# What are the dimensions of density?

1. [M1L-3T0]
2. [M1L-1T0]
3. [M-1L3T0]
4. [M-1L-3T0]

Option 1 : [M1L-3T0]

## Dimensions of physical quantities MCQ Question 1 Detailed Solution

CONCEPT:

• Density: The mass per unit volume is called a density of a material.

$$Density(ρ)=\frac{Mass \,(m)}{Volume \,(V)}$$

• Density is a scalar quantity.

EXPLANATION:

• Density is defined as mass per unit volume i.e.,

$$Density(ρ)=\frac{Mass \,(m)}{Volume \,(V)}$$

As we know, the dimension of mass (M) = [M]

The dimension of volume (V) = [L3]

• Therefore, the dimension of density (ρ) is

$$Density(ρ)=\frac{[M]}{[L^3]}=[M^1L^{-3}T^0]$$

• Some Basic Dimensional formula:
 Sl. No. Quantity Common Symbol SI Unit Dimension 1 Velocity v, u ms-1 LT-1 2 Acceleration a ms-2 LT-2 3 Force F Newton (N) M L T-2 4 Momentum p Kg-ms-1 M L T-1 5 Gravitational constant G N-m2Kg-2 L3 M-1 T-2 6 Gravitational potential P J-kg-1 M0 L2 T-2 7 Bulk Modulus B Nm-2 M L-1 T-2 8 Energy E, U, K joule (J) M L2 T-2 9 Heat Q joule (J) M L2 T-2 10 Pressure P Nm-2 (Pa) M L-1 T-2

# Pressure is measured in terms ofA. Mass & DensityB. Work doneC. Force and AreaD. Force and Distance

1. D
2. A
3. C
4. B

Option 3 : C

## Dimensions of physical quantities MCQ Question 2 Detailed Solution

The Correct Answer is Option 3 i.e C.

CONCEPT:

• Pressure: The pressure is the force applied per unit area in a direction perpendicular to the surface of the object.
• SI unit of pressure is the pascal (Pa).

Pressure (P) = Force (F)/Area (A)

• Work:  Work is measured by the product of force and displacement of the body along the direction of the force.
• SI unit of work is Joule (J).
• Energy: The capacity of doing work by a body is called its energy.
• SI unit of energy is Joule (J).
• DensityDensity is the mass per unit volume.
• SI unit of volume is the cubic meter.
• SI unit of density is kilogram per cubic meter (kg/m3).

EXPLANATION:

• Since the force per unit area is called pressure. So pressure is measured in terms of force and area. Hence option 3 is correct.

# The dimensional formula of speed

1. [ML0T-1]
2. [MLT-1]
3. [M0LT-1]
4. [MLT-2]

Option 3 : [M0LT-1]

## Dimensions of physical quantities MCQ Question 3 Detailed Solution

CONCEPT:

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

EXPLANATION:

• Speed is defined as the rate of change of distance i.e.,

$$\Rightarrow Speed (v) = \frac{Distance (d)}{time (t)}$$

As we know, the dimension formula of distance (d) = [L]

The dimension formula of time (t) = [t]

∴ The dimension formula of speed is

$$\Rightarrow v =\frac{[L]}{[T]}=[M^0LT^{-1}]$$

 Quantity Unit Dimension Pressure Pascal [ML-1T-2] Stress Pascal [ML-1T-2] Velocity m/s [LT-1] Speed m/s [LT-1] Force Newton [MLT-2] Impulse Newton-second [MLT-1] Work Joule [ML2T-2] Energy Joule [ML2T-2] Capacitance (C) Coulomb/volt or Farad [M-1L-2T4A2] Resistivity or Specific resistance (ρ) Ohm-meter [ML3T-3A-2] Electric current (I) Ampere [A] Electric charge (q) Coulomb [AT] Inductance (H) henry [ML2T-2A-2]

# If v = at + bt2v is velocity and t in seconds, then the dimension of b is:

1. LT0
2. LT-1
3. LT-2
4. LT-3

Option 4 : LT-3

## Dimensions of physical quantities MCQ Question 4 Detailed Solution

CONCEPT:

Principle of homogeneity of dimensions:

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

EXPLANATION

Given - v = at + bt2

• From the principle of dimensional homogeneity, the left-hand side of the equation dimensionally equal to the right-hand side of the equation.
• The dimension formula of velocity (v) = [LT-1]

∴ [LT-1] = [a] [T]

$$⇒ \left[ a \right] =\frac{[LT^{-1}]}{[T]}=[LT^{-2}]$$

• Therefore the dimension of 'a' is [LT-2].

