# Physical quantities

**Physical Quantities:** The quantities which can be compared and measured are physical quantities. Eg, Time, weight

**Measurement: **The comparison of unknown quantity with known quantity is measurement.

__Types of Physical Quantities__

**a. Fundamental Quantities: **Those quantities which do not depend on other quantities (independent) are Fundamental quantities. They are: Mass, length, time, temperature, current, luminous intensity and amount of substance . ** Fundamental unit:** The unit that is used to measure fundamental quantity is fundamental unit,

**b. Derived Quantities:** Those quantities which depend on other quantities are Derived quantities. Eg, speed, area, volume, density acceleration, force etc. ** Derived unit:** The unit that is used to measure derived quantity is derived unit.

__System of units:__

- FPS: Foot Pound Second
- CGS: Centimeter Gram Second
- MKS: Meter Kilogram Second
- SI: System International de Unit

__Dimension, Dimensional formula and Dimensional equation.__

** Dimension:** Dimension of physical quantity is defined as a power raised to fundamental quantities involved in a physical quantity. All physical quantity in mechanics can be expressed in terms a mass, length and time represents as [M], [L] and [T] respectively. So, they are called mechanical quantities. For example: The dimension of velocity is find out as follows

^{3}] = [M

^{0}L

^{-3}T

^{0}]

Density has dimension of 1 in mass, -3 in length and 0 in time.

[Velocity] = [Displacement/Time] = [L^{1}/T

^{1}] = [M

^{0}L

^{1}T

^{-1}]

Velocity has dimension of 0 in mass, 1 in length and -1 in time.

**Q1. Find the dimension of following physical quantities. (with conclusion)**

Area, Volume, Density, Acceleration

- [Area] = [Length*Breadth] = [L*L] = [L
^{2}] = [M^{0}L^{2}T^{0}]

Thus, Area has dimension of 0 in mass, 2 in length and 0 in time.

2. [Volume] = [Length*Breadth*Height] = [L*L*L] = [L^{3}] = [M^{0}L^{3}T^{0}]

Thus, Area has dimension of 0 in mass, 3 in length and 0 in time.

3. [Density] = [Mass/Volume] = [M/L^{3}] = [M^{0}L^{-3}T^{0}]

Thus, Density has dimension of 1 in mass, -3 in length and 0 in time.

4. [Acceleration] = [velocity/time] = [M^{0}L^{1}T^{-1}/T^{1}] = [M^{0}L^{1}T^{-2}]

Thus, Acceleration has dimension of 0 in mass, 1 in length and -2 in time.

** Dimensional Formula:** An expression of a physical quantity in terms of its fundamental quantities along with their dimensions is called dimensional formula It is written by using square bracket i.e For eg the dimensional formula a velocity is [M

^{0}L

^{1}T

^{-1}]

** Dimensional Equation:** The equation obtained by equating a physical quantity with its dimensional formula is called dimensional equation.

For eg, The dimensional equation a velocity is given as, [velocity] = [M^{0}L^{1}T^{-1}]

__Q. Find the dimension of the following:__

**Force:**[Force] = [Mass x Acceleration] = [M] x [M^{0}L^{1}T^{-2}] = [M^{1}L^{1}T^{-2}] . Thus, Force has dimension of 1 in mass, 1 in length and -2 in time.**Work:**[Work] = [Force x displacement] = [F] x [d] = [M^{1}L^{1}T^{-2}] x [L] = [M^{1}L^{2}T^{-2}] . Thus, Work has dimension of 1 in mass 2 in length and -2 in time.**Power:**[Power] = [Work/Time] = [M^{1}L^{2}T^{-2}]/[T] = [M^{1}L^{2}T^{-3}] . Thus, Power has dimension of 1 in mass, 2 in length and -3 in time.**Pressure:**[Pressure] = [Force/Area] = [M^{1}L^{1}T^{-2}]/[L^{2}] = [M^{1}L^{-1}T^{-2}] . Thus, Pressure has dimension of 1 in mass, -1 length and -2 in time.**Gravitational constant:**[Gravitational constant] = [Force x distance^{2}]/[Mass^{2}] = [M^{1}L^{1}T^{-2}x L^{2}]/[M^{2}] = [M^{-1}L^{3}T^{-2}] . Thus, Gravitational constant has dimension of -1 in mass, 3 in length and -2 in time.**Specific Heat capacity:**[Specific Heat capacity] = [Heat/Mass x change in temp.] = [M^{1}L^{2}T^{-2}]/[M x K] = [M^{0}L^{2}T^{-2}K^{-1}] .Thus, Specific Heat Capacity has dimension 0 in mass, 2 in length, -2 in time and -1 in temperature.**Potential Difference:**[Potential Difference] = [Power/current] = [M^{1}L^{2}T^{-3}/I] = [M^{1}L^{2}T^{-3}I^{-1}] . Thus, P.D. has dimension of 1 in mass, 2 in length, -3 in time and -1 in current.**Resistance:**[Resistance] = [Potential Difference/Current] = [M^{1}L^{2}T^{-3}I^{-1}/I] = [M^{1}L^{2}T^{-3}I^{-2}] . Thus, Resistance has dimension of 1 in mass, 2 in length, -2 in time and -2 in current.**Charge:**[charge ]= [Current x time] = [IxT] = [M^{0}L^{0}T^{1}I^{1}] . Thus, Charge has dimension of ) in mass, 0 in length, 1 in current and 1 in time.**Impulse:**[Impulse] = [Force x Time] = [M^{1}L^{1}T^{-2 }x T] = [M^{1}L^{1}T^{-1}] . Thus, impulse has dimension of 1 in mass, 1 in length and -1 in time.**Frequency:**[Frequency] = [1/Time Period] = 1/[T] = [T^{-1}] = [M^{0}L^{0}T^{-1}] . Thus, Frequency has dimension of 0 in mass, 0 in length and -1 in time.

