔT: [Θ][\Theta][Θ]
c=QmΔTc = \frac{Q}{m \Delta T}c=mΔTQ [c]=[ML2T−2][M]⋅[Θ][c] = \frac{[ML^2T^{-2}]}{[M] \cdot [\Theta]}[c]=[M]⋅[Θ][ML2T−2] [c]=[L2T−2Θ−1][c] = [L^2T^{-2}\Theta^{-1}][c]=[L2T−2Θ−1]
Answer: The dimensional formula for specific heat is [L2T−2Θ−1][L^2T^{-2}\Theta^{-1}][L2T−2Θ−1].
Problem 122: Dimensional Formula of Power Factor
Question: Find the dimensional formula for the power factor cosϕ\cos \phicosϕ.
Solution: The power factor cosϕ\cos \phicosϕ is the ratio of the real power to the apparent power. It is a dimensionless quantity since it is a ratio of two quantities with the same dimensions.
Answer: The dimensional formula for the power factor is dimensionless (i.e., no dimensions).
Problem 123: Dimensional Formula of Efficiency
Question: Determine the dimensional formula for efficiency η\etaη.
Solution: Efficiency η\etaη is defined as the ratio of useful output energy to the input energy. Since both input and output energy have the same dimensions, efficiency is a dimensionless quantity.
Answer: The dimensional formula for efficiency is dimensionless (i.e., no dimensions).
Problem 124: Dimensional Formula of Mechanical Advantage
Question: Find the dimensional formula for mechanical advantage MAMAMA.
Solution: Mechanical advantage MAMAMA is the ratio of output force to input force. Since both forces have the same dimensions, mechanical advantage is dimensionless.
Answer: The dimensional formula for mechanical advantage is dimensionless (i.e., no dimensions).
Problem 125: Dimensional Formula of Velocity of Sound
Question: Determine the dimensional formula for the velocity of sound vvv.
Solution: The velocity of sound is given by the formula v=Eρv = \sqrt{\frac{E}{\rho}}v=ρE, where EEE is the modulus of elasticity and ρ\rhoρ is density.
Modulus of elasticity EEE: [ML−1T−2][ML^{-1}T^{-2}][ML−1T−2]
Density ρ\rhoρ: [ML−3][ML^{-3}][ML−3]
v=Eρv = \sqrt{\frac{E}{\rho}}v=ρE [v] = \sqrt{\frac{[ML^{-1}T^{-2}]}{[ML^{-3}]} [v]=[L2T−2][v] = \sqrt{[L^2T^{-2}]}[v]=[L2T−2] [v]=[LT−1][v] = [LT^{-1}][v]=[LT−1]
Answer: The dimensional formula for the velocity of sound is [LT−1][LT^{-1}][LT−1].
Problem 126: Dimensional Formula of Wavelength
Question: Find the dimensional formula for wavelength λ\lambdaλ.
Solution: Wavelength λ\lambdaλ is the distance between successive crests or troughs of a wave.
Wavelength λ\lambdaλ: [L][L][L]
Answer: The dimensional formula for wavelength is [L][L][L].
Problem 127: Dimensional Formula of Magnetic Field Strength
Question: Determine the dimensional formula for magnetic field strength HHH.
Solution: Magnetic field strength HHH is defined by the formula H=BμH = \frac{B}{\mu}H=μB, where BBB is magnetic flux density and μ\muμ is magnetic permeability.
Magnetic flux density BBB: [MLT−2A−1][MLT^{-2}A^{-1}][MLT−2A−1]
Magnetic permeability μ\muμ: [MLT−2A−2][MLT^{-2}A^{-2}][MLT−2A−2]
H=BμH = \frac{B}{\mu}H=μB [H]=[MLT−2A−1][MLT−2A−2][H] = \frac{[MLT^{-2}A^{-1}]}{[MLT^{-2}A^{-2}]}[H]=[MLT−2A−2][MLT−2A−1] [H]=[A][H] = [A][H]=[A]
Answer: The dimensional formula for magnetic field strength is [A][A][A].
Problem 128: Dimensional Formula of Impedance
Question: Find the dimensional formula for impedance ZZZ.
Solution: Impedance ZZZ is defined as the ratio of voltage to current.
Voltage VVV: [ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1]
Current III: [A][A][A]
Z=VIZ = \frac{V}{I}Z=IV [Z]=[ML2T−3A−1][A][Z] = \frac{[ML^2T^{-3}A^{-1}]}{[A]}[Z]=[A][ML2T−3A−1] [Z]=[ML2T−3A−2][Z] = [ML^2T^{-3}A^{-2}][Z]=[ML2T−3A−2]
Answer: The dimensional formula for impedance is [ML2T−3A−2][ML^2T^{-3}A^{-2}][ML2T−3A−2].
Problem 129: Dimensional Formula of Electric Charge
Question: Determine the dimensional formula for electric charge QQQ.
Solution: Electric charge QQQ is the product of current III and time ttt.
Current III: [A][A][A]
Time ttt: [T][T][T]
Q=I⋅tQ = I \cdot tQ=I⋅t [Q]=[A]⋅[T][Q] = [A] \cdot [T][Q]=[A]⋅[T] [Q]=[AT][Q] = [AT][Q]=[AT]
Answer: The dimensional formula for electric charge is [AT][AT][AT].
Problem 130: Dimensional Formula of Force Constant
Question: Find the dimensional formula for the force constant kkk of a spring.
Solution: The force constant kkk (spring constant) is defined as F=kxF = kxF=kx, where FFF is force and xxx is displacement.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Displacement xxx: [L][L][L]
k=Fxk = \frac{F}{x}k=xF [k]=[MLT−2][L][k] = \frac{[MLT^{-2}]}{[L]}[k]=[L][MLT−2] [k]=[ML−1T−2][k] = [ML^{-1}T^{-2}][k]=[ML−1T−2]
Answer: The dimensional formula for the force constant is [ML−1T−2][ML^{-1}T^{-2}][ML−1T−2].
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Problem 131: Dimensional Formula of Heat Capacity
Question: Determine the dimensional formula for heat capacity CCC.
Solution: Heat capacity CCC is defined as the ratio of heat QQQ to the change in temperature ΔT\Delta TΔT.
Heat QQQ: [ML2T−2][ML^2T^{-2}][ML2T−2]
Temperature change ΔT\Delta TΔT: [Θ][\Theta][Θ]
C=QΔTC = \frac{Q}{\Delta T}C=ΔTQ [C]=[ML2T−2][Θ][C] = \frac{[ML^2T^{-2}]}{[\Theta]}[C]=[Θ][ML2T−2] [C]=[ML2T−2Θ−1][C] = [ML^2T^{-2}\Theta^{-1}][C]=[ML2T−2Θ−1]
Answer: The dimensional formula for heat capacity is [ML2T−2Θ−1][ML^2T^{-2}\Theta^{-1}][ML2T−2Θ−1].
Problem 132: Dimensional Formula of Thermal Conductivity
Question: Find the dimensional formula for thermal conductivity kkk.
Solution: Thermal conductivity kkk is defined by the formula Q=k⋅A⋅ΔTdQ = k \cdot A \cdot \frac{\Delta T}{d}Q=k⋅A⋅dΔT, where QQQ is heat, AAA is area, ΔT\Delta TΔT is temperature difference, and ddd is thickness.