For the second term,

⇒ [LT-1] = [b] [T2]

$$⇒ \left[ b \right] =\frac{[LT^{-1}]}{[T^2]}=[LT^{-3}]$$

• Therefore the dimension of 'b' is [LT-3].

# The dimensions of EMF are

1. ML2T-3A-1
2. ML2T2A3
3. M-1T3
4. ML3T1A3

Option 1 : ML2T-3A-1

## Dimensions of physical quantities MCQ Question 5 Detailed Solution

CONCEPT:

Emf of the cell (E):

• The potential difference across the terminals of a cell when it is not supplying any current is called its emf.
• Equation of cell can be written as

⇒ E = V + Ir

Where E = emf of the cell, V = Potential difference, I = current and r = internal resistance

EXPLANATION:

• When no current is passing through the cell, then I = 0 A, hence

⇒ E = V

As we know electric potential can be is written as

$$\Rightarrow E=V = \frac{{{\rm{Work\;done\;}}\left( {\rm{W}} \right)}}{{{\rm{Charge\;}}\left( {\rm{q}} \right)}}$$

The dimension of work done (W) = [ML2T-2]

The dimension of charge (q) = [IT]

∴ The dimension of emf is given by

$$\Rightarrow E=\frac{ML^2T^{-2}}{AT}=[ML^2A^{-1}T^{-3}]$$

 Quantity Unit Dimension Pressure Pascal [ML-1T-2] Stress Pascal [ML-1T-2] Young Modulus Pascal [ML-1T-2] Speed m/s [LT-1] Momentum kg⋅m/s [MLT-1] Potential difference Volt [ML2T-3A-1] Torque Newton meter [M1L2T-2] Work Joule [ML2T-2] Energy Joule [ML2T-2] Weight Newton [MLT-2] Capacitance (C) Coulomb/volt or Farad [M-1L2T4A2] Resistivity or Specific resistance (ρ) Ohm-meter [ML3T-3A-2] Electric current (I) Ampere [A]

# [M0L2T-­2] are the dimensions of _______.

1. Gravitational potential energy
2. Gravitational potential
3. Gravitational constant
4. Gravitational intensity

Option 2 : Gravitational potential

## Dimensions of physical quantities MCQ Question 6 Detailed Solution

CONCEPT:

• Gravitational potential is defined as potential energy associated with its unit mass due to its position in the gravitational field of other body or it is defined as the amount of work done in bringing a body of unit mass from infinity to that point.
• Gravitational potential is a scalar quantity.
• Its SI unit is Jkg-1 or Nm/kg and the cgs unit is erg g-1.
• Gravitational potential is given by

$$V =\frac{-GM}{r}$$

Where G = Gravitational constant, M = mass , and r = Distance

• The variation of gravitational potential with distance is given by

EXPLANATION:

• Mathematically, Gravitational potential can be written as

$$\Rightarrow P = \frac{W}{m}$$

As we know, the dimension of work done (W) = [ML2T-2]

The dimension of mass (m) = [M]

∴ The dimension of Gravitational potential is

$$\Rightarrow P = \frac{[ML^2T^{-2}]}{[M]}=[M^0L^2T^{-2}]$$

• Some Basic Dimensional formula:
 Sl. No. Quantity Common Symbol SI Unit Dimension 1 Velocity v, u ms-1 LT-1 2 Acceleration a ms-2 LT-2 3 Force F Newton (N) M L T-2 4 Momentum p Kg-ms-1 M L T-1 5 Gravitational constant G N-m2Kg-2 L3 M-1 T-2 6 Gravitational potential P J-kg-1 M0 L2 T-2 7 Bulk Modulus B Nm2 M L-1 T-2 8 Energy E, U, K joule (J) M L2 T-2 9 Heat Q joule (J) M L2 T-2 10 Pressure P Nm-2 (Pa) M L-1 T-2 11 Electric Field E Vm-1, NC-1 M L I-1 T-3 12 Potential (voltage) V V, JC-1 M L2 I-1 T-3 13 Magnetic Field B Tesla (T), Wb m-1 M I-1T-2 14 Magnetic Flux ΦB Wb M L2 I-1 T-2 15 Resistance R Ohm (Ω) M L2 I-2 T-3 16 Electromotive Force E Volt (V) M L2 I-1 T-3 17 Density ρ, d kg/m3 ML-3 T0

