Dimensions of Physical Quantities: Formula, Equations & Applications

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Dimensions of Physical Quantity are defined as the power raised to the fundamental units to obtain one unit of that physical quantity. These quantities are expressed using the ‘[ ]’ symbol (i.e., square brackets) and help in determining its nature.

There are seven fundamental units, namely mass, length, time, temperature, electric current, luminous intensity and amount of substance.
Every physical quantity can be expressed using these fundamental units and their dimensions.
The dimension of a physical quantity is the same as that of the dimensions of its unit.
Unit is a physical quantity used in the measurement of constant quantities like mass and length.
Physical Quantities are divided into two categories, namely fundamental quantity and derived quantity.
The process of determining how small or large a physical quantity is compared to a basic standard is referred to as measurement.

The dimensional formula is a combined expression of fundamental units involved in deriving the physical quantity. Physical quantities are represented in terms of dimension equations.The dimension equation is a relationship between a physical quantity and its dimensional formula.

Fundamental Quantities and Units

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Fundamental quantities are defined as quantities that are independent of other quantities. The units that are used to measure these fundamental quantities are defined as fundamental units.

They cannot be derived from other units and are not related to each other.
There are seven fundamental quantities in the physical world.
Every physical quantity can be derived and expressed using these fundamental quantities.
Units like CGS, MKS, FPS and SI are used to measure a fundamental quantity.

The table below highlights seven fundamental quantities along with their unit and symbols.

Physical Quantity	Dimensional Formula	Unit (SI)	Symbol
Mass	[M]	Kilogram	Kg
Length	[L]	Metre	m
Time	[T]	Second	s
Temperature	[K]	Kelvin	K
Electric Current	[I]	Ampere	A
Intensity of Light	[cd]	Candela	cd
Amount of Substance	[mol]	Mole	mol

Significant Figures Video Lecture

Dimensions of Physical Quantity

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Dimension of Physical Quantity is referred to as a physical quantity which is derived from the fundamental quantity represented by some power raised to these fundamental physical quantities.

If A is the derived physical quantity represented by the dimensional formula M^aL^bT^c then the power (exponents) a, b, and c are called the dimensions of the physical quantity.
Here, [M] is the dimension of mass, [L] is the dimension of length and [T] is the dimension of time.

To better understand, let’s consider some examples.

The area of any surface is the product of its length. Thus, A = length * Breadth
We know, the dimension of length is given by [L]

Therefore, [A] = [L] * [L] = [L²]

[L²] is the dimensional formula of quantity A and we can say quantity
Area has 2 dimensions in length and 0 dimensions in time and mass.
The volume of any surface is the product of its three lengths which is length, breadth, and height.

V = length x breadth x height

Therefore, [V] = [L] * [L] * [L] = [L³]

We have [V] = [L³] = [M⁰L⁰T³]

The dimension of volume is [M⁰L⁰T³] and we say volume has three dimensions in length and zero dimensions in time and mass.

Note: In the representation of the dimensional formula, the magnitude is not considered. Thus, change in any magnitude like velocity, average velocity, initial velocity, and final velocity is similar in context. All these quantities can be represented as length/ time and dimension is [L]/ [T] or [LT^-1].

Dimensional Formula and their Derivation

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Dimensional Formula of any physical quantity is an expression that shows which base (fundamental) quantities are used and how they are combined to form one unit of the physical quantity. For example, Dimensional Force is calculated as F=[MLT⁻²] as the unit of Force is N(newton) or kg×m/s².
As discussed above, we have seven fundamental quantities. The dimensional Formula of fundamental quantities is:

Physical Quantity	Dimension Formula
Length	[L]
Mass	[M]
Time	[T]
Temperature	[K]

Using these fundamental dimensional formulas, we derive the dimensional formula of any physical quantity.