Heat QQQ: [ML2T−2][ML^2T^{-2}][ML2T−2]
Area AAA: [L2][L^2][L2]
Temperature difference ΔT\Delta TΔT: [Θ][\Theta][Θ]
Thickness ddd: [L][L][L]
k=Q⋅dA⋅ΔTk = \frac{Q \cdot d}{A \cdot \Delta T}k=A⋅ΔTQ⋅d [k]=[ML2T−2]⋅[L][L2]⋅[Θ][k] = \frac{[ML^2T^{-2}] \cdot [L]}{[L^2] \cdot [\Theta]}[k]=[L2]⋅[Θ][ML2T−2]⋅[L] [k]=[MLT−2Θ−1][k] = [MLT^{-2}\Theta^{-1}][k]=[MLT−2Θ−1]
Answer: The dimensional formula for thermal conductivity is [MLT−2Θ−1][MLT^{-2}\Theta^{-1}][MLT−2Θ−1].
Problem 133: Dimensional Formula of Capacitance
Question: Determine the dimensional formula for capacitance CCC.
Solution: Capacitance CCC is defined by the formula C=QVC = \frac{Q}{V}C=VQ, where QQQ is charge and VVV is voltage.
Charge QQQ: [AT][AT][AT]
Voltage VVV: [ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1]
C=QVC = \frac{Q}{V}C=VQ [C]=[AT][ML2T−3A−1][C] = \frac{[AT]}{[ML^2T^{-3}A^{-1}]}[C]=[ML2T−3A−1][AT] [C]=[M−1L−2T4A2][C] = [M^{-1}L^{-2}T^4A^2][C]=[M−1L−2T4A2]
Answer: The dimensional formula for capacitance is [M−1L−2T4A2][M^{-1}L^{-2}T^4A^2][M−1L−2T4A2].
Problem 134: Dimensional Formula of Magnetic Flux
Question: Find the dimensional formula for magnetic flux Φ\PhiΦ.
Solution: Magnetic flux Φ\PhiΦ is given by Φ=B⋅A\Phi = B \cdot AΦ=B⋅A, where BBB is magnetic flux density and AAA is area.
Magnetic flux density BBB: [MLT−2A−1][MLT^{-2}A^{-1}][MLT−2A−1]
Area AAA: [L2][L^2][L2]
Φ=B⋅A\Phi = B \cdot AΦ=B⋅A [Φ]=[MLT−2A−1]⋅[L2][\Phi] = [MLT^{-2}A^{-1}] \cdot [L^2][Φ]=[MLT−2A−1]⋅[L2] [Φ]=[ML2T−2A−1][\Phi] = [ML^2T^{-2}A^{-1}][Φ]=[ML2T−2A−1]
Answer: The dimensional formula for magnetic flux is [ML2T−2A−1][ML^2T^{-2}A^{-1}][ML2T−2A−1].
Problem 135: Dimensional Formula of Electric Permittivity
Question: Determine the dimensional formula for electric permittivity ϵ\epsilonϵ.
Solution: Electric permittivity ϵ\epsilonϵ is related to the capacitance CCC, distance ddd, and area AAA by C=ϵAdC = \frac{\epsilon A}{d}C=dϵA.
Capacitance CCC: [M−1L−2T4A2][M^{-1}L^{-2}T^4A^2][M−1L−2T4A2]
Distance ddd: [L][L][L]
Area AAA: [L2][L^2][L2]
ϵ=C⋅dA\epsilon = \frac{C \cdot d}{A}ϵ=AC⋅d [ϵ]=[M−1L−2T4A2]⋅[L][L2][\epsilon] = \frac{[M^{-1}L^{-2}T^4A^2] \cdot [L]}{[L^2]}[ϵ]=[L2][M−1L−2T4A2]⋅[L] [ϵ]=[M−1L−1T4A2][\epsilon] = [M^{-1}L^{-1}T^4A^2][ϵ]=[M−1L−1T4A2]
Answer: The dimensional formula for electric permittivity is [M−1L−1T4A2][M^{-1}L^{-1}T^4A^2][M−1L−1T4A2].
Problem 136: Dimensional Formula of Conductivity
Question: Find the dimensional formula for electrical conductivity σ\sigmaσ.
Solution: Electrical conductivity σ\sigmaσ is the reciprocal of resistivity ρ\rhoρ, and resistivity is given by ρ=R⋅AL\rho = \frac{R \cdot A}{L}ρ=LR⋅A.
Resistance RRR: [ML2T−3A−2][ML^2T^{-3}A^{-2}][ML2T−3A−2]
Area AAA: [L2][L^2][L2]
Length LLL: [L][L][L]
ρ=R⋅AL\rho = \frac{R \cdot A}{L}ρ=LR⋅A [ρ]=[ML2T−3A−2]⋅[L2][L][\rho] = \frac{[ML^2T^{-3}A^{-2}] \cdot [L^2]}{[L]}[ρ]=[L][ML2T−3A−2]⋅[L2] [ρ]=[MLT−3A−2][\rho] = [MLT^{-3}A^{-2}][ρ]=[MLT−3A−2]
Since conductivity σ=1ρ\sigma = \frac{1}{\rho}σ=ρ1,
[σ]=[M−1L−1T3A2][\sigma] = [M^{-1}L^{-1}T^3A^2][σ]=[M−1L−1T3A2]
Answer: The dimensional formula for electrical conductivity is [M−1L−1T3A2][M^{-1}L^{-1}T^3A^2][M−1L−1T3A2].
Problem 137: Dimensional Formula of Luminous Flux
Question: Determine the dimensional formula for luminous flux ΦL\Phi_LΦL.
Solution: Luminous flux ΦL\Phi_LΦL is measured in lumens, which is a derived unit with no fundamental dimension. It is related to luminous intensity and solid angle.
Answer: The dimensional formula for luminous flux is dimensionless (i.e., no dimensions).
Problem 138: Dimensional Formula of Sound Intensity
Question: Find the dimensional formula for sound intensity III.
Solution: Sound intensity III is defined as power per unit area.
Power PPP: [ML2T−3][ML^2T^{-3}][ML2T−3]
Area AAA: [L2][L^2][L2]
I=PAI = \frac{P}{A}I=AP [I]=[ML2T−3][L2][I] = \frac{[ML^2T^{-3}]}{[L^2]}[I]=[L2][ML2T−3] [I]=[MT−3][I] = [MT^{-3}][I]=[MT−3]
Answer: The dimensional formula for sound intensity is [MT−3][MT^{-3}][MT−3].
Problem 139: Dimensional Formula of Electric Potential
Question: Determine the dimensional formula for electric potential VVV.
Solution: Electric potential VVV is defined as the work done per unit charge.
Work done WWW: [ML2T−2][ML^2T^{-2}][ML2T−2]
Charge QQQ: [AT][AT][AT]
V=WQV = \frac{W}{Q}V=QW [V]=[ML2T−2][AT][V] = \frac{[ML^2T^{-2}]}{[AT]}[V]=[AT][ML2T−2] [V]=[ML2T−3A−1][V] = [ML^2T^{-3}A^{-1}][V]=[ML2T−3A−1]
Answer: The dimensional formula for electric potential is [ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1].
Problem 140: Dimensional Formula of Magnetic Energy
Question: Find the dimensional formula for magnetic energy UBU_BUB.
Solution: Magnetic energy UBU_BUB stored in an inductor is given by UB=12LI2U_B = \frac{1}{2} L I^2UB=21LI2, where LLL is inductance and III is current.
Inductance LLL: [ML2T−2A−2][ML^2T^{-2}A^{-2}][ML2T−2A−2]
Current III: [A][A][A]
UB=12LI2U_B = \frac{1}{2} L I^2UB=21LI2 [UB]=[ML2T−2A−2]⋅[A2][U_B] = [ML^2T^{-2}A^{-2}] \cdot [A^2][UB]=[ML2T−2A−2]⋅[A2] [UB]=[ML2T−2][U_B] = [ML^2T^{-2}][UB]=[ML2T−2]
Answer: The dimensional formula for magnetic energy is [ML2T−2][ML^2T^{-2}][ML2T−2].