** Principle of Homogeneity: **According to this principle the dimensions fundamental quantities on the left hand side of an equation must be equal to the dimensions a fundamental quantities on the right hand side of that equation.

Let X = Y+Z be an equation.

According to the principle o homogeneity.

D[X] = D[Y] = D[Z]

__Application (uses) of Dimensional Equation__

**1. To check the correctness of a physical relation. (V.Imp)**

For eg Check the correctness of given physical relation.

**a) T = 2**π�** (l/g) ^{1/2}**

Where, T = Time period of simple pendulum, l = length of pendulum and g = acceleration due to gravity.

Given, T = 2π� (l/g)^{1/2} ———(i)

The dimensional equation of each quantity are

[T] = [M^{0}L

^{0}T

^{1}], [L] = [M

^{0}L

^{1}T

^{0}], [g] = [M

^{0}L

^{1}T

^{-2}]

Writing eq (i) in dimensional form, we get,

[M^{0}L

^{0}T

^{1}] = [M

^{0}L

^{0}T

^{1}/M

^{0}L

^{1}T

^{-2}]

^{1/2}[M

^{0}L

^{0}T

^{1}] = [M

^{0}L

^{0}T

^{1}]

Here, 2π� is a dimensionless quantity. So, LHS = RHS. Hence, the given physical relation is dimensionally correct.

**b) F = mv ^{2}/r**

Given, F = mv^{2}/r ————(i)

The dimensional equation of each quantity are

[F] = [M^{1}L

^{1}T

^{-2}], [M] = [M

^{1}], [V] = [M

^{0}L

^{1}T

^{-1}], r = [L]

Writing eq (i) in dimensional form we get

[M^{1}L

^{1}T

^{-2}] = [M

^{1}L

^{2}T

^{-2}/L

^{1}] [M

^{1}L

^{1}T

^{-2}] = [M

^{1}L

^{1}T

^{-2}]

Here, LHS = RHS, the given physical relation is dimensionally correct.

**c) V = (2GM/R) ^{1/2}**

Given, V = (2GM/R)^{1/2 }———(i)

The dimensional equation of each physical quantity are

[V] = [M^{0}L

^{1}T

^{-1}], [G] = [M

^{-1}L

^{3}T

^{-2}], [M] = [M

^{1}L

^{0}T

^{0}], [R] = [L]

Writing eq (i) in dimensional form, we get,

[M^{0}L

^{1}T

^{-1}] = [M

^{1}L

^{3}T

^{-2}/M

^{0}L

^{1}T

^{0}]

^{1/2 }= [M

^{1}L

^{2}T

^{-2}]

^{1/2}[M

^{0}L

^{1}T

^{-1}] = [M

^{0}L

^{1}T

^{-1}]

Hence, 2 is a dimensionless quantity, LHS = RHS, Hence, the given physical relation is dimensionally correct.

**d) V = (R/2GM) ^{1/2}**

Given, V = (R/2GM)^{1/2 }———(i)

The dimension a each physical quantity is

[V] = [M^{0}L

^{1}T

^{-1}], [G] = [M

^{-1}L

^{3}T

^{-2}], [M] = [M

^{1}L

^{0}T

^{0}], [R] = [L]

Writing eq (i) in dimensional form, we get,

[M^{0}L

^{1}T

^{-1}] = [L/M

^{-1}L

^{3}T

^{-2 }xM

^{1}L

^{0}T

^{0}]

^{1/2 }= [L/M

^{0}L

^{3}T

^{-2 }]

^{1/2 }= [M

^{0}L

^{-2}T

^{-1 }]

^{1/2}

Hence, 2 is a dimensionless quantity, LHS is not equal to RHS. Hence the given physical quantity relation isn’t dimensionally correct.

**e)v² = u²+2as**

Given, v² = u²+2as ————(i)

The dimensional equation of each physical quantity are:

[V] = [M^{0}L

^{1}T

^{-1}], [U] = [M

^{0}L

^{1}T

^{-1}], [a] = [M

^{0}L

^{1}T

^{-2}], [S] = [M

^{0}L

^{1}T

^{-1}]

Writing eq (i) in dimensional form we get,

[M^{0}L

^{1}T

^{-1}]

^{2 }= [M

^{0}L

^{1}T

^{-1}]

^{2 }+ [M

^{0}L

^{1}T

^{-2 }x M

^{0}L

^{1}T

^{-1}] [M

^{0}L

^{2}T

^{-2}] = [M

^{0}L

^{2}T

^{-2}]+ [M

^{0}L

^{2}T

^{-2}] [M

^{0}L

^{2}T

^{-2}] = [M

^{0}L

^{2}T

^{-2}]

Hence, 2 is a dimensionless quantity, LHS=RHS, Hence, the given physical relation is dimensionally correct.

**Q1) A students writes an expression of the force that helps a body to move in a circular motion as F=mv ^{2}. Check the correctness of given formula using dimensional method analysis. Find the relationship between various physical quantities. F= mv^{2}/r**

A force (f) causing a body to move in circular motion depends on its mass (m), velocity(v) and radius(r) of the path. Find an expression of the force using dimensional method.

Here, f=force, m=mass, v= velocity, r= radius

According to question.

Fααm^{a}

Fααv^{b}

Fααr^{c}

Combining we get,

F αα m^{a}v^{b}r^{c}

F = k m^{a}v^{b}r^{c }————(i)

where k is dimensionless constant, The dimensional equation of each quantity in eq (i) is

[F] = [M^{1}L

^{1}T

^{-2}], [V] = [M

^{0}L

^{1}T

^{-1}], [m] = [M

^{1}L

^{0}T

^{0}], [r] = [M

^{0}L

^{1}T

^{0}]

Thus, dimensional form of eq (i) is

[M^{1}L

^{1}T

^{-2}] = [M

^{1}L

^{0}T

^{0}]

^{a}[M

^{0}L

^{1}T

^{-1}]

^{b}[M

^{0}L

^{1}T

^{0}]

^{c}[M

^{1}L

^{1}T

^{-2}] = [M

^{a}L

^{0}T

^{0}] [M

^{0}L

^{b}T

^{-b}] [M

^{0}L

^{c}T

^{0}] [M

^{1}L

^{1}T

^{-2}] = = [M

^{a}L

^{b+c}T

^{-b}]

Comparing dimensions on both sides, we get

a=1, -b = -2 ,

b+c =1

2+c=1 or, c=-1

Putting the values of a, b and c in equation (i) we get,

F= k m^{1}v^{2}r^{-1}

F = kmv^{2}/r

Experimentally value of k is found to be 1.