# The dimensional formula of force:

1. [ML-3T2]
2. [MLT-2]
3. [ML-2T-2]
4. [ML-1T-1]

Option 2 : [MLT-2]

## Dimensions of physical quantities MCQ Question 7 Detailed Solution

CONCEPT:

• Force: The interaction which after applying on a body changes or try to change the state of motion or state of rest is called force.
• Force is denoted by F and the SI unit is Newton (N).
• Force (F) = Mass (m) × acceleration (a)

EXPLANATION:

• Force can be written mathematically as

⇒ F = m × a

As we know, the dimension of mass (m) = [M]

The dimension of acceleration (a) = [LT-2]

∴ The dimension of force (F) is

⇒ F = [M] ×  [LT-2] = [MLT-2]

# The dimensions of Young's Modulus match with the dimensions of ___________.

1. Pressure
2. Torque
3. Momentum
4. Weight

Option 1 : Pressure

## Dimensions of physical quantities MCQ Question 8 Detailed Solution

Concept:

Young’s modulus:

• Young's modulus of elasticity, applicable to the stretching of wire, etc., is equal to the ratio of the applied load per unit area of the cross-section to the increase in length per unit length.
• It is denoted as E.
• The unit of Young’s modulus is N/m2.

$$\text{E}=\frac{\text{ }\!\!\sigma\!\!\text{ }}{\epsilon }$$

where σ = stress, ϵ = strain in wire.

Young’s Modulus Formula by using other quantities:

$$\sigma=\frac{F}{A},\;\;\;\epsilon=\frac{\Delta{L}}{L_o}$$

$${\rm{E}} = \frac{{{\rm{F}}{{\rm{L}}_0}}}{{{\rm{A\Delta L}}}}$$

where F = force exerted under tension, A = actual cross-sectional area, L0 = actual length, ΔL = change in length.

Explanation:

Young’s Modulus:

$${\rm{E}} = \frac{{{\rm{F}}{{\rm{L}}_0}}}{{{\rm{A\Delta L}}}}$$

The dimension of Force (F) = [MLT-2]

The dimension of Area (A) = [L2]

∴ The dimension of Young’s Modulus (E) is

$$\Rightarrow E=\frac{FL_o}{A\Delta{L}}\Rightarrow\frac{MLT^{-2}}{L^2}=[ML^{-1}T^{-2}]\;\;\;(∵ length\;unit\;gets\;cancelled)$$

Pressure:

The pressure is defined as force per unit area i.e.

$$P = \frac{F}{A}$$

As Force = mass × acceleration

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

The dimension of Area (A) = [L2]

Hence the dimensional formula of pressure is:

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

Important Points

 Quantity Unit Dimension Pressure Pascal [ML-1T-2] Stress Pascal [ML-1T-2] Young Modulus Pascal [ML-1T-2] Speed m/s [LT-1] Momentum kg⋅m/s [MLT-1] Potential difference Volt [ML2T-3A-1] Torque Newton meter [ML2T-2] Work Joule [ML2T-2] Energy Joule [ML2T-2] Weight Newton [MLT-2] Capacitance (C) Coulomb/volt or Farad [M-1L2T4A2] Resistivity or Specific resistance (ρ) Ohm-meter [ML3T-3A-2] Electric current (I) Ampere [A]

# What is the SI unit of amplitude?

1. Kilogram
2. Metre
3. Hertz
4. Second

Option 2 : Metre

## Dimensions of physical quantities MCQ Question 9 Detailed Solution

The Correct Answer is Option 2 i.e Metre.

• Amplitude: It is defined as a maximum displacement of the vibrating particle on either side from the equilibrium position.
• SI unit of Amplitude is Metre
 SI Unit Physical Quantity Amplitude metre Force newton Velocity Metre per second Time Sec Mass Kilogram Frequency Hertz

# [M-1L3T-2] are the dimensions of?

1. Gravitational constant
2. Gravitational potential energy
3. Gravitational potential
4. Gravitational intensity

Option 1 : Gravitational constant

## Dimensions of physical quantities MCQ Question 10 Detailed Solution

CONCEPT:

• Newton's law of Gravitation: It states that everybody in this universe attracts every other body with a force, which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
• The direction of the force is along the line joining the particles.

The magnitude of the gravitational force F is given by:

$$F = G\frac{{{M_1}{M_2}}}{{{R^2}}}$$

Where G = universal gravitational constant, M1 = mass of 1st body, M2 = mass of 2nd body and R = distance between the two bodies.