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Problem 141: Dimensional Formula of Inductance
Question: Determine the dimensional formula for inductance LLL.
Solution: Inductance LLL is defined using the formula UB=12LI2U_B = \frac{1}{2} L I^2UB=21LI2, where UBU_BUB is magnetic energy and III is current.
Magnetic energy UBU_BUB: [ML2T−2][ML^2T^{-2}][ML2T−2]
Current III: [A][A][A]
L=UBI2L = \frac{U_B}{I^2}L=I2UB [L]=[ML2T−2][A2][L] = \frac{[ML^2T^{-2}]}{[A^2]}[L]=[A2][ML2T−2] [L]=[ML2T−2A−2][L] = [ML^2T^{-2}A^{-2}][L]=[ML2T−2A−2]
Answer: The dimensional formula for inductance is [ML2T−2A−2][ML^2T^{-2}A^{-2}][ML2T−2A−2].
Problem 142: Dimensional Formula of Magnetic Permeability
Question: Find the dimensional formula for magnetic permeability μ\muμ.
Solution: Magnetic permeability μ\muμ is related to the magnetic field BBB and magnetic field strength HHH by B=μHB = \mu HB=μH.
Magnetic field BBB: [MLT−2A−1][MLT^{-2}A^{-1}][MLT−2A−1]
Magnetic field strength HHH: [A][A][A]
μ=BH\mu = \frac{B}{H}μ=HB [μ]=[MLT−2A−1][A][\mu] = \frac{[MLT^{-2}A^{-1}]}{[A]}[μ]=[A][MLT−2A−1] [μ]=[MLT−2A−2][\mu] = [MLT^{-2}A^{-2}][μ]=[MLT−2A−2]
Answer: The dimensional formula for magnetic permeability is [MLT−2A−2][MLT^{-2}A^{-2}][MLT−2A−2].
Problem 143: Dimensional Formula of Planck’s Constant
Question: Determine the dimensional formula for Planck's constant hhh.
Solution: Planck's constant hhh relates energy EEE and frequency ν\nuν through E=hνE = h \nuE=hν.
Energy EEE: [ML2T−2][ML^2T^{-2}][ML2T−2]
Frequency ν\nuν: [T−1][T^{-1}][T−1]
h=Eνh = \frac{E}{\nu}h=νE [h]=[ML2T−2][T−1][h] = \frac{[ML^2T^{-2}]}{[T^{-1}]}[h]=[T−1][ML2T−2] [h]=[ML2T−1][h] = [ML^2T^{-1}][h]=[ML2T−1]
Answer: The dimensional formula for Planck's constant is [ML2T−1][ML^2T^{-1}][ML2T−1].
Problem 144: Dimensional Formula of Gravitational Constant
Question: Find the dimensional formula for the gravitational constant GGG.
Solution: The gravitational constant GGG is defined by Newton's law of gravitation: F=Gm1m2r2F = G \frac{m_1 m_2}{r^2}F=Gr2m1m2, where FFF is force, m1m_1m1 and m2m_2m2 are masses, and rrr is distance.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Mass mmm: [M][M][M]
Distance rrr: [L][L][L]
G=Fr2m1m2G = \frac{F r^2}{m_1 m_2}G=m1m2Fr2 [G]=[MLT−2]⋅[L2][M2][G] = \frac{[MLT^{-2}] \cdot [L^2]}{[M^2]}[G]=[M2][MLT−2]⋅[L2] [G]=[ML−1T−2M−1][G] = [ML^{-1}T^{-2}M^{-1}][G]=[ML−1T−2M−1] [G]=[M−1L3T−2][G] = [M^{-1}L^3T^{-2}][G]=[M−1L3T−2]
Answer: The dimensional formula for the gravitational constant is [M−1L3T−2][M^{-1}L^3T^{-2}][M−1L3T−2].
Problem 145: Dimensional Formula of Electron Mass
Question: Determine the dimensional formula for the mass of an electron mem_eme.
Solution: The mass of an electron is simply a mass quantity.
Mass mem_eme: [M][M][M]
Answer: The dimensional formula for the mass of an electron is [M][M][M].
Problem 146: Dimensional Formula of Young’s Modulus
Question: Find the dimensional formula for Young’s modulus YYY.
Solution: Young’s modulus YYY is defined as stress divided by strain. Stress is force per unit area, and strain is dimensionless.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Area AAA: [L2][L^2][L2]
Y=StressStrain=F/AdimensionlessY = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\text{dimensionless}}Y=StrainStress=dimensionlessF/A [Y]=[MLT−2][L2][Y] = \frac{[MLT^{-2}]}{[L^2]}[Y]=[L2][MLT−2] [Y]=[ML−1T−2][Y] = [ML^{-1}T^{-2}][Y]=[ML−1T−2]
Answer: The dimensional formula for Young’s modulus is [ML−1T−2][ML^{-1}T^{-2}][ML−1T−2].
Problem 147: Dimensional Formula of Bulk Modulus
Question: Determine the dimensional formula for bulk modulus KKK.
Solution: Bulk modulus KKK is defined as the ratio of pressure to fractional change in volume. Pressure is force per unit area.
Pressure PPP: [ML−1T−2][ML^{-1}T^{-2}][ML−1T−2]
Volume change: Dimensionless
K=PressureFractional change in volumeK = \frac{\text{Pressure}}{\text{Fractional change in volume}}K=Fractional change in volumePressure [K]=[ML−1T−2][K] = [ML^{-1}T^{-2}][K]=[ML−1T−2]
Answer: The dimensional formula for bulk modulus is [ML−1T−2][ML^{-1}T^{-2}][ML−1T−2].
Problem 148: Dimensional Formula of Frequency
Question: Find the dimensional formula for frequency ν\nuν.
Solution: Frequency ν\nuν is the number of cycles per unit time.
Frequency ν\nuν: [T−1][T^{-1}][T−1]
Answer: The dimensional formula for frequency is [T−1][T^{-1}][T−1].
Problem 149: Dimensional Formula of Current Density
Question: Determine the dimensional formula for current density JJJ.
Solution: Current density JJJ is the current per unit area.
Current III: [A][A][A]
Area AAA: [L2][L^2][L2]
J=IAJ = \frac{I}{A}J=AI [J]=[A][L2][J] = \frac{[A]}{[L^2]}[J]=[L2][A] [J]=[AL−2][J] = [A L^{-2}][J]=[AL−2]
Answer: The dimensional formula for current density is [AL−2][A L^{-2}][AL−2].
Problem 150: Dimensional Formula of Permittivity
Question: Find the dimensional formula for permittivity ϵ\epsilonϵ.
Solution: Permittivity ϵ\epsilonϵ is related to capacitance CCC, area AAA, and distance ddd by C=ϵAdC = \frac{\epsilon A}{d}C=dϵA.
Capacitance CCC: [M−1L−2T4A2][M^{-1}L^{-2}T^4A^2][M−1L−2T4A2]
Area AAA: [L2][L^2][L2]
Distance ddd: [L][L][L]
ϵ=C⋅dA\epsilon = \frac{C \cdot d}{A}ϵ=AC⋅d [ϵ]=[M−1L−2T4A2]⋅[L][L2][\epsilon] = \frac{[M^{-1}L^{-2}T^4A^2] \cdot [L]}{[L^2]}[ϵ]=[L2][M−1L−2T4A2]⋅[L] [ϵ]=[M−1L−1T4A2][\epsilon] = [M^{-1}L^{-1}T^4A^2][ϵ]=[M−1L−1T4A2]
Answer: The dimensional formula for permittivity is [M−1L−1T4A2][M^{-1}L^{-1}T^4A^2][M−1L−1T4A2].