Read more:

Relevant Concepts
Measurement of Length	Measurement of Mass	Unit of Velocity
Unit of Acceleration	Unit of Momentum	Unit of Length
Unit of Time	Accuracy and Precision	Micrometer

Derivation of Dimensional Formula

To derive the dimensional formula of any quantity, we first need to calculate its formula. If we know the formula, we will simplify the formula in the basic quantities (like mass, length, time, etc.).

Let’s consider an example

Dimensional Formula of Density

Density of a substance is defined as the quantity’s mass per unit volume. It is defined as the ratio of mass over volume.

Thus, Density = Mass/ Volume
The dimension formula of mass is [M]
Volume is the product of its three lengths. Thus, Volume = length x breadth x height
The dimensional formula of length is [L]

Therefore, [V] = [L] * [L] * [L] = [L³]

So, [Density] = [Mass]/ [Volume]

[Density] = [M]/ [L³]

[ρ] = [ML^-3]

Thus, [ML^-3] is the dimensional formula of density.
Here, time raised to zero represents that the density does not have a dimension of time or simply we can say, density is time-dimensional less quantity.

Dimensional formula of Velocity

Velocity is defined as rate of change of object position with respect to time. In other, velocity is the ratio of distance traveled per unit of time.

So, Velocity = Distance/ Time

Distance is the length whose dimensional formula is [L]. The dimensional formula of time is [T].

[Velocity] = [Distance]/ [Time]

[Velocity] = [L]/ [T]

[v] = [L]/ [T] or [LT^-1] or [M⁰LT^-1]

Here, velocity has zero dimension in mass, one dimension in length, and -1 dimension in time.

Dimensional Formula of Some Physical Quantity

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The below table showcases the Dimensional Formula of some Physical Quantity which are as follows:

Physical Quantity	General Formula	Units	Dimensional Formula
Angle	arc/ radius	rad	M⁰L⁰T⁰
Acceleration/ acceleration due to gravity (g)	Change in velocity/ time	ms^-2	LT^-2
Angular Displacement	s = rθ	rad	M⁰I⁰T⁰
Angular Momentum (L)	L = I x ω	kgm²s^-1	ML²T^-1
Area (A)	A = l x b	m²	M⁰L²T⁰
Boltzmann’s Constant	k	JK^-1	ML²T^-2K^-1
Coefficient of thermal conductivity	Kappa	Wm^-1K^-1	MLT^-3K^-1
Density (ρ)	ρ = mass/ volume	kg m^-3	ML^-3
Displacement/ Wavelength	d / λ	m	L
Electric Capacitance	C = Q/ V	CV^-1	M^-2L^-2T⁴I²
Electric Conductivity	1/ resistivity	siemens/ metre or Sm^-1	M^-1L^-3T³I²
Electric Current	I = V/ R	Ampere	I
Electric Field/ Electric Field Strength	E = charge x voltage	Vm^-1, NC^-1	MLT^-3I^-1
Energy	E = Power x Time	Joule	ML²T^-2
Force	F = ma	Newton (N)	MLT^-2
Frequency	f = 1/Time Period	Hz	T^-1
Heat	ΔQ = C/ mT	J or calorie	ML²T^-2
Impulse	I = F x t	Kgs^-1 or Ns	MLT^-1
Gravitational Constant	F = G. (m₁m₂/r₂)	Nkg^-2/m^-2	M^-1L³T^-2
Latent Heat (L)	L = Q/ M	JKg^-1	M⁰L²T^-2
Magnetic Dipole Moment	T = m x B	Am²	L²I
Modulus of Elasticity (γ )	γ = stress/ strain	Pa, Nm^-2	ML^-1T^-2
Moment of Inertia (I)	I = mass x radius²	Kgm²	ML²
Permeability of free space	μ_o = 4πFd²/ m₁m₂	NA^-2, Hm^-1	MLT^-2I^-2
The permittivity of free space	ε_o = Q₁Q₂/ 4πFd²	Fm^-1 or C²N^-1m^-2	M^-1L^-3T4I²
Power	P= work/time	Watt or Js^-1	ML²T^-3
Pressure	Pr = Force/ Area	Pa, Nm^-2	ML^-1T^-2
Refractive Index	n = c / v	no specific unit	M⁰L⁰T⁰
Specific Conductance/ Conductivity	1/ specific conductance	siemens/metre or Sm^-1	M^-1L^-3T³I²
Specific Heat	mass x specific heat x temperature	Jkg^-1θ^-1	M⁰L²T^-2K^-1
Specific Volume	1/ density	m³kg^-1	M^-1L³
Speed	distance/ time	ms^-1	LT^-1
Stress	restoring force/ area	Pa, Nm^-2	ML^-1T^-2
Temperature	K = C + 273.15 or C = K − 273.15	^oC or K	K
Torque/ Moment of Force	Force x Distance	Nm	ML²T^-2
Velocity	Displacement/ Time	ms^-1	LT^-1
Velocity Gradient	dv/ dx	s^-1	T^-1
Volume	v = l x b x h	m³	L³
Work	Force x Displacement	J	ML²T^-2