F = mv^{2}/r

This is required relation.

**Q2) The time period of bob of a simple pendulum depends on the length of pendulum and acceleration due to gravity. Find the expression for the time period by dimensional method. hint: T=2**π�**√(l/g)**

Here,

T=Time, l=Length, g=acceleration due to gravity

According to question,

T αα l^{a}

T αα g^{b}

Combining we get,

T αα l^{a}g^{b}

T = kl^{a}g^{b}————(i)

Where , k is dimensional constant,

The dimensional equation of each quantity in eq(i) is

[T] = [M^{0}L

^{0}T

^{1}], [L] = [M

^{0}L

^{1}T

^{0}]

^{a}, [G] = [M

^{0}L

^{1}T

^{-2}]

^{b}

The dimensional form an eq (i) is

[T] = [L][G] [M^{0}L

^{0}T

^{1}] = [M

^{0}L

^{1}T

^{0}]

^{a }[M

^{0}L

^{1}T

^{-2}]

^{b}[M

^{0}L

^{0}T

^{1}] = [M

^{0}L

^{a}T

^{0}]

^{ }[M

^{0}L

^{b}T

^{-2b}] [M

^{0}L

^{0}T

^{1}] = [M

^{0}L

^{a+b}T

^{-2b}]

Comparing dimensions on both sides, we get,

a+b=0 or, a=1/2

1=-2b or, b =-1/2

Putting the Value a and b in eq(i), we get,

T = l^{1/2}g^{-1/2}

Experimentally, k is found to be 2π�

T = 2π�**√**(l/g)

**Q3) The viscous force depends on the coefficient of viscosity (n), radius of sphere and terminal Velocity. Find the expression for the viscous force given that the value of dimensionless constant is 6p. (hint: F=6**π�ηη**rv)**

Here,

n = coefficient of viscosity, F=force, r=radius, v=terminal velocity

According to question,

Fααηη^{a}

Fααr^{b}

Fααv^{c}

Combining we get,

F αα ηη^{a}r^{b}v^{c}

F = k ηη^{a}r^{b}v^{c}————(i)

where, k is dimensionless quantity.

The dimensional equation of each quantity in eq(i) is

[F] = [M^{1}L

^{1}T

^{-2}], [ηη] = [M

^{1}L

^{-1}T

^{-1}], [R] = [M

^{0}L

^{1}T

^{0}], [V] = [M

^{0}L

^{1}T

^{-1}]

The dimensional form of eq(i) is

[M^{1}L

^{1}T

^{-2}] = [M

^{1}L

^{-1}T

^{-1}]

^{a}[M

^{0}L

^{1}T

^{0}]

^{b}[M

^{0}L

^{1}T

^{-1}]

^{c}[M

^{1}L

^{1}T

^{-2}] = [M

^{a}L

^{-a}T

^{-a}][M

^{0}L

^{b}T

^{0}][M

^{0}L

^{c}T

^{-c}] [M

^{1}L

^{1}T

^{-2}] = [M

^{a}L

^{-a+b+c}T

^{-a-c}]

Comparing dimensions on both sides, we get

a = 1, -a+b+c=1 , -a-c=-2 or, c=1

-1+b+1=1 or, b=1

Putting the value of a, b and c in eq(i) we get,

F = k ηη^{a}r^{b}v^{c}

F = k ηηrv

F = 6π�ηηrv

__To find the dimensions of Constants in a Physical Equation.__

For eg: the displacement travelled by a body is expressed as S = A+Bt+Ct^{2} . Find the dimensions of A, B and C.

Given equations is S = A+Bt+Ct^{2} ———(i) where A,B,C are constants

The dimensional equation of S and t are

[S] = [M^{0}L

^{1}T

^{0}] and [t] = [M

^{0}L

^{0}T

^{1}]

According to the principle of homogeneity. We can write

D[S]= D[A]
[M^{0}L^{1}T^{0}] = [A]

Therefore, [A] = [M^{0}L^{1}T^{0}]

Again,

D[S]= D[Bt]
[M^{0}L^{1}T^{0}] = [B][M^{0}L^{0}T^{1}]
[B] = [M^{0}L^{1}T^{-1}]

And,

D[S]= D[Ct^{2}]
[M^{0}L^{1}T^{0}] = [C][M^{0}L^{0}T^{1}]^{2}

^{0}L

^{1}T

^{0}] / [M

^{0}L

^{0}T

^{2}] [C] = [M

^{0}L

^{1}T

^{-2}]

Hence, dimensions of A = [M^{0}L^{1}T^{0}], B = [M^{0}L^{1}T^{-1}] and [C] = [M^{0}L^{1}T^{-2}]

**Q) If a particle travels distance x in time t sec given by x = a + bt + ct ^{2}+ dt^{2} then find the dimensions of a, b, c and d.**

Given equation is x= a + bt + ct^{2}+ dt^{2 }————(i)

The dimensional equation of x and t are

[x] = [M^{0}L

^{1}T

^{1}] and [t] = [M

^{0}L

^{0}T

^{1}]

According to the principle of homogeneity, we can write ,

D[x] = D[a] = D[bt] = D[ct^{2}] = D[ct^{2}]

D[x] = D[a]
[M^{0}L^{1}T^{1}] = [a]

Therefore, [a] = [M^{0}L^{1}T^{1}]