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Problem 151: Dimensional Formula of Luminous Intensity
Question: Determine the dimensional formula for luminous intensity ILI_LIL.
Solution: Luminous intensity ILI_LIL is measured in candela (cd), which is a base unit in the International System of Units and does not have a fundamental dimensional formula.
Answer: The dimensional formula for luminous intensity is dimensionless (i.e., no dimensions).
Problem 152: Dimensional Formula of Pressure
Question: Find the dimensional formula for pressure PPP.
Solution: Pressure PPP is defined as force per unit area.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Area AAA: [L2][L^2][L2]
P=FAP = \frac{F}{A}P=AF [P]=[MLT−2][L2][P] = \frac{[MLT^{-2}]}{[L^2]}[P]=[L2][MLT−2] [P]=[ML−1T−2][P] = [ML^{-1}T^{-2}][P]=[ML−1T−2]
Answer: The dimensional formula for pressure is [ML−1T−2][ML^{-1}T^{-2}][ML−1T−2].
Problem 153: Dimensional Formula of Specific Heat
Question: Determine the dimensional formula for specific heat ccc.
Solution: Specific heat ccc is defined as the heat capacity per unit mass.
Heat capacity CCC: [ML2T−2Θ−1][ML^2T^{-2}\Theta^{-1}][ML2T−2Θ−1]
Mass mmm: [M][M][M]
c=Cmc = \frac{C}{m}c=mC [c]=[ML2T−2Θ−1][M][c] = \frac{[ML^2T^{-2}\Theta^{-1}]}{[M]}[c]=[M][ML2T−2Θ−1] [c]=[L2T−2Θ−1][c] = [L^2T^{-2}\Theta^{-1}][c]=[L2T−2Θ−1]
Answer: The dimensional formula for specific heat is [L2T−2Θ−1][L^2T^{-2}\Theta^{-1}][L2T−2Θ−1].
Problem 154: Dimensional Formula of Electric Field
Question: Find the dimensional formula for electric field EEE.
Solution: Electric field EEE is defined as the force per unit charge.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Charge QQQ: [AT][AT][AT]
E=FQE = \frac{F}{Q}E=QF [E]=[MLT−2][AT][E] = \frac{[MLT^{-2}]}{[AT]}[E]=[AT][MLT−2] [E]=[ML−1T−3A−1][E] = [ML^{-1}T^{-3}A^{-1}][E]=[ML−1T−3A−1]
Answer: The dimensional formula for electric field is [ML−1T−3A−1][ML^{-1}T^{-3}A^{-1}][ML−1T−3A−1].
Problem 155: Dimensional Formula of Heat Transfer Coefficient
Question: Determine the dimensional formula for the heat transfer coefficient hhh.
Solution: Heat transfer coefficient hhh is defined by Q=hAΔT⋅tQ = hA\Delta T \cdot tQ=hAΔT⋅t, where QQQ is heat, AAA is area, ΔT\Delta TΔT is temperature change, and ttt is time.
Heat QQQ: [ML2T−2][ML^2T^{-2}][ML2T−2]
Area AAA: [L2][L^2][L2]
Temperature change ΔT\Delta TΔT: [Θ][\Theta][Θ]
Time ttt: [T][T][T]
h=QAΔT⋅th = \frac{Q}{A \Delta T \cdot t}h=AΔT⋅tQ [h]=[ML2T−2][L2]⋅[Θ]⋅[T][h] = \frac{[ML^2T^{-2}]}{[L^2] \cdot [\Theta] \cdot [T]}[h]=[L2]⋅[Θ]⋅[T][ML2T−2] [h]=[T−1Θ−1][h] = [T^{-1}\Theta^{-1}][h]=[T−1Θ−1]
Answer: The dimensional formula for the heat transfer coefficient is [T−1Θ−1][T^{-1}\Theta^{-1}][T−1Θ−1].
Problem 156: Dimensional Formula of Rotational Inertia
Question: Find the dimensional formula for rotational inertia III.
Solution: Rotational inertia III is defined similarly to mass inertia but for rotational motion.
Rotational inertia III: [ML2][ML^2][ML2]
Answer: The dimensional formula for rotational inertia is [ML2][ML^2][ML2].
Problem 157: Dimensional Formula of Magnetic Susceptibility
Question: Determine the dimensional formula for magnetic susceptibility χm\chi_mχm.
Solution: Magnetic susceptibility χm\chi_mχm is a dimensionless quantity, representing the ratio of magnetization MMM to the magnetic field strength HHH.
Magnetization MMM: [ML−1A−1][M L^{-1} A^{-1}][ML−1A−1]
Magnetic field strength HHH: [AL−1][A L^{-1}][AL−1]
χm=MH\chi_m = \frac{M}{H}χm=HM [χm]=[ML−1A−1][AL−1][\chi_m] = \frac{[M L^{-1} A^{-1}]}{[A L^{-1}]}[χm]=[AL−1][ML−1A−1] [χm]=dimensionless[\chi_m] = \text{dimensionless}[χm]=dimensionless
Answer: The dimensional formula for magnetic susceptibility is dimensionless (i.e., no dimensions).
Problem 158: Dimensional Formula of Electromotive Force
Question: Find the dimensional formula for electromotive force E\mathcal{E}E.
Solution: Electromotive force E\mathcal{E}E is the work done per unit charge.
Work done WWW: [ML2T−2][ML^2T^{-2}][ML2T−2]
Charge QQQ: [AT][AT][AT]
E=WQ\mathcal{E} = \frac{W}{Q}E=QW [E]=[ML2T−2][AT][\mathcal{E}] = \frac{[ML^2T^{-2}]}{[AT]}[E]=[AT][ML2T−2] [E]=[ML2T−3A−1][\mathcal{E}] = [ML^2T^{-3}A^{-1}][E]=[ML2T−3A−1]
Answer: The dimensional formula for electromotive force is [ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1].
Problem 159: Dimensional Formula of Planck’s Radiation Constant
Question: Determine the dimensional formula for Planck's radiation constant bbb.
Solution: Planck's radiation constant bbb is used in Planck’s law of blackbody radiation: E=bT4exp(hνkT)−1E = \frac{bT^4}{\exp{\left(\frac{h \nu}{kT}\right)} - 1}E=exp(kThν)−1bT4.
Planck's radiation constant bbb: [ML3T−4Θ−3][ML^3T^{-4}\Theta^{-3}][ML3T−4Θ−3]
Answer: The dimensional formula for Planck's radiation constant is [ML3T−4Θ−3][ML^3T^{-4}\Theta^{-3}][ML3T−4Θ−3].
Problem 160: Dimensional Formula of Wavelength
Question: Find the dimensional formula for wavelength λ\lambdaλ.
Solution: Wavelength λ\lambdaλ is the distance over which the wave's shape repeats.
Wavelength λ\lambdaλ: [L][L][L]
Answer: The dimensional formula for wavelength is [L][L][L].
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Problem 161: Dimensional Formula of Magnetic Flux
Question: Determine the dimensional formula for magnetic flux Φ\PhiΦ.
Solution: Magnetic flux Φ\PhiΦ is the product of the magnetic field BBB and the area AAA perpendicular to the field.
Magnetic field BBB: [MLT−2A−1][MLT^{-2}A^{-1}][MLT−2A−1]
Area AAA: [L2][L^2][L2]
Φ=B⋅A\Phi = B \cdot AΦ=B⋅A [Φ]=[MLT−2A−1]⋅[L2][\Phi] = [MLT^{-2}A^{-1}] \cdot [L^2][Φ]=[MLT−2A−1]⋅[L2] [Φ]=[ML2T−2A−1][\Phi] = [ML^2T^{-2}A^{-1}][Φ]=[ML2T−2A−1]
Answer: The dimensional formula for magnetic flux is [ML2T−2A−1][ML^2T^{-2}A^{-1}][ML2T−2A−1].