Read More: Matrix addition

Dimensional Equation

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The equation obtained by any physical quantity with its dimensional formula is known as the dimensional equation of that physical quantity. The dimensional equation represents the dimensions of a physical quantity in terms of fundamental quantities.

Thus, dimensional equations of force [F], energy [E], density [ρ], and volume [v] are given below:

[F] = [MLT^-2]

[E] = [ML²T^-2]

[ρ] = [ML^-3T⁰]

[v] = [M⁰L³T⁰]

The dimensional equations can be determined by the equations representing the relationship between required physical quantities.

Read more: Associative law

Dimensional Analysis and its Applications

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Dimensional Analysis is defined as the study of the concept of dimensions, which helps in understanding the physical behaviour of a quantity. For instance, two physical quantities can be added or subtracted only if they have dimensions.

It cannot add or subtract physical quantities having different dimensions.
Dimensional Analysis helps us check the derivation, deducing a relationship between different physical quantities, dimensional consistency, accuracy, and homogeneity of dimensional expression.
When magnitudes of two (or more) physical quantities are multiplied, then their units should be treated as an ordinary algebraic symbol.
Like, we can cancel out similar units in the numerator and denominator of the expression.
The same is followed for dimensions of the physical quantity.
Thus, mathematical equations in which symbols on both sides represent physical quantities must have the same dimensions.

Checking Dimensional Consistency of Equations

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As discussed, magnitudes of two or more quantities can be added or subtracted only if they have dimensions. In simple words, one can add or subtract only similar physical quantities. For example, velocity cannot be added to the energy.

The principle of adding or subtracting similar quantities is called the principle of homogeneity of dimensions.
Principle: In the mathematical equation, if the dimensions on both sides are not the same, then the equation is incorrect.
For example, if one derives an equation for length, then regardless of the symbols appearing in the equation, when all the dimensions of each term are simplified, it must be equivalent to the dimension of length that is [L].

To check the equation's dimensional consistency (homogeneity), the equation must satisfy these points.

The dimensional formula on both sides of an equation must be the same.
The dimensional formula must be the same on both sides of a plus sign (+) and a minus sign (-).
Any constant in the equation will be considered a dimensionless quantity.

Let us check the dimensional consistency of the equation.

s = ut + at²

In the equation, s ⇒ displacement of an object

u ⇒ initial velocity of an object

a ⇒ acceleration

t ⇒ time taken by the object

The dimensions of each term are,

[s] = [L]

[u] = [LT^-1]

[t] = [T]

[a] = [LT^-2]

Individual dimensions in the equation can be written as:

[L] = [LT^-1] * [T] + [LT^-2] * [T²]

[L] = [L] + [L]

As seen, all the terms on the right side of the equation have the same dimension, that is length, which is the same as the dimension of the left side, hence the equation is dimensionally consistent (valid).

Deducing Relationship between Physical Quantities

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Relationships between physical quantities can be derived using dimensional analysis. The basic requirement for the analysis is we should know the dependence of the required physical quantity on other quantities. Consider this as a product of dependence.