Again,

D[x] = D[bt]
[M^{0}L^{1}T^{1}] = [b] [M^{0}L^{0}T^{1}]
[b] = [M^{0}L^{1}T^{1}] / [M^{0}L^{0}T^{1}]
[b] = [M^{0}L^{1}T^{0}]

Again,

D[x] = D[ct^{2}]
[M^{0}L^{1}T^{1}] = [c] [M^{0}L^{0}T^{1}]^{2}

^{0}L

^{1}T

^{1}] / [M

^{0}L

^{0}T

^{2}] [c] = [M

^{0}L

^{1}T

^{1}] / [M

^{0}L

^{0}T

^{2}] [c] = [M

^{0}L

^{1}T

^{-1}]

Again,

D[x] = D[dt^{2}]
[M^{0}L^{1}T^{1}] = [d] [M^{0}L^{0}T^{1}]^{2}

^{0}L

^{1}T

^{1}] / [M

^{0}L

^{0}T

^{2}] [d] = [M

^{0}L

^{1}T

^{1}] / [M

^{0}L

^{0}T

^{2}] [d] = [M

^{0}L

^{1}T

^{-1}]

Hence, [a] = [M^{0}L^{1}T^{1}], [b] = [M^{0}L^{1}T^{0}], [c] = [M^{0}L^{1}T^{-1}] and [d] = [M^{0}L^{1}T^{-1}]

**Q) To convert the value of physical quantity from one system of unit to another.**

^{0}L

^{1}T

^{0}] m , [b] = [M

^{0}L

^{1}T

^{-1}] m/s , [c] = [M

^{0}L

^{1}T

^{-2}] m/s

^{2}and [d] = [M

^{0}L

^{1}T

^{-3}] m/s

^{3}

Let us consider a physical quantity x which expressed in terms of its magnitude (m) and unit (u) given as,

x = nu

In the first system of unit

x = n_{1}u_{1}

x = n_{1} [M_{1}^{a}L_{1}^{b}T_{1}^{c}]

In the second system of unit

x = n_{2}u_{2}

x = n_{2} [M_{2}^{a}L_{2}^{b}T_{2}^{c}]

Hence, M_{1}L_{1}T_{1}^{ }and M_{2}L_{2}T_{2} are the units of mass, length and time respectively in this systems of unit. Also a, b, and c are the dimension of mass length and time respectively.

Now we can write

n_{1}u_{1 }= n_{2}u_{2}

or, n_{1} [M_{1}^{a}L_{1}^{b}T_{1}^{c}] = n_{2} [M_{2}^{a}L_{2}^{b}T_{2}^{c}]

or, n_{2} = n_{1} [M_{1}/M_{2}]^{a }[L_{1}/L_{2}]^{b }[T_{1}/T_{2}]^{c}

Using this relations we can convert one system unit to another.

**Q) Convert 1 dyne into Newton**

Force = 1 dyne = 1 [M_{1}L_{1}T_{1}^{-2}] = 1 [gcms^{-2}]

Force = n_{2} Newton = n_{2} [M_{2}L_{2}T_{2}^{-2}] = n_{2} [kgms^{-2}]

n_{1}u_{1 }= n_{2}u_{2}

n_{2} = n_{1} [M_{1}/M_{2}]^{a }[L_{1}/L_{2}]^{b }[T_{1}/T_{2}]^{c}

n_{2} = 1 [g/kg] [cm/m] [s/s]^{-2}

n_{2} = [g/1000g] [cm/100cm] [1/1]^{-2}

n_{2} = [1/100000]

n_{2} = [1×10^{-5}] N

Therefore, 1 dyne = [1×10^{-5}]N

**Q) Convert 1N Into dyne.**

1N = n_{2} dyne

As we know, Newton and dyne are the units of force in CGS and SI system respectively and the dimensional formula of force is [MLT^{-2}]

CGS system n_{1 }= 1, M_{1 }= 1kg, L_{1 }= 1m, T_{1 }= 1s

SI system n_{2 }= ?, M_{2 }= 1g, L_{2 }= 1cm, T_{2 }= 1s

And dimensions of mass, length and time are a=1, b=1 and c = -2 respectively,

According to conversion formula, we have

n_{2} = n_{1} [M_{1}/M_{2}]^{a }[L_{1}/L_{2}]^{b }[T_{1}/T_{2}]^{c}

n_{2} = 1 [1kg/1g]^{1 }[1m/1cm]^{1 }[1s/1s]^{–} n_{2} = [100000/1]

n_{2} = 1 [1000g/1g]^{1 }[100cm/1cm]^{1 }[1s/1s]^{-2}

n_{2} = [100000/1]

n_{2} = 100000 dyne = 10^{5}dyne

Therefore, 1N = 10^{5}dyne .

**Q) Convert 10 ergs into Joule.**

Let, 10 ergs = n_{2} Joule

As we know, ergs and Joule are the units of energy in CGS and SI system respectively and the dimensional formula of energy is [ML^{2}T^{-2}]

CGS system n_{1 }= 10, M_{1 }= 1g, L_{1 }= 1cm, T_{1 }= 1s

SI system n_{2 }= ?, M_{2 }= 1kg, L_{2 }= 1m, T_{2 }= 1s

And dimensions of mass, length and time are a=1, b=2 and c = -2 respectively,

According to conversion formula, we have

n_{2} = n_{1} [M_{1}/M_{2}]^{a }[L_{1}/L_{2}]^{b }[T_{1}/T_{2}]^{c}

n_{2} = 10 [1g/1kg]^{1 }[1cm/1m]^{2 }[1s/1s]^{-2}

n_{2} = 10 [1g/1000g]^{1 }[1cm/100cm]^{2 }[1s/1s]^{-2}

n_{2} = [10/10000000]

n_{2} = 10^{-6 }Joule

Therefore, 10 ergs = 10^{-6 }Joule

__Practice questions: [Dimensional analysis]__

**Q1) The density of lead is 11.3 g/cm ^{3}. What is its value in kg/m^{3}?**

let 11.3 g/cm^{3} = n_{2} kg/m^{3}

As we know, g/cm^{3} and kg/m^{3} are the units of density in CGS and SI system respectively and the dimensional formula of density is [M^{1}L^{-3}T^{0}]