Problem 162: Dimensional Formula of Magnetic Field Strength
Question: Find the dimensional formula for magnetic field strength HHH.
Solution: Magnetic field strength HHH is related to magnetic flux density BBB and magnetic permeability μ\muμ by B=μHB = \mu HB=μH.
Magnetic flux density BBB: [MLT−2A−1][MLT^{-2}A^{-1}][MLT−2A−1]
Magnetic permeability μ\muμ: [MLT−2A−2][MLT^{-2}A^{-2}][MLT−2A−2]
H=BμH = \frac{B}{\mu}H=μB [H]=[MLT−2A−1][MLT−2A−2][H] = \frac{[MLT^{-2}A^{-1}]}{[MLT^{-2}A^{-2}]}[H]=[MLT−2A−2][MLT−2A−1] [H]=[AL−1][H] = [A L^{-1}][H]=[AL−1]
Answer: The dimensional formula for magnetic field strength is [AL−1][A L^{-1}][AL−1].
Problem 163: Dimensional Formula of Capacitance
Question: Determine the dimensional formula for capacitance CCC.
Solution: Capacitance CCC is defined by C=QVC = \frac{Q}{V}C=VQ, where QQQ is charge and VVV is voltage.
Charge QQQ: [AT][AT][AT]
Voltage VVV: [ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1]
C=QVC = \frac{Q}{V}C=VQ [C]=[AT][ML2T−3A−1][C] = \frac{[AT]}{[ML^2T^{-3}A^{-1}]}[C]=[ML2T−3A−1][AT] [C]=[M−1L−2T4A2][C] = [M^{-1}L^{-2}T^4A^2][C]=[M−1L−2T4A2]
Answer: The dimensional formula for capacitance is [M−1L−2T4A2][M^{-1}L^{-2}T^4A^2][M−1L−2T4A2].
Problem 164: Dimensional Formula of Electric Potential Energy
Question: Find the dimensional formula for electric potential energy UUU.
Solution: Electric potential energy UUU is given by U=12CV2U = \frac{1}{2} C V^2U=21CV2, where CCC is capacitance and VVV is voltage.
Capacitance CCC: [M−1L−2T4A2][M^{-1}L^{-2}T^4A^2][M−1L−2T4A2]
Voltage VVV: [ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1]
U=12CV2U = \frac{1}{2} C V^2U=21CV2 [U]=[M−1L−2T4A2]⋅[ML2T−3A−1]2[U] = [M^{-1}L^{-2}T^4A^2] \cdot [ML^2T^{-3}A^{-1}]^2[U]=[M−1L−2T4A2]⋅[ML2T−3A−1]2 [U]=[M−1L−2T4A2]⋅[M1L4T−6A−2][U] = [M^{-1}L^{-2}T^4A^2] \cdot [M^1L^4T^{-6}A^{-2}][U]=[M−1L−2T4A2]⋅[M1L4T−6A−2] [U]=[ML2T−2][U] = [ML^2T^{-2}][U]=[ML2T−2]
Answer: The dimensional formula for electric potential energy is [ML2T−2][ML^2T^{-2}][ML2T−2].
Problem 165: Dimensional Formula of Thermal Conductivity
Question: Determine the dimensional formula for thermal conductivity kkk.
Solution: Thermal conductivity kkk is defined by Q=kAΔTdtQ = kA\frac{\Delta T}{d}tQ=kAdΔTt, where QQQ is heat, AAA is area, ΔT\Delta TΔT is temperature difference, ddd is distance, and ttt is time.
Heat QQQ: [ML2T−2][ML^2T^{-2}][ML2T−2]
Area AAA: [L2][L^2][L2]
Temperature difference ΔT\Delta TΔT: [Θ][\Theta][Θ]
Distance ddd: [L][L][L]
Time ttt: [T][T][T]
k=QA⋅ΔTd⋅tk = \frac{Q}{A \cdot \frac{\Delta T}{d} \cdot t}k=A⋅dΔT⋅tQ [k]=[ML2T−2][L2]⋅[Θ]⋅[L][T][k] = \frac{[ML^2T^{-2}]}{[L^2] \cdot [\Theta] \cdot \frac{[L]}{[T]}}[k]=[L2]⋅[Θ]⋅[T][L][ML2T−2] [k]=[ML−1T−3Θ−1][k] = [ML^{-1}T^{-3}\Theta^{-1}][k]=[ML−1T−3Θ−1]
Answer: The dimensional formula for thermal conductivity is [ML−1T−3Θ−1][ML^{-1}T^{-3}\Theta^{-1}][ML−1T−3Θ−1].
Problem 166: Dimensional Formula of Magnetic Moment
Question: Find the dimensional formula for magnetic moment μ\muμ.
Solution: Magnetic moment μ\muμ is defined as the product of current III and area AAA.
Current III: [A][A][A]
Area AAA: [L2][L^2][L2]
μ=I⋅A\mu = I \cdot Aμ=I⋅A [μ]=[A]⋅[L2][\mu] = [A] \cdot [L^2][μ]=[A]⋅[L2] [μ]=[AL2][\mu] = [AL^2][μ]=[AL2]
Answer: The dimensional formula for magnetic moment is [AL2][AL^2][AL2].
Problem 167: Dimensional Formula of Rotational Kinetic Energy
Question: Determine the dimensional formula for rotational kinetic energy ErE_rEr.
Solution: Rotational kinetic energy ErE_rEr is given by Er=12Iω2E_r = \frac{1}{2} I \omega^2Er=21Iω2, where III is rotational inertia and ω\omegaω is angular velocity.
Rotational inertia III: [ML2][ML^2][ML2]
Angular velocity ω\omegaω: [T−1][T^{-1}][T−1]
Er=12Iω2E_r = \frac{1}{2} I \omega^2Er=21Iω2 [Er]=[ML2]⋅[T−2][E_r] = [ML^2] \cdot [T^{-2}][Er]=[ML2]⋅[T−2] [Er]=[ML2T−2][E_r] = [ML^2T^{-2}][Er]=[ML2T−2]
Answer: The dimensional formula for rotational kinetic energy is [ML2T−2][ML^2T^{-2}][ML2T−2].
Problem 168: Dimensional Formula of Magnetic Field Intensity
Question: Find the dimensional formula for magnetic field intensity HHH.
Solution: Magnetic field intensity HHH is related to magnetic flux density BBB and permeability μ\muμ by B=μHB = \mu HB=μH.
Magnetic flux density BBB: [MLT−2A−1][MLT^{-2}A^{-1}][MLT−2A−1]
Permeability μ\muμ: [MLT−2A−2][MLT^{-2}A^{-2}][MLT−2A−2]
H=BμH = \frac{B}{\mu}H=μB [H]=[MLT−2A−1][MLT−2A−2][H] = \frac{[MLT^{-2}A^{-1}]}{[MLT^{-2}A^{-2}]}[H]=[MLT−2A−2][MLT−2A−1] [H]=[AL−1][H] = [A L^{-1}][H]=[AL−1]
Answer: The dimensional formula for magnetic field intensity is [AL−1][A L^{-1}][AL−1].
Problem 169: Dimensional Formula of Frequency of Light
Question: Determine the dimensional formula for the frequency of light ν\nuν.
Solution: Frequency ν\nuν is the number of cycles per unit time.
Frequency ν\nuν: [T−1][T^{-1}][T−1]
Answer: The dimensional formula for the frequency of light is [T−1][T^{-1}][T−1].
Problem 170: Dimensional Formula of Emissivity
Question: Find the dimensional formula for emissivity ε\varepsilonε.