Let’s take an example to better understand this concept.

The period (T) of a simple pendulum depends on the mass of the bob (m), length of string (l), and acceleration due to gravity (g). We have to derive an expression of the period using the dimension homogeneity method.

Then, the dependence of the period of a pendulum on m, l, and g can be written as:

T = km^x l^y g^z

where, k is a dimensionless quantity, and x, y, and z are the powers (dimensions).

From the principle of homogeneity of the dimensional equation, we know dimensions of both sides of the equation are the same then,

[M⁰L⁰T¹] = [M¹] ^x [L¹]^y [L¹T^-2] ^z

= M^x L^y + zT – 2z

On comparing the dimensions of both sides, we have

x = 0; y + z = 0; – 2z = 1

So, z = – 1/2, y = ½

Then, substituting these values in the equation

T = kl¹/2g-1/2

Or simply can rewrite the expression as;

T = k√l/g

Note: The value of constant k cannot be determined by this dimensional method. But whatever the value of k is, it does not affect the dimension of the expression.

Here, k = 2π Hence, T = 2π√l/g

Things to Remember

Dimension of Physical Quantity is defined as a relationship between derived quantity with respect to fundamental quantities.
They are divided into seven fundamental categories, namely mass, length, time, temperature, current, intensity of light, and amount of substance.
Using these quantities, we can derive any physical quantity and its dimensions.
Dimensional Formula is a combined expression of fundamental units involved in deriving the physical quantity.
The dimensional equation is an expression that equates physical quantity with its dimensional Formula.
Dimensional Analysis states that only two or more physical quantities can be added or subtracted if they have the exact dimensions.

Sample Questions

Ques. Check whether Energy Density and Young’s Modulus have the same dimensions (5 marks)

Ans. Given,

Energy Density = Energy/ Volume = Work Done/ Volume

Energy Density = Force x Displacement/ Volume

Thus, the dimension of energy is given by,

[Energy Density] = [M] x [LT^-2] x [L]/ [L³]

⇒[Energy Density] = [ML^-1T^-2]

Now, for Young’s Modulus,

Young’s Modulus = Linear Stress/ Linear Strain

⇒ Strain is a ratio, hence it does not have any dimensions.

Linear Stress = Force/ Area = Mass x Acceleration/ Area

Hence, the dimensions of stress can be written as,

⇒ [Linear Stress] = [M] x [LT^-2]/ [L²]

⇒ [Linear Stress] = [ML^-1T^-2]

Therefore, Dimensions of Energy Density and Young’s Modulus are the same.

Ques 2. Find the dimension of the following(s). (4 marks)
(a) Kinetic Energy
(b) Efficiency

Ans. To determine the dimension of the following, we must know the dependency of these quantities on fundamental quantities.

Kinetic Energy

Given,

Kinetic Energy = ½ (mass) x (velocity)²

Therefore, the dimension of kinetic energy can be written as,

⇒ [Kinetic Energy] = [M] x [LT^-1]²

⇒ [Kinetic Energy] = [M] x [L²T^-2]

⇒ [Kinetic Energy] = [ML²T^-2]

Note Here: ½ is constant and hence considered a dimensionless quantity.

Efficiency

Known, efficiency is defined as the ratio of output value over input value of the same quantity like power.

∴ Efficiency = Output Power/ Input Power

Since the dimensions of output and input power are the same, they cancel out each other.

∴ Efficiency is a dimensionless quantity.

Ques 3. If velocity (V), force (F), and time (T) are considered fundamental units, then find the dimension of mass. (5 marks)

Ans. To determine the dimension of mass, we have to deduce the relationship between mass and these fundamental quantities.