CGS system n_{1 }= 11.3, M_{1 }= 1g, L_{1 }= 1cm, T_{1 }= 1s

SI system n_{2 }= ?, M_{2 }= 1kg, L_{2 }= 1m, T_{2 }= 1s

And dimensions of mass, length and time are a=1, b=-3 and c=0 respectively,

According to conversion formula, we have

n_{2} = n_{1} [M_{1}/M_{2}]^{a }[L_{1}/L_{2}]^{b }[T_{1}/T_{2}]^{c}

n_{2} = 11.3 [1g/1kg]^{1 }[1cm/1m]^{-3 }[1s/1s]^{0}

n_{2} = 11.3 [1g/1000g]^{1 }[1cm/100cm]^{-3 }[1s/1s]^{0}

n_{2} = 11300 kg/m^{3}

Therefore, 11.3 g/cm^{3} = 11300 kg/m^{3}

**Q2) Convert a velocity 72 km/hr into m/s with the help of dimensional analysis.**

let 72 km/hr = n_{2} m/s

As we know, km/hr and m/s are the units of velocity in SI and MKS system respectively and the dimensional formula of velocity is [M^{0}L^{1}T^{-1}]

CGS system n_{1 }= 72, M_{1 }= 1kg, L_{1 }= 1km, T_{1 }= 1hr

SI system n_{2 }= ?, M_{2 }= 1g, L_{2 }= 1m, T_{2 }= 1s

And dimensions of mass, length and time are a=1, b=1 and c=-1 respectively,

According to conversion formula, we have

n_{2} = n_{1} [M_{1}/M_{2}]^{a }[L_{1}/L_{2}]^{b }[T_{1}/T_{2}]^{c}

n_{2} = 72 [1kg/1g]^{0 }[1km/1m]^{1 }[1hr/1s]^{-1}

n_{2} = 72 [1000g/1g]^{0 }[1000m/1m]^{1 }[60x60s/1s]^{-1}

n_{2} = 72 x 1000 x [1/3600] = 20 m/s

Therefore, 72 km/hr= 20 m/s

**Q3) The value of Planck’s constant (h) is 6.62×10 ^{-34 }kgm^{2}s^{-2}. Write down its value in CGS unit.**

let 6.62×10^{-34 }kgm^{2}s^{-2} = n_{2} erg (gmcm^{2})

As we know, kgm^{2}s^{-2} and erg are the units of work in SI and CGS system respectively and the dimensional formula of velocity is [M^{1}L^{2}T^{-2}]

CGS system n_{1 }= 6.62×10^{-34}, M_{1 }= 1kg, L_{1 }= 1m, T_{1 }= 1s

SI system n_{2 }= ?, M_{2 }= 1g, L_{2 }= 1cm, T_{2 }= 1s

And dimensions of mass, length and time are a=1, b=2 and c=-2 respectively,

According to conversion formula, we have

n_{2} = n_{1} [M_{1}/M_{2}]^{a }[L_{1}/L_{2}]^{b }[T_{1}/T_{2}]^{c}

n_{2} = 6.62×10^{-34} [1kg/1g]^{1 }[1m/1cm]^{2 }[1s/1s]^{-2}

n_{2} = 6.62×10^{-34} [1000g/1g]^{1 }[100cm/1cm]^{2 }[1s/1s]^{-2}

n_{2} = 6.62×10^{-34}x10^{7}

n_{2} = 6.62×10^{-27}

Therefore, 6.62×10^{-34 }kgm^{2}s^{-2 }= 6.62×10^{-27 }erg

**Limitation of Dimensional Analysis:**

- It does not give information about dimensionless constant (quantity).
- If a physical quantity depends on more than three factors having dimension, the formula cannat be found by using this method.
- We cannot derive the formula containing trigonometric functions, exponential functions, etc
- If a physical relation consists of more than one part, the relation cannot be derived by this method.
- It does not tell whether a physical quantity is scalar or vector.

**Precision And Accuracy**

The degree to which observed values are least scattered is called precision. It shows how close observed values are. The measurement became precise if the instrument used has smaller value of least count. The degree to which observed value approaches the true value is called accuracy. The observed value is called accurate value if it is very close to true value.

**Significant Figures (Digits)**

The meaningful digits in a given number are called significant figures. The number of meaningful digits in a given number is called number of significant figures.

__Rules of finding number of significant figures.__

- All the non-zero digits in the given number are significant. for eg: 329(3), 78912(5) etc.
- All the zeros between two non-zero digits are significant figures. For eg: 1003(4), 502(3) etc
- All zeros after non-zero digit are not significant for eg 2000(1), 1500(2) etc
- All zeros to the right of a decimal point are significant if they are not followed by non-zero digits. For eg: 13.00(4), 5.0 (2) etc All zeros to the right of a non-zero digit after the decimal point are significant. For eg, 0.0760 (3), 0.004300(4) etc
- All zeros to the right of a decimal point but to the left of a non-zero digit are not significant For eg: 0.00356(3), 0.052(2) etc

**Percentage Error: **The percentage error can be defined by the following formula,

Percentage error = [(Standard value – observed value) / standard value] x 100%

Percentage error = [least count / standard value] x 100%

__Some important Questions:__

**Q1. Is dimensionally correct equation necessary to be a correct physical relation? What about dimensionally wrong equation?**

No, Dimensionally correct equations is not necessarily be a physical relation. For eg, v = u + at is dimensional correct but not correct physical relation. This is due to the fact that the constant numerical quantities can’t be expressed dimensionally. The dimensionally wrong equations cannot be correct physically. For eg, s = u + at is dimensionally wrong equation and it is incorrect physical relation as well.