Solution: Emissivity ε\varepsilonε is a dimensionless quantity representing the ratio of emitted radiation to that emitted by a black body.
Answer: The dimensional formula for emissivity is dimensionless (i.e., no dimensions).
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Problem 171: Dimensional Formula of Stefan-Boltzmann Constant
Question: Determine the dimensional formula for the Stefan-Boltzmann constant σ\sigmaσ.
Solution: The Stefan-Boltzmann law states that the power radiated per unit area of a black body is P=σT4P = \sigma T^4P=σT4.
Power PPP: [ML2T−3][ML^2T^{-3}][ML2T−3]
Temperature TTT: [Θ][\Theta][Θ]
Area AAA: [L2][L^2][L2]
σ=PAT4\sigma = \frac{P}{A T^4}σ=AT4P [σ]=[ML2T−3][L2]⋅[Θ]4[\sigma] = \frac{[ML^2T^{-3}]}{[L^2] \cdot [\Theta]^4}[σ]=[L2]⋅[Θ]4[ML2T−3] [σ]=[MT−3Θ−4][\sigma] = [M T^{-3} \Theta^{-4}][σ]=[MT−3Θ−4]
Answer: The dimensional formula for the Stefan-Boltzmann constant is [MT−3Θ−4][M T^{-3} \Theta^{-4}][MT−3Θ−4].
Problem 172: Dimensional Formula of Heat Capacity
Question: Find the dimensional formula for heat capacity CCC.
Solution: Heat capacity CCC is defined as the amount of heat required to change the temperature of a body by one degree.
Heat QQQ: [ML2T−2][ML^2T^{-2}][ML2T−2]
Temperature ΔT\Delta TΔT: [Θ][\Theta][Θ]
C=QΔTC = \frac{Q}{\Delta T}C=ΔTQ [C]=[ML2T−2][Θ][C] = \frac{[ML^2T^{-2}]}{[\Theta]}[C]=[Θ][ML2T−2] [C]=[ML2T−2Θ−1][C] = [ML^2T^{-2}\Theta^{-1}][C]=[ML2T−2Θ−1]
Answer: The dimensional formula for heat capacity is [ML2T−2Θ−1][ML^2T^{-2}\Theta^{-1}][ML2T−2Θ−1].
Problem 173: Dimensional Formula of Young’s Modulus
Question: Determine the dimensional formula for Young’s modulus YYY.
Solution: Young’s modulus YYY is defined as the ratio of stress to strain.
Stress: [ML−1T−2][ML^{-1}T^{-2}][ML−1T−2]
Strain (dimensionless): No dimensions
Y=StressY = \text{Stress}Y=Stress [Y]=[ML−1T−2][Y] = [ML^{-1}T^{-2}][Y]=[ML−1T−2]
Answer: The dimensional formula for Young’s modulus is [ML−1T−2][ML^{-1}T^{-2}][ML−1T−2].
Problem 174: Dimensional Formula of Gravitational Potential Energy
Question: Find the dimensional formula for gravitational potential energy UgU_gUg.
Solution: Gravitational potential energy UgU_gUg is given by Ug=mghU_g = mghUg=mgh, where mmm is mass, ggg is acceleration due to gravity, and hhh is height.
Mass mmm: [M][M][M]
Acceleration due to gravity ggg: [LT−2][LT^{-2}][LT−2]
Height hhh: [L][L][L]
Ug=mghU_g = mghUg=mgh [Ug]=[M]⋅[LT−2]⋅[L][U_g] = [M] \cdot [LT^{-2}] \cdot [L][Ug]=[M]⋅[LT−2]⋅[L] [Ug]=[ML2T−2][U_g] = [ML^2T^{-2}][Ug]=[ML2T−2]
Answer: The dimensional formula for gravitational potential energy is [ML2T−2][ML^2T^{-2}][ML2T−2].
Problem 175: Dimensional Formula of Surface Tension
Question: Determine the dimensional formula for surface tension γ\gammaγ.
Solution: Surface tension γ\gammaγ is defined as the force per unit length.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Length LLL: [L][L][L]
γ=FL\gamma = \frac{F}{L}γ=LF [γ]=[MLT−2][L][\gamma] = \frac{[MLT^{-2}]}{[L]}[γ]=[L][MLT−2] [γ]=[ML−1T−2][\gamma] = [ML^{-1}T^{-2}][γ]=[ML−1T−2]
Answer: The dimensional formula for surface tension is [ML−1T−2][ML^{-1}T^{-2}][ML−1T−2].
Problem 176: Dimensional Formula of Torque
Question: Find the dimensional formula for torque τ\tauτ.
Solution: Torque τ\tauτ is defined as the product of force and the distance from the pivot point.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Distance ddd: [L][L][L]
τ=F⋅d\tau = F \cdot dτ=F⋅d [τ]=[MLT−2]⋅[L][\tau] = [MLT^{-2}] \cdot [L][τ]=[MLT−2]⋅[L] [τ]=[ML2T−2][\tau] = [ML^2T^{-2}][τ]=[ML2T−2]
Answer: The dimensional formula for torque is [ML2T−2][ML^2T^{-2}][ML2T−2].
Problem 177: Dimensional Formula of Density
Question: Determine the dimensional formula for density ρ\rhoρ.
Solution: Density ρ\rhoρ is defined as mass per unit volume.
Mass mmm: [M][M][M]
Volume VVV: [L3][L^3][L3]
ρ=mV\rho = \frac{m}{V}ρ=Vm [ρ]=[M][L3][\rho] = \frac{[M]}{[L^3]}[ρ]=[L3][M] [ρ]=[ML−3][\rho] = [ML^{-3}][ρ]=[ML−3]
Answer: The dimensional formula for density is [ML−3][ML^{-3}][ML−3].
Problem 178: Dimensional Formula of Impulse
Question: Find the dimensional formula for impulse JJJ.
Solution: Impulse JJJ is defined as the change in momentum.
Momentum ppp: [MLT−1][MLT^{-1}][MLT−1]
J=pJ = pJ=p [J]=[MLT−1][J] = [MLT^{-1}][J]=[MLT−1]
Answer: The dimensional formula for impulse is [MLT−1][MLT^{-1}][MLT−1].
Problem 179: Dimensional Formula of Permeability
Question: Determine the dimensional formula for magnetic permeability μ\muμ.
Solution: Magnetic permeability μ\muμ relates the magnetic field BBB to the magnetic field strength HHH.
Magnetic field BBB: [MLT−2A−1][MLT^{-2}A^{-1}][MLT−2A−1]
Magnetic field strength HHH: [AL−1][A L^{-1}][AL−1]
B=μHB = \mu HB=μH μ=BH\mu = \frac{B}{H}μ=HB [μ]=[MLT−2A−1][AL−1][\mu] = \frac{[MLT^{-2}A^{-1}]}{[A L^{-1}]}[μ]=[AL−1][MLT−2A−1] [μ]=[MLT−2A−2][\mu] = [MLT^{-2}A^{-2}][μ]=[MLT−2A−2]
Answer: The dimensional formula for magnetic permeability is [MLT−2A−2][MLT^{-2}A^{-2}][MLT−2A−2].
Problem 180: Dimensional Formula of Magnetic Flux Density
Question: Find the dimensional formula for magnetic flux density BBB.
Solution: Magnetic flux density BBB is related to magnetic field strength HHH and permeability μ\muμ by B=μHB = \mu HB=μH.