EXPLANATION:

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

$$\Rightarrow G = \frac{{F \times {R^2}}}{{{M_1}{M_2}}}$$

Now,

Force = mass × acceleration

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

Dimensional formula of radius (R2) = [L2]

Dimensional formula of radius (M) = [M]

• Dimensional formula of universal gravitational constant (G)

$$\Rightarrow G = \frac{{ML{T^{ - 2}} \times {L^2}}}{{{M^2}}} = \frac{{M{L^3}{T^{ - 2}}}}{{{M^2}}} = {M^{ - 1}}{L^3}{T^{ - 2}}$$

∴ The dimensional formula of universal gravitational constant G is [M-1L3T-2].

• Some Basic Dimensional formula:
 Sl. No. Quantity Common Symbol SI Unit Dimension 1 Velocity v, u ms-1 LT-1 2 Acceleration a ms-2 LT-2 3 Force F Newton (N) M L T-2 4 Momentum p Kg-ms-1 M L T-1 5 Gravitational constant G N-m2Kg-2 L3 M-1 T-2 6 Gravitational potential P J-kg-1 M0 L2 T-2 7 Bulk Modulus B Nm2 M L-1 T-2 8 Energy E, U, K joule (J) M L2 T-2 9 Heat Q joule (J) M L2 T-2 10 Pressure P Nm-2 (Pa) M L-1 T-2 11 Electric Field E Vm-1, NC-1 M L I-1 T-3 12 Potential (voltage) V V, JC-1 M L2 I-1 T-3 13 Magnetic Field B Tesla (T), Wb m-1 M I-1T-2 14 Magnetic Flux ΦB Wb M L2 I-1 T-2 15 Resistance R Ohm (Ω) M L2 I-2 T-3 16 Electromotive Force E Volt (V) M L2 I-1 T-3 17 Density ρ, d kg/m3 ML-3 T0

# Candella is a unit of ____________.

1. Density
2. Volume
3. Light Intensity
4. Sound

Option 3 : Light Intensity

## Dimensions of physical quantities MCQ Question 11 Detailed Solution

The Correct Answer is Option 3 i.e Light Intensity.

 SI Units Physical Quantity Candela Light Intensity Cubic meter Volume Decibel Sound kilogram per cubic metre (kg/m3) Density Pascal Pressure Poise Kinematic viscosity

# Which of the following is the SI-derived unit for measuring electric conductance?

1. Tesla
3. Henry
4. Siemens

Option 4 : Siemens

## Dimensions of physical quantities MCQ Question 12 Detailed Solution

Key Points

• Siemens is the SI unit for measuring electric conductance.
• Siemens per unit is the unit of electrical conductivity.
• Mho is also known as siemens.
• The reciprocal of ohm is siemens.
• The reciprocal of resistance is siemens.
• In the 14th general conference on weights and measures, the unit of conductance known as siemens was adopted.
• The unit was named after Earner won Siemens.
• The ratio of current density to the electric field strength gives the value of electrical conductivity.
• If the conductivity of a substance is higher, then the resistance of that substance will be lower and the flow of electric current will be smoother.
• The ability of electrons or charge carriers decides the value of electrical conductivity.

•  Henry is the SI-derived unit of electrical inductance.
•  Farad is the SI-derived unit of electrical capacitance.
• Tesla  is a unit of the magnetic induction.

# [ML2T-2] are the dimensions of __________.

1. Gravitational constant
2. Gravitational potential
3. Gravitational potential energy
4. Gravitational in tensity

Option 3 : Gravitational potential energy

## Dimensions of physical quantities MCQ Question 13 Detailed Solution

CONCEPT:

• Dimensions: When a derived quantity is expressed in terms of fundamental quantities, it is written as a product of different powers of the fundamental quantities.
• The powers to which fundamental quantities must be raised in order to express the given physical quantity are called its dimensions.