**Q2. Check the correctness of formula PV=RT where P = Pressure, V=Volume, R= gas constant and T = temperature.**

Given, PV=RT

The dimensional equation of each quantity are

[P] = [M^{1}L

^{-1}T

^{-2}], [V] = [M

^{0}L

^{3}T

^{0}], [R] = [M

^{1}L

^{2}T

^{-2}K

^{-1}] and [T] = [M

^{0}L

^{0}T

^{0}K

^{1}]

Writing eq(i) in dimensional form,

[M^{1}L

^{-1}T

^{-2}][M

^{0}L

^{3}T

^{0}] = [M

^{1}L

^{2}T

^{-2}K

^{-1}][M

^{0}L

^{0}T

^{0}K

^{1}] [M

^{1}L

^{2}T

^{-2}] = [M

^{1}L

^{2}T

^{-2}]

Since, LHS = RHS, the given physical relation is dimensionally correct.

**Q3. A student writes an expression for the momentum (p) of a body of mass (m) with total energy (E) and considering the duration of time (t) as p=(2mE/t) ^{1/2} Check the correctness on dimensional analysis.**

Given, p=(2mE/t)^{1/2 }————(i)

Where, p = Momentum, m = mass, E = Energy, t = Time

The dimensional equation of each quantity are

[p] = [M^{1}L

^{1}T

^{-1}], [m] = [M

^{1}L

^{0}T

^{0}], [E] = [M

^{1}L

^{2}T

^{-2}] and [t] = [M

^{0}L

^{0}T

^{1}]

Writing eq (i) in dimensional form,

[M^{1}L

^{1}T

^{-1}] = [M

^{2}L

^{2}T

^{-2}/ M

^{0}L

^{0}T

^{-1}]

^{1/2}[M

^{1}L

^{1}T

^{-1}] = [M

^{1}L

^{1}T

^{-3/2}]

Since, 2 is a dimensionless quantity, LHS ≠ RHS, the given physical relation is not dimensionally correct.

**Q4. The length of a rod is exactly 1cm. An observer records the readings as 1.0cm, 1.00cm and 1.000cm, which is the most accurate measurement?**

1.000cm is the most accurate measurement because it consists of more no. of significant figures and it is measured by Instrument which have smaller value of least count. (0.001)

**Q5. What is the difference between accurate and precise measurement?**

The difference between accurate and precise measurements is that in precise measurement, observed values are very near/close but in accurate measurement, observed values are very far.

**Q6. In one of the printed documents the unit of universal gravitational constant is given as a Nmkg ^{-2} check its Correctness from dimensional analysis.**

Given, Universal Constant (G)= Nm/kg^{-2} ————(i)

Dimensional equation of each quantity involved are:

[G] = [M^{-1}L

^{3}T

^{-2}], [N] = [M

^{1}L

^{1}T

^{-2}], [m] = [M

^{0}L

^{1}T

^{0}] and [kg] = [M

^{1}L

^{0}T

^{0}]

Now, replacing values in equation (i), we get,

[M^{-1}L

^{3}T

^{-2}] = [M

^{1}L

^{1}T

^{-2}M

^{0}L

^{1}T

^{0}] / [M

^{1}L

^{0}T

^{0}]

^{2}[M

^{-1}L

^{3}T

^{-2}] = [M

^{1}L

^{2}T

^{-2}] / [M

^{2}L

^{0}T

^{0}] [M

^{-1}L

^{3}T

^{-2}] = [M

^{-1}L

^{2}T

^{-2}]

Since, LHS ≠ RHS the given physical relation is not dimensionally correct.

**Q7. Name any two physical quantities which have the same dimensions. Can a quantity have unit but no dimension?**

Any two physical quantities which have same dimension are work done and Energy. Yes a quantity can have a unit but no dimension. For eg, plane angle have a unit radian but no dimension.

**Q8. The force is given in terms of time (t) and displacement (x) by the equation F = A sin BX+C sin Dt. What is the dimension of D/B? [Angle has no dimensions]**

Given F = A sin BX+C sin Dt

In this equation, Bx and Dt are dimensionless as they are angles but dimensional formula of x is [L] and t is [T]. So, dimensional formula of B is [L^{-1}] and D is [T^{-1}].

D/B = [T^{-1}]/[L^{-1}] = [LT^{-1}]

**Q9. Find the dimensions of Planck’s (**η�**) constant from the given equation **λ�**=**η�**/p where **λ�** is wavelength and p is the momentum of photon.**

Given, equation is λ� = η�/p or, η�=λ�p

Where λ� = Wavelength, p = Momentum and η� = Planck’s constant

H = [M^{0}L^{1}T^{0}] x [M^{1}L^{1}T^{-1}] = [M^{1}L^{2}T^{-1}]

**Q10. Taking force, length and time to be fundamental quantities, find the dimensional formula for density.**

Here, [density] = [mass/volume] = [force/(acceleration x volume)]

The dimensional formula of force, acceleration and volume are,

[Force] = [F] [Acceleration] = [LT^{-2}] [Volume] = [L

^{3}]

Now, [Density] = [F]/[LT^{-2}][L^{3}] = [F]/[L^{4}T^{-2}] = [FL^{-4}T^{2}]

The dimensional formula density is [FL^{-4}T^{2}]

Chapter 2: Vectors

** Scalar quantities:** Those quantities which have only magnitude but no direction are called scalar quantities. Eg. mass, time, distance, speed etc. They can be explained simply by stating their magnitude. They can be added, subtracted, multiplied or divided by using simple rules of algebra.

** Vector quantities:** Those quantities which have both magnitude and direction are called vector quantities. For eg: displacement, velocity, acceleration, momentum etc. They can be added, subtracted or multiplied by using the laws of vector algebra.

__Graphical representation of a vector:__

A vector is represented by a straight line with arrow. The length of line gives magnitude of the vector and the arrow indicates its direction. The starting point of a vector is called origin or initial paint or tail whereas its ending paint is called tip or terminal point or head.