We know,

F = m*a …(i)

Here, F is a force which is considered a fundamental quantity, m is mass and a is acceleration which is not a fundamental quantity. Therefore, simplify the acceleration as;

Acceleration = Velocity/ time

Now, rearrange the eq (i)

F = m (V/T)

m = F*T/ v

⇒ [m] = [F] x [T]/ [V]

If force (F), Velocity (V), and time (T) are considered fundamental quantities then the dimension of mass is given by;

[m] = [FTV^-1]

Ques 4. The energy E is given by the expression E = at + bt² where t is the time. Find the dimensions of ‘a’ and ‘b’. (10 marks)

Ans. The energy is given by the expression E = at + bt²

We can derive the dimension of energy using its relationship with mass, acceleration, and displacement.

Energy = Work Done

⇒ Work Done = Force x Displacement

⇒ Work Done = Mass x Acceleration x Displacement (F = ma)

In this formula, the unit of mass is kg, the unit of acceleration is meter per second-square, and the unit of displacement is meter. Hence, use a dimensional formula of these units in the formula.

E = [M] [LT^-2] [L]

E = [ML²T^-2] …(i)

The above is the dimensional formula of energy. The expression of energy is E = at + bt²

According to the homogeneity principle of dimensions, the dimensional formula of both sides of the equation must be the same.

Therefore, [at] = [E]

⇒ a [T] = [E]

⇒ a [T] = [ML²T^-2]

Simplifying the above equation,

⇒ a = [ML²T^-2] x [T^-1]

⇒ a = [ML²T^-3]

Hence, the dimension of ‘a’ is a = [ML²T^-3].

Similarly, for the term bt²

Equating the dimensional formula of energy with the expression considering only the ‘b’ term.

[bt²] = [E]

b [T²] = [E]

⇒ b [T²] = [ML²T^-2]

Simplifying the above equation,

⇒ b = [ML²T^-2] x [T^-2]

⇒ b = [ML²T^-4]

Hence, the dimension of the term b is b = [ML²T^-4].

Ques 5. The speed of a wave traveling in a string (v), depends on the tension (T) in the string and mass density (μ). Find the expression of velocity relating to tension and mass density. (5 marks)

Ans. As given in the question, the velocity depends on the tension in the string and mass density. We don't have the power to be raised to Tension and mass density. So, let’s assume the power is ‘x’ and ‘y’ respectively. Then,

v = kTxμy

k is the dimensionless quantity.

From the principle of consistency (homogeneity), we know dimensions on both sides of the equation are the same. Thus,

[v] = [T]x [μ]y

[T] = [MLT^-2]

[μ] = [Mass]/ [Length] = [ML^-1]

Thus, [LT^-1] = [MLT^-2]x [ML^-1]y

⇒ [LT^-1] = [Mx+y Lx-y T^-2x]

Now, compare each dimension on the sides,

x + y = 0;

x - y = 1

-2x = -1

x = ½ and y = -½

Substituting these values in the expression, we have

v = k T1/2 μ-1/2

v = k√t/μ

Hence, the expression of the velocity of a string is v = k√t/μ.

Ques 6. Check the correctness of equation v² = u² + 2as using dimensional analysis. (5 marks)

Ans. Any equation is said to be dimensionally correct if the dimension of both sides' expressions is the same.

Here,

u ⇒ initial velocity

v ⇒ final velocity

s ⇒ displacement

a ⇒ acceleration

The dimension of LHS is

v² = [MLT^-1]²

⇒ [M⁰L²T^-2]

The dimension of R.H.S is given by

u² + 2as = [M⁰LT^-1]² + 2 [M⁰LT^-2][M⁰LT⁰]

⇒ [M⁰L²T^-2] + 2[M⁰L²T^-2]

Here, the dimension on both sides of the (+) sign can be added. Thus,

⇒ 3 [M⁰L²T^-2]

⇒ [M⁰L²T^-2] (Because 3, is a dimensionless quantity)

Thus, the dimension of L.H.S is the same as the dimension of R.H.S of the equation.

Therefore, The equation v² = u² + 2as is correct.

Ques 7. Convert 5 ergs into joules by using dimensional analysis. (5 marks)

Ans. Here, ‘ergs’ is the unit of energy in the CGS system and ‘joule’ in the SI system.