EXPLANATION:

Gravitational Potential Energy (U):

• The gravitational potential energy of a body is the energy associated with it due to its position in the gravitational field of another body and is measured by the amount of work done in bringing a body from infinity to a given point in the gravitational field of the other.
• Mathematically gravitational potential energy can be written as:

⇒ U = m × g × h

As we know, the dimension of mass (m) = [M]

The dimension of acceleration due to gravity (g) is = [LT-2]

The dimension of height (h) = [L]

∴ The dimension of  gravitational potential energy is

⇒ U = [M] ×  [LT-2] × [L]

⇒ U = [ML2T-2]

• Some Basic Dimensional formula:
 Sl. No. Quantity Common Symbol SI Unit Dimension 1 Velocity v, u ms-1 LT-1 2 Acceleration a ms-2 LT-2 3 Force F Newton (N) M L T-2 4 Momentum p Kg-ms-1 M L T-1 5 Gravitational constant G N-m2Kg-2 L3 M-1 T-2 6 Torque τ N-m M L2 T-2 7 Bulk Modulus B Nm2 M L-1 T-2 8 Energy E, U, K joule (J) M L2 T-2 9 Heat Q joule (J) M L2 T-2 10 Pressure P Nm-2 (Pa) M L-1 T-2 11 Electric Field E Vm-1, NC-1 M L I-1 T-3 12 Potential (voltage) V V, JC-1 M L2 I-1 T-3

# Which dimensions match with the unit Pa?

1. [M1L1T-2]
2. [M1L-1T-2]
3. [M1L-1T2]
4. [M1L1T2]

Option 2 : [M1L-1T-2]

## Dimensions of physical quantities MCQ Question 14 Detailed Solution

CONCEPT:

• Pressure: The force per unit area is called pressure. It is denoted by P.
• The SI unit of pressure is the pascal (Pa).
• The basic formula for pressure is given by:

$${\rm{Pressure (P)}} = \frac{{{\rm{Force (F)}}}}{{{\rm{Area (A)}}}}$$

EXPLANATION:

The pressure is defined as force per unit area i.e.

$$P = \frac{F}{A}$$

As force = mass × acceleration

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

• Hence the dimensional formula of pressure is

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

• Some Basic Dimensional formula:
 Sl. No. Quantity Common Symbol SI Unit Dimension 1 Velocity v, u ms-1 LT-1 2 Acceleration a ms-2 LT-2 3 Force F Newton (N) M L T-2 4 Momentum p Kg-ms-1 M L T-1 5 Gravitational constant G N-m2Kg-2 L3 M-1 T-2 6 Gravitational potential P J-kg-1 M0 L2 T-2 7 Bulk Modulus B Nm2 M L-1 T-2 8 Energy E, U, K joule (J) M L2 T-2 9 Heat Q joule (J) M L2 T-2 10 Pressure P Nm-2 (Pa) M L-1 T-2

# Magnetic field intensity has the dimensions

1. IL
2. I2L
3. IL-1
4. IL-2

Option 3 : IL-1

## Dimensions of physical quantities MCQ Question 15 Detailed Solution

CONCEPT:

The intensity of magnetization:

• When a magnetic material is placed in a magnetizing field, it gets magnetized.
• The magnetic moment developed per unit volume of a material when placed in a magnetizing field is called intensity of magnetization or simply magnetization. Thus

$$I = \frac{{ m}}{V}$$

EXPLANATION:

• Mathematically, magnetic field intensity can be written as,

$$\Rightarrow I = \frac{{ m}}{V}$$

• As we know, the magnetic moment (m) is equal to the current time's area, therefore,

The dimension of magnetic moment (I) = [IL2]

The dimension of volume (V) = [L3]

∴ The dimension of magnetic field intensity is,

$$\Rightarrow I=\frac{[IL^2]}{L^3}=[IL^{-1}]$$

 Quantity Unit Dimension Pressure Pascal [ML-1T-2] Stress Pascal [ML-1T-2] Young Modulus Pascal [ML-1T-2] Speed m/s [LT-1] Momentum kg⋅m/s [MLT-1] Potential difference Volt [ML2T-3A-1] Torque Newton meter [M1L2T-2] Work Joule [ML2T-2] Energy Joule [ML2T-2] Weight Newton [MLT-2] Capacitance (C) Coulomb/volt or Farad [M-1L2T4A2] Resistivity or Specific resistance (ρ) Ohm-meter [ML3T-3A-2] Electric current (I) Ampere [A]

# [M0LT-2] are the dimensions of _______.

1. Gravitational potential energy
2. Gravitational intensity
3. Gravitational potential
4. Gravitational constant

Option 2 : Gravitational intensity

## Dimensions of physical quantities MCQ Question 16 Detailed Solution

CONCEPT:

• Dimensions: When a derived quantity is expressed in terms of fundamental quantities, it is written as a product of different powers of the fundamental quantities.
• The powers to which fundamental quantities must be raised in order to express the given physical quantity are called its dimensions.
 Gravitational potential energy (U) Gravitational intensity (EG) Gravitational potential (VG) Gravitational constant (G) Formula: U = m × g × h $${E_G} = \frac{{GM}}{{{r^2}}}$$ $${V_G} = - \frac{{GM}}{r}$$ $$F = \frac{{G{m_1}{m_2}}}{{{r^2}}}$$ $$G = \frac{{F{r^2}}}{{{m_1}{m_2}}} = \frac{{(ma){r^2}}}{{{m_1}{m_2}}},$$ Dimensions: As we know, the dimension of mass (m) = [M] The dimension of acceleration due to gravity (g) is = [LT-2] The dimension of height (h) = [L] ∴ The dimension of  gravitational potential energy is  $$\left[ U \right] = \left[ M \right]\;\left[ {L{T^{ - 2}}} \right]\;\left[ L \right] = M\;{L^2}\;{T^{ - 2}}$$ Units: $${E_G} = \left( {k{g^{ - 1}}{m^3}{{\sec }^{ - 2}}.} \right)\left( {Kg} \right)\left( {{r^{ - 2}}} \right)\;$$ $$dimensions:{E_G} = M^0{L^1}{T^{ - 2}}$$ Units:   $${V_G} = \left( {k{g^{ - 1}}{m^3}{{\sec }^{ - 2}}.} \right)\left( {kg} \right)\left( {{m^{ - 1}}} \right)$$ $$dimensions:{V_G} = {L^2}{T^{ - 2}}$$ Units:     $$G = kg\;m{s^{ - 2}}\frac{{{m^2}}}{{k{g^2}}} = k{g^{ - 1}}{m^3}{\sec ^{ - 2}}.$$ $$Dimensions:G = {M^{ - 1}}{L^3}{T^{ - 2}}$$

EXPLANATION:

• [M0LT-2] are the dimensions of the gravitational field/intensity. So option 2 is correct.

# What is the SI unit for measuring the luminous intensity?

1. Ampere
2. Mole
3. Candela
4. Kelvin

Option 3 : Candela

## Dimensions of physical quantities MCQ Question 17 Detailed Solution

• In the field of physics the Candela can be defined as the unit of luminous intensity a measure of the wavelength weighted power emitted by a source of light.
• The Candela is a base unit which can be defined by taking the fixed value of the luminous efficacy of the monochromatic radiation having a frequency of 540 * 1012 Hertz. Alternately, it can be described as the luminous power corresponding to one unit solid angle that is emitted by a point source of light.
• Unit system SI base unit.
• Unit of  Luminous intensity.
• International candles  ≈ 1.02 cp
• Candela symbol is cd.

 Ampere Ampere or amp is a unit of electric current denoted by A. It is named after the French Physicist and Mathematician Andre-Marie Ampere, who is considered as the Father of Electrodynamics. Ampere = 1 Coloumb/second. Mole The mole is the unit of measurement for amount of substance in the International System of Units. Symbol - mol. Unit of  Amount of substance. SI base units 1000 mmol Unit system SI base unit. Kelvin The kelvin is a base SI unit of measurement, defined as the fraction 1/273.16 of the temperature of the triple point of water, which is the temperature at which water in solid, liquid, and gaseous state coexist in equilibrium. Unit of  Temperature. Unit system SI base unit.

# Which of the follwing is dimensionless?

1. Velocity
2. Angle
3. Mass
4. Area

Option 2 : Angle

## Dimensions of physical quantities MCQ Question 18 Detailed Solution

CONCEPT:

• The powers to which fundamental quantities must be raised in order to express the given physical quantity are called its dimensions.
• A quantity without dimension will usually be a ratio of two quantities with similar dimensions and hence, will cancel out. Thus, they will have no units.

EXPLANATION:

• Velocity: It is defined as the rate of change of displacement i.e.

$$\Rightarrow Velocity = \frac{Displacemnt}{Time}=\frac{[L]}{[T]}=[LT^{-1}]$$

∴ Velocity is not the dimensionless quantity.

• Angle: It is defined as the ratio of the length of arc to the radius, i.e.,

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

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

∴ An angle is a dimensionless quantity

• Mass is a not dimensionless quantity. as the dimension of mass is [M].
• The area is a not dimensionless quantity. as the dimension of the area is [L2].

 Quantity Unit Dimension Pressure Pascal [ML-1T-2] Stress Pascal [ML-1T-2] Young Modulus Pascal [ML-1T-2] Speed m/s [LT-1] Momentum kg⋅m/s [MLT-1] Potential difference Volt [ML2T-3A-1] Torque Newton meter [M1L2T-2] Work Joule [ML2T-2]
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