The figure shows a vector represented by straight line with initial point O and terminal point P.

Thus, we can write A →A →= OP−→−OP→ . The magnitude of vector is expressed in the form of modulus as:

Magnitude of vector A →A → = ∣∣∣A→∣∣∣�→= ∣∣∣OP−→−∣∣∣��→= A = OP (Mod of vector A →A →or OP−→−OP→)

__Types of Vector__

**Unit Vector:**The vector having magnitude 1 is called unit vector. It is represented by nˆ�^(n cap). Any vector A →A →can be represented/given as the product of its magnitude and unit vector i.e

A →A →= Anˆ�^

nˆ�^ = A→A�→�

nˆ�^ = A→|A|�→�

Thus, unit vector can be calculated by dividing a vector by its modulus. In a rectangular co-ordinate system iˆ�^, jˆ�^ and kˆ�^ represent three mutually perpendicular vectors along X, Y, Z axis respectively.

**2. Zero or null vector:** The vector which have magnitude zero is called zero or null vector. Its initial point coincides with terminal point. It is represented by O −→O →. The resultant obtained by adding two negative vectors give zero vector i.e. A →A →+ (-A →A →) = O −→O →

**3. Equal vectors: **Two vectors A →A → and B→�→ are said to be equal if they have same magnitude and direction. They can be represented as A →A → = B→�→

**4. Negative vectors:** Two vectors are said to be negative to each other if they have same magnitude but opposite direction. The negative vector of A →A →is –A →A →. Since, →A→→�→and ←B→←�→ Here, A →A →and B→�→ have same magnitude but they are opposite in direction so they are negative vectors to each other.

**5. Collinear vectors:** Two vectors are said to be collinear vectors if they act along the line or parallel lines. They are of two types:

**Parallel vector: **Two collinear vectors are said to be parallel if they act in same direction. For eg, →A→→�→→B→→�→

**Anti Parallel vector:** ** **Two collinear vectors are said to be anti-parallel if they act in opposite direction. For eg, →A→→�→←B→←�→

**6. Coplanar vectors:** Two or more vectors are said to be coplanar. If they lie in same plane. Here, A →A → and B→�→ are coplanar vectors

**7. Position vector:** The vector which represent the position of a point with reference to the origin of the co-ordinate system is called position vector. It is represented by a straight line from origin (O) to that point i.e. OP in the given figure is a position vector.

For every point in a plane, there is a position vector from origin to that point. For eg, OQ−→−OQ→ in figure is also a position vector.

__Addition of Vectors (Composition of Vectors)__

The addition of two or more vectors gives single vector which is called resultant vector. The process of finding resultant vector is called composition of vector. If we are adding two vectors, we follow:

- Triangle law of vectors
- Parallelogram law of vectors. But if we add more than two vectors, we follow
- Polygon law of vectors.

__Triangle Law of Vectors [V.important]__

It states that, “If two vectors are represented in magnitude and direction by two sides triangle taken in same order then the third side of the triangle taken in opposite order represents magnitude and direction of resultant vector.

let two vectors A →A → and B →B →are acting simultaneously on a particle which are represented both in magnitude and direction by two sides OP and PQ of Δ∆OPQ. The angle between them be θθ. The third side OQ of the triangle gives magnitude and the direction of resultant vector R →R → with one of the vectors A →A →

**Magnitude of **R→R→

Let us produce OP to N and draw QN⊥⊥ON

Now, in right angled Δ∆QNO.

OQ^{2} = ON^{2} + QN^{2}

OQ^{2} = (OP+PN)^{2} + QN^{2}

OQ^{2} = OP^{2} + 2.OP.PN + PN^{2} + QN^{2} ——-(i)

from figure, OP=A, PQ=B and OQ=R

In right angled triangle QNP,

Cosθθ = PN/PQ

or, PN = PQ.Cosθθ

or, PN = B.Cosθθ

Also, Sinθθ = QN/PQ

or, QN = PQ.Sinθθ

or, QN = B.Sinθθ

Putting all these values in equation (i), we get

OQ^{2} = OP^{2} + 2.OP.PN + PN^{2} + QN^{2}

R^{2} = A^{2} + 2.A.B.Cosθθ + B^{2}.Cos^{2}θθ + B^{2}.Sin^{2}θθ

R^{2} = A^{2} + 2.A.B.Cosθθ + B^{2}.(Cos^{2}θθ + Sin^{2}θθ)

R^{2} = A^{2} + 2.A.B.Cosθθ + B^{2}.(1)

R^{2} = A^{2} + B^{2 }+ 2.A.B.Cosθθ

R = A2 + B2 + 2.A.B.Cosθ−−−−−−−−−−−−−−−−−−−−√�2 + �2 + 2.�.�.����

This gives the magnitude of resultant vector R →R →

**Direction of **R →R →

The direction of resultant vector R →R →is given by

tanβ� = QNON���� (In Δ∆QNO)

tanβ� = QNOP+PN����+��

tanβ� = BsinθA+BCosθ�sin��+�����

β� = tan−|tan-|(BsinθA+BCosθ�sin��+�����)

This gives the direction of resultant vector R →R →

__Parallelogram Law of Vector Addition (Sure question)__

It states that, “If the Two vectors are represented in both magnitude and direction by the two adjacent sides of parallelogram drawn from a point then their resultant vector is represented both in magnitude and direction by the diagonal of the parallelogram passing through the same point.

Let two vectors A →A → and B→B→ are represented in magnitude and direction by two adjacent sides OP and OS of parallelogram OPQS. The angle between them be ‘θθ‘. Their resultant vector R →R → is represented in magnitude and direction by the diagonal OQ of the parallelogram. Suppose the resultant vector makes an angle ‘β�’ with one of the vectors i.e. A →A →

**Magnitude of **R →R →

let us produce OP to N and draw QN perpendicular to PN.