The dimension formula of energy is [ML²T^-2]

In the CGS system	In the SI system
n₁ = 5	n₂ =?
M₁ = 1g	M₂ = 1Kg
L₁ = 1 cm	L₂ = 1m
T₁ = 1 sec	T₂ = 1 sec

Given,

n₂ = n₁ [M₁/M₂]a [L₁/L₂]b [T₁/T₂]c

where a = 1, b = 2, c = -2; from the dimension of energy [ML²T^-2]

Substitute the values in the expression,

n₂ = 5 [1g/1kg] [1 cm/1 m]² [1 sec/1 sec]^-2

n₂ = 5 [1g/1000g]1 [1 cm/100 cm]² [1]^-2

⇒ n₂ = 5 x (1/1000) x (1/10000)

⇒ n₂ = 5 x 10^-7

Known,

n₁u₁ = n₂u₂

Therefore, 5 ergs = 5 x 10^-7 Joules

Ques 8. In the Van der Waals equation, (P + a/V²)(V - b) = RT, find the dimensions of constants ‘a’ and ‘b’. (5 marks)

Ans. Van der Waals equation is

(P + a/V²)(V - b) = RT

Since the equation is correct it follows the rule of dimension consistency i.e.,

the Dimensional formula of a/V²= Dimensional formula of ‘P’

Or,

Dimensional formula of ‘a’ = Dimensional formula of P x V²

Dimensional formula of ‘a’ = Dimensional formula of (F/A) x V²

Thus,

[a] = [MLT^-2]/ [L²] x [L³]²

Therefore, [a] = [ML⁵T^-2]

Similarly,

Dimension formula of ‘b’ = Dimension formula of ‘V’

[b] = [L³]

Therefore, [b] = [M⁰L⁰T³]

Ques 9. A student derives a formula of escape velocity as √(R/2GM). Check if the formula is correct using dimensional analysis. (3 marks)

Ans. The formula of escape velocity is given as,

Ve = √(R/2GM) (R - Radius, G- Gravitational Constant, M - Mass)

Dimension formula of L.H.S of the formula is,

Ve = [LT^-1] …(1)

The dimensional formula of R.H.S formula is,

√(R/2GM) = √[L]/ ([M^-1L³T^-2] [M]) {√1/2 is dimensionless quantity}

⇒√(R/2GM) = √([T²]/ [L²])

⇒ [L^-1T]

Here, the dimensional formula of L.H.S ≠ Dimensional formula of R.H.S

Therefore, The given formula for escape velocity is incorrect.

Ques 10. A force ‘F’ depends on time ‘t’ and displacement ‘x’, given by the expression F = AcosBx + CsinDt. Find the dimension of B/D. (3 marks)

Ans. The force is given by,

F = AcosBx + CsinDt

Known, an angle is a dimensionless quantity i.e. [M⁰L⁰T⁰].

So, if the given equation is correct.

Dimensional formula of Bx = Dimensional formula of Dt

Or,

Dimension Formula of B/D = Dimension formula of t/x

⇒ [B/D] = [L]/ [T]

⇒ [M⁰L¹T^-1]

Ques 11. What are the limitations of dimensional analysis? (3 marks)

Ans. The following are the limitations of dimensional analysis

If dimensions are given, the physical quantity may not be unique as many physical quantities have the same dimensions.
Numerical constants have no dimensions such as (1/2), 1, or 2π, etc. cannot be deduced by the methods of dimensions.
The method of dimensions can not be used to derive relations other than the product of power functions.
The method of dimensions cannot be applied to derive the formula if in mechanics a physical quantity depends on more than 3 physical quantities as then there will be less number (= 3) of equations than the unknowns (>3).
Even if a physical quantity depends on 3 physical quantities, out of which two have the same dimensions, the formula cannot be derived from the theory of dimensions.

Ques 12. Which one of the following is a dimensionless quantity? (2 marks)
(a) Weight
(b) Mass
(c) Reynold’s number
(d) Specific weight

Ans. The correct answer is c. Reynolds number

Explanation: Reynold's number is a number with no dimensions. It has the formula Re = ρvD/µ. When the related dimensional equation is solved, it comes out that Reynold's number is dimensionless.