Permeability μ\muμ: [MLT−2A−2][MLT^{-2}A^{-2}][MLT−2A−2]
Magnetic field strength HHH: [AL−1][A L^{-1}][AL−1]
B=μHB = \mu HB=μH [B]=[MLT−2A−2]⋅[AL−1][B] = [MLT^{-2}A^{-2}] \cdot [A L^{-1}][B]=[MLT−2A−2]⋅[AL−1] [B]=[MLT−2A−1][B] = [MLT^{-2}A^{-1}][B]=[MLT−2A−1]
Answer: The dimensional formula for magnetic flux density is [MLT−2A−1][MLT^{-2}A^{-1}][MLT−2A−1].
Problem 181: Dimensional Formula of Specific Gravity
Question: Determine the dimensional formula for specific gravity SGSGSG.
Solution: Specific gravity SGSGSG is the ratio of the density of a substance to the density of a reference substance (usually water), which is dimensionless.
Answer: The dimensional formula for specific gravity is dimensionless (i.e., no dimensions).
Problem 182: Dimensional Formula of Moment of Inertia
Question: Find the dimensional formula for the moment of inertia III.
Solution: Moment of inertia III is defined as the sum of m⋅r2m \cdot r^2m⋅r2, where mmm is mass and rrr is distance from the axis.
Mass mmm: [M][M][M]
Distance rrr: [L][L][L]
I=∑m⋅r2I = \sum m \cdot r^2I=∑m⋅r2 [I]=[M]⋅[L2][I] = [M] \cdot [L^2][I]=[M]⋅[L2] [I]=[ML2][I] = [ML^2][I]=[ML2]
Answer: The dimensional formula for moment of inertia is [ML2][ML^2][ML2].
Problem 183: Dimensional Formula of Gravitational Constant
Question: Determine the dimensional formula for the gravitational constant GGG.
Solution: The gravitational constant GGG appears in Newton's law of gravitation F=Gm1m2r2F = G \frac{m_1 m_2}{r^2}F=Gr2m1m2.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Mass mmm: [M][M][M]
Distance rrr: [L][L][L]
G=F⋅r2m1⋅m2G = \frac{F \cdot r^2}{m_1 \cdot m_2}G=m1⋅m2F⋅r2 [G]=[MLT−2]⋅[L2][M2][G] = \frac{[MLT^{-2}] \cdot [L^2]}{[M^2]}[G]=[M2][MLT−2]⋅[L2] [G]=[M−1L3T−2][G] = [M^{-1}L^3T^{-2}][G]=[M−1L3T−2]
Answer: The dimensional formula for the gravitational constant is [M−1L3T−2][M^{-1}L^3T^{-2}][M−1L3T−2].
Problem 184: Dimensional Formula of Relative Permittivity
Question: Find the dimensional
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Problem 184: Dimensional Formula of Relative Permittivity
Question: Find the dimensional formula for relative permittivity εr\varepsilon_rεr.
Solution: Relative permittivity εr\varepsilon_rεr is a dimensionless quantity representing the ratio of the permittivity of a substance to the permittivity of free space.
Answer: The dimensional formula for relative permittivity is dimensionless (i.e., no dimensions).
Problem 185: Dimensional Formula of Electric Field Strength
Question: Determine the dimensional formula for electric field strength EEE.
Solution: Electric field strength EEE is defined as the force per unit charge.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Charge QQQ: [AT][AT][AT]
E=FQE = \frac{F}{Q}E=QF [E]=[MLT−2][AT][E] = \frac{[MLT^{-2}]}{[AT]}[E]=[AT][MLT−2] [E]=[ML−1T−3A−1][E] = [ML^{-1}T^{-3}A^{-1}][E]=[ML−1T−3A−1]
Answer: The dimensional formula for electric field strength is [ML−1T−3A−1][ML^{-1}T^{-3}A^{-1}][ML−1T−3A−1].
Problem 186: Dimensional Formula of Capacitance
Question: Find the dimensional formula for capacitance CCC.
Solution: Capacitance CCC is defined by C=QVC = \frac{Q}{V}C=VQ, where QQQ is charge and VVV is voltage.
Charge QQQ: [AT][AT][AT]
Voltage VVV: [ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1]
C=QVC = \frac{Q}{V}C=VQ [C]=[AT][ML2T−3A−1][C] = \frac{[AT]}{[ML^2T^{-3}A^{-1}]}[C]=[ML2T−3A−1][AT] [C]=[M−1L−2T4A2][C] = [M^{-1}L^{-2}T^4A^2][C]=[M−1L−2T4A2]
Answer: The dimensional formula for capacitance is [M−1L−2T4A2][M^{-1}L^{-2}T^4A^2][M−1L−2T4A2].
Problem 187: Dimensional Formula of Magnetic Flux
Question: Determine the dimensional formula for magnetic flux Φ\PhiΦ.
Solution: Magnetic flux Φ\PhiΦ is the product of the magnetic field BBB and the area AAA perpendicular to the field.
Magnetic field BBB: [MLT−2A−1][MLT^{-2}A^{-1}][MLT−2A−1]
Area AAA: [L2][L^2][L2]
Φ=B⋅A\Phi = B \cdot AΦ=B⋅A [Φ]=[MLT−2A−1]⋅[L2][\Phi] = [MLT^{-2}A^{-1}] \cdot [L^2][Φ]=[MLT−2A−1]⋅[L2] [Φ]=[ML2T−2A−1][\Phi] = [ML^2T^{-2}A^{-1}][Φ]=[ML2T−2A−1]
Answer: The dimensional formula for magnetic flux is [ML2T−2A−1][ML^2T^{-2}A^{-1}][ML2T−2A−1].
Problem 188: Dimensional Formula of Electric Dipole Moment
Question: Find the dimensional formula for electric dipole moment ppp.
Solution: Electric dipole moment ppp is defined as the product of charge QQQ and distance ddd between the charges.
Charge QQQ: [AT][AT][AT]
Distance ddd: [L][L][L]
p=Q⋅dp = Q \cdot dp=Q⋅d [p]=[AT]⋅[L][p] = [AT] \cdot [L][p]=[AT]⋅[L] [p]=[ATL][p] = [ATL][p]=[ATL]
Answer: The dimensional formula for electric dipole moment is [ATL][ATL][ATL].
Problem 189: Dimensional Formula of Magnetic Susceptibility
Question: Determine the dimensional formula for magnetic susceptibility χm\chi_mχm.
Solution: Magnetic susceptibility χm\chi_mχm is a dimensionless quantity representing the ratio of the magnetization MMM to the magnetic field strength HHH.
Answer: The dimensional formula for magnetic susceptibility is dimensionless (i.e., no dimensions).
Problem 190: Dimensional Formula of Inductive Reactance
Question: Find the dimensional formula for inductive reactance XLX_LXL.
Solution: Inductive reactance XLX_LXL is given by XL=ωLX_L = \omega LXL=ωL, where ω\omegaω is angular frequency and LLL is inductance.
Angular frequency ω\omegaω: [T−1][T^{-1}][T−1]
Inductance LLL: [ML2T−2A−2][ML^2T^{-2}A^{-2}][ML2T−2A−2]
XL=ωLX_L = \omega LXL=ωL [XL]=[T−1]⋅[ML2T−2A−2][X_L] = [T^{-1}] \cdot [ML^2T^{-2}A^{-2}][XL]=[T−1]⋅[ML2T−2A−2] [XL]=[ML2T−3A−2][X_L] = [ML^2T^{-3}A^{-2}][XL]=[ML2T−3A−2]
Answer: The dimensional formula for inductive reactance is [ML2T−3A−2][ML^2T^{-3}A^{-2}][ML2T−3A−2].
Problem 191: Dimensional Formula of Capacitive Reactance
Question: Determine the dimensional formula for capacitive reactance XCX_CXC.
Solution: Capacitive reactance XCX_CXC is given by XC=1ωCX_C = \frac{1}{\omega C}XC=ωC1, where ω\omegaω is angular frequency and CCC is capacitance.