Ques 13. In the formula X = 5YZ², X and Z have dimensions of capacitance and magnetic field respectively. What are the dimensions of Y in SI units? (3 marks)

Ans. The given formula is

X = 5YZ²

⇒ Y = X/5Z²

Therefore

Dimensions of Y = Dimensions of X/(Dimensions of Z)²

⇒ [Y] = [X]/[Z]²

But dimensions of X, [X] = [Capacitance] = [A² M^-1 L^-2 T⁴]

Also, dimensions of Z, [Z] = [Magnetic field] = [M A^-1 T^-2]

On substituting, we get

Dimensions of Y, [Y] = [A² M^-1 L^-2 T^-1]/[M A^-1 T^-2]²

⇒ [Y] = [M^-3 L^-2 T⁸ A⁴]

Ques 14. The dimension whose unit does not depend on any other dimension’s unit is known as (2 marks)
(a) Dependent dimension
(b) Fundamental dimension
(c) Independent dimension
(d) Absolute dimension

Ans. The correct answer is b. Fundamental dimension

Explanation: A fundamental dimension is one whose units are independent of all other dimensions. The SI system has seven fundamental dimensions from which all others are derived.

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Dimensions of Physical Quantities: Formula, Equations & Applications

Related Links

Fundamental Quantities and Units

Significant Figures Video Lecture

Dimensions of Physical Quantity

Dimensional Formula and their Derivation

Derivation of Dimensional Formula

Dimensional Formula of Density

Dimensional formula of Velocity

Dimensional Formula of Some Physical Quantity

Dimensional Equation

Dimensional Analysis and its Applications

Checking Dimensional Consistency of Equations

Deducing Relationship between Physical Quantities

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.
If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

2.
Write any two features of nuclear forces.

3.
Suppose a pure Si crystal has \( 5 \times 10^{28} \) atoms per \( \text{m}^3 \). It is doped with \( 5 \times 10^{22} \) atoms per \( \text{m}^3 \) of Arsenic. Calculate majority and minority carrier concentration in the doped silicon. (Given: \( n_i = 1.5 \times 10^{16} \, \text{m}^{-3} \))

4.
A tank is filled with a liquid to a height of \( 12.5 \, \text{m} \). The apparent depth of a needle lying at the bottom of the tank is measured to be \( 9.0 \, \text{m} \). Calculate the speed of light in the liquid.

5.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

6.
Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

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CBSE CLASS XII Related Questions

1. If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

2. Write any two features of nuclear forces.

3.Suppose a pure Si crystal has \( 5 \times 10^{28} \) atoms per \( \text{m}^3 \). It is doped with \( 5 \times 10^{22} \) atoms per \( \text{m}^3 \) of Arsenic. Calculate majority and minority carrier concentration in the doped silicon. (Given: \( n_i = 1.5 \times 10^{16} \, \text{m}^{-3} \))

4.A tank is filled with a liquid to a height of \( 12.5 \, \text{m} \). The apparent depth of a needle lying at the bottom of the tank is measured to be \( 9.0 \, \text{m} \). Calculate the speed of light in the liquid.

5.Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

6.Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

Similar Physics Concepts

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1.
If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

2.
Write any two features of nuclear forces.

3.
Suppose a pure Si crystal has \( 5 \times 10^{28} \) atoms per \( \text{m}^3 \). It is doped with \( 5 \times 10^{22} \) atoms per \( \text{m}^3 \) of Arsenic. Calculate majority and minority carrier concentration in the doped silicon. (Given: \( n_i = 1.5 \times 10^{16} \, \text{m}^{-3} \))

4.
A tank is filled with a liquid to a height of \( 12.5 \, \text{m} \). The apparent depth of a needle lying at the bottom of the tank is measured to be \( 9.0 \, \text{m} \). Calculate the speed of light in the liquid.

5.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

6.
Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]