Angular frequency ω\omegaω: [T−1][T^{-1}][T−1]
Capacitance CCC: [M−1L−2T4A2][M^{-1}L^{-2}T^4A^2][M−1L−2T4A2]
XC=1ωCX_C = \frac{1}{\omega C}XC=ωC1 [XC]=1[T−1]⋅[M−1L−2T4A2][X_C] = \frac{1}{[T^{-1}] \cdot [M^{-1}L^{-2}T^4A^2]}[XC]=[T−1]⋅[M−1L−2T4A2]1 [XC]=[ML2T−3A−2][X_C] = [M L^2 T^{-3} A^{-2}][XC]=[ML2T−3A−2]
Answer: The dimensional formula for capacitive reactance is [ML2T−3A−2][ML^2T^{-3}A^{-2}][ML2T−3A−2].
Problem 192: Dimensional Formula of Electromotive Force (EMF)
Question: Find the dimensional formula for electromotive force (EMF) E\mathcal{E}E.
Solution: Electromotive force (EMF) E\mathcal{E}E is equivalent to voltage.
Voltage VVV: [ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1]
Answer: The dimensional formula for electromotive force (EMF) is [ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−1].
Problem 193: Dimensional Formula of Conductance
Question: Determine the dimensional formula for conductance GGG.
Solution: Conductance GGG is the reciprocal of resistance RRR.
Resistance RRR: [ML2T−3A−2][ML^2T^{-3}A^{-2}][ML2T−3A−2]
G=1RG = \frac{1}{R}G=R1 [G]=1[ML2T−3A−2][G] = \frac{1}{[ML^2T^{-3}A^{-2}]}[G]=[ML2T−3A−2]1 [G]=[M−1L−2T3A2][G] = [M^{-1}L^{-2}T^3A^2][G]=[M−1L−2T3A2]
Answer: The dimensional formula for conductance is [M−1L−2T3A2][M^{-1}L^{-2}T^3A^2][M−1L−2T3A2].
Problem 194: Dimensional Formula of Electric Charge
Question: Find the dimensional formula for electric charge QQQ.
Solution: Electric charge QQQ is the product of current III and time ttt.
Current III: [A][A][A]
Time ttt: [T][T][T]
Q=I⋅tQ = I \cdot tQ=I⋅t [Q]=[A]⋅[T][Q] = [A] \cdot [T][Q]=[A]⋅[T] [Q]=[AT][Q] = [AT][Q]=[AT]
Answer: The dimensional formula for electric charge is [AT][AT][AT].
Problem 195: Dimensional Formula of Force Constant (Spring Constant)
Question: Determine the dimensional formula for force constant kkk of a spring.
Solution: Force constant kkk is defined by Hooke's law as F=kxF = kxF=kx, where FFF is force and xxx is displacement.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Displacement xxx: [L][L][L]
k=Fxk = \frac{F}{x}k=xF [k]=[MLT−2][L][k] = \frac{[MLT^{-2}]}{[L]}[k]=[L][MLT−2] [k]=[ML−1T−2][k] = [ML^{-1}T^{-2}][k]=[ML−1T−2]
Answer: The dimensional formula for the force constant is [ML−1T−2][ML^{-1}T^{-2}][ML−1T−2].
Problem 196: Dimensional Formula of Angular Momentum
Question: Find the dimensional formula for angular momentum LLL.
Solution: Angular momentum LLL is defined as L=r⋅pL = r \cdot pL=r⋅p, where rrr is the distance and ppp is momentum.
Distance rrr: [L][L][L]
Momentum ppp: [MLT−1][MLT^{-1}][MLT−1]
L=r⋅pL = r \cdot pL=r⋅p [L]=[L]⋅[MLT−1][L] = [L] \cdot [MLT^{-1}][L]=[L]⋅[MLT−1] [L]=[ML2T−1][L] = [ML^2T^{-1}][L]=[ML2T−1]
Answer: The dimensional formula for angular momentum is [ML2T−1][ML^2T^{-1}][ML2T−1].
Problem 197: Dimensional Formula of Planck’s Constant
Question: Determine the dimensional formula for Planck’s constant hhh.
Solution: Planck’s constant hhh appears in the relation E=hνE = h \nuE=hν, where EEE is energy and (
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Problem 197: Dimensional Formula of Planck’s Constant
Question: Determine the dimensional formula for Planck’s constant hhh.
Solution: Planck’s constant hhh is used in the equation E=hνE = h \nuE=hν, where EEE is energy and ν\nuν (nu) is frequency.
Energy EEE: [ML2T−2][ML^2T^{-2}][ML2T−2]
Frequency ν\nuν: [T−1][T^{-1}][T−1]
h=Eνh = \frac{E}{\nu}h=νE [h]=[ML2T−2][T−1][h] = \frac{[ML^2T^{-2}]}{[T^{-1}]}[h]=[T−1][ML2T−2] [h]=[ML2T−1][h] = [ML^2T^{-1}][h]=[ML2T−1]
Answer: The dimensional formula for Planck’s constant is [ML2T−1][ML^2T^{-1}][ML2T−1].
Problem 198: Dimensional Formula of Rotational Kinetic Energy
Question: Find the dimensional formula for rotational kinetic energy KKK.
Solution: Rotational kinetic energy KKK is given by K=12Iω2K = \frac{1}{2} I \omega^2K=21Iω2, where III is the moment of inertia and ω\omegaω is the angular velocity.
Moment of inertia III: [ML2][ML^2][ML2]
Angular velocity ω\omegaω: [T−1][T^{-1}][T−1]
K=12Iω2K = \frac{1}{2} I \omega^2K=21Iω2 [K]=[ML2]⋅[T−2][K] = [ML^2] \cdot [T^{-2}][K]=[ML2]⋅[T−2] [K]=[ML2T−2][K] = [ML^2T^{-2}][K]=[ML2T−2]
Answer: The dimensional formula for rotational kinetic energy is [ML2T−2][ML^2T^{-2}][ML2T−2].
Problem 199: Dimensional Formula of Specific Heat Capacity
Question: Determine the dimensional formula for specific heat capacity ccc.
Solution: Specific heat capacity ccc is defined as the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius (or Kelvin).
Heat QQQ: [ML2T−2][ML^2T^{-2}][ML2T−2]
Mass mmm: [M][M][M]
Temperature ΔT\Delta TΔT: [Θ][\Theta][Θ]
c=Qm⋅ΔTc = \frac{Q}{m \cdot \Delta T}c=m⋅ΔTQ [c]=[ML2T−2][M]⋅[Θ][c] = \frac{[ML^2T^{-2}]}{[M] \cdot [\Theta]}[c]=[M]⋅[Θ][ML2T−2] [c]=[L2T−2Θ−1][c] = [L^2T^{-2}\Theta^{-1}][c]=[L2T−2Θ−1]
Answer: The dimensional formula for specific heat capacity is [L2T−2Θ−1][L^2T^{-2}\Theta^{-1}][L2T−2Θ−1].
Problem 200: Dimensional Formula of Work Done
Question: Find the dimensional formula for work done WWW.
Solution: Work done WWW is given by W=F⋅dW = F \cdot dW=F⋅d, where FFF is force and ddd is distance.
Force FFF: [MLT−2][MLT^{-2}][MLT−2]
Distance ddd: [L][L][L]
W=F⋅dW = F \cdot dW=F⋅d [W]=[MLT−2]⋅[L][W] = [MLT^{-2}] \cdot [L][W]=[MLT−2]⋅[L] [W]=[ML2T−2][W] = [ML^2T^{-2}][W]=[ML2T−2]
Answer: The dimensional formula for work done is [ML2T−2][ML^2T^{-2}][ML2T−2].