Now, in right angled Δ∆QNO,

or, OQ² = ON² + QN²

or, OQ² = (OP+PN)² + QN²

or, OQ² = OP²+2.OP.PN+PN² + QN² ———(i)

From figure,

OP = SQ = A, OS = PQ = B and OQ = R

Also, In right angled Δ∆QNP,

cosθθ = PNPQ����

or, PN = PQ.cosθθ

or, PN = B.cosθθ

and, sinθθ = QNPQ����

or, QN = PQ.sinθθ

or, QN = B.sinθθ

putting all these value in equation (i) we get,

R^{2} = A^{2} + 2.A.B.Cosθθ + B^{2}.Cos^{2}θθ + B^{2}.Sin^{2}θθ

R^{2} = A^{2} + 2.A.B.Cosθθ + B^{2}.(Cos^{2}θθ + Sin^{2}θθ)

R^{2} = A^{2} + 2.A.B.Cosθθ + B^{2}.(1)

R^{2} = A^{2} + B^{2 }+ 2.A.B.Cosθθ

R = A2 + B2 + 2.A.B.Cosθ−−−−−−−−−−−−−−−−−−−−√�2 + �2 + 2.�.�.����

This gives the magnitude of resultant vector R →R →

**Direction of **R →R →

The direction of resultant vector R →R → of two vectors A →A → and B→B→ is given by

tanβ� = QNON���� (In Δ∆QNO)

tanβ� = QNOP+PN����+��

tanβ� = BsinθA+BCosθ�sin��+�����

β� = tan−|tan-|(BsinθA+BCosθ�sin��+�����)

This gives the direction of resultant vector R →R →with A →A →.

a) When θθ = 0°° i.e. the two vectors act along the same direction then,

R = A2 + B2 + 2.A.B.Cosθ−−−−−−−−−−−−−−−−−−−−√�2 + �2 + 2.�.�.����

R = A2 + B2 + 2.A.B.Cos0−−−−−−−−−−−−−−−−−−−−−√�2 + �2 + 2.�.�.���0

R = A2 + B2 + 2.A.B−−−−−−−−−−−−−−−−√�2 + �2 + 2.�.�

R = (A + B)2−−−−−−−−√(� + �)2

R = (A+B)

and , β� = tan−|tan-|(BsinθA+BCosθ�sin��+�����)

β� = tan−|tan-|(Bsin0A+BCos0�sin0�+����0)

β� = 0

Thus, when two vectors act along the same direction, then the magnitude of their resultant vector is equal to the sum of magnitude of the two vectors and act along the same direction.

b) When θθ = 90°° i.e. the two vectors act at right angle to each other then,

R = A2 + B2 + 2.A.B.Cosθ−−−−−−−−−−−−−−−−−−−−√�2 + �2 + 2.�.�.����

R = A2 + B2 + 2.A.B.Cos90−−−−−−−−−−−−−−−−−−−−−√�2 + �2 + 2.�.�.���90

R = A2 + B2 −−−−−−−−√�2 + �2

and , β� = tan−|tan-|(BsinθA+BCosθ�sin��+�����)

β� = tan−|tan-|(Bsin90A+BCos90�sin90�+����90)

β� = tan−|tan-|(BA��)

c) When θθ = 180°° i.e. the two vectors act at right angle to each other then,

R = A2 + B2 + 2.A.B.Cosθ−−−−−−−−−−−−−−−−−−−−√�2 + �2 + 2.�.�.����

R = A2 + B2 + 2.A.B.Cos180−−−−−−−−−−−−−−−−−−−−−−√�2 + �2 + 2.�.�.���180

R = A2 + B2 −2.A.B−−−−−−−−−−−−−−−√�2 + �2 -2.�.�

R = (A − B)2−−−−−−−−√(� – �)2

R = ±±(A-B)

and , β� = tan−|tan-|(BsinθA+BCosθ�sin��+�����)

β� = tan−|tan-|(Bsin180A+BCos180�sin180�+����180)

β� = 0°° or 180°°

Thus , when two vectors act along opposite direction then the magnitude of resultant vector is equal to the difference of the magnitude of two vectors and act along the same direction (along A →A → for β� = 0°°) or opposite direction (along B→�→ for β�=180°).

__Polygon law of vector addition__

It states that “if a number of vectors are represented in magnitude and direction by the sides of a polygon taken in same order then their resultant vector is represented in magnitude and direction by the closing side of the polygon taken in opposite order.

Let four vectors A →A →, B→�→, C→C→ and D→D→ are arranged using tail head method to form a polygon as in the figure above. The four vectors A →A →, B→�→, C→C→ and D→D→ are represented in magnitude and direction by sides of OP, PQ & QS and ST of the given polygon taken in same order. Their resultant vector A →A → represented in magnitude and direction by the closing side of OT of the polygon taken in opposite order.

Now using triangle law of vector in Δ∆OPQ

OQ −→−OQ →=OP −→−OP →+PQ −→−PQ →

OQ −→−OQ →=A →A →+B→�→ ————(i)

Again using triangle law of vector in Δ∆OQS.

OS−→��→=OQ −→−OQ →+QS−→��→

OS−→��→=A →A →+B→�→+C→C→ [using equation (i)]

And, using triangle how of vector in OST

OT −→−OT →=OS−→��→+ST−→��→

R→�→=A →A →+B→�→+C→C→+D→D→

__Properties of vector Addition__

a) Vector addition follows commutative law i.e.

A →A →+B→�→+C→C→ = C→C→+A →A →+B→�→ = B→�→+C→C→+A →A →

b) Vector addition follows distributive law i.e.

α�(A →A →+B→�→+C→C→) = α�A →A →+α�B→�→+α�C→C→. where α� is scalar

c) Vector addition follows associative law

(A →A →+B →B →)+C→C→ = A →A →+(B→�→+C→C→)

__Subtraction of vectors__

The subtraction of B vector from A is defined as the addition of -B (negative vector B) with A.

__Resolution of Vectors (Decomposition of Vectors)__

The process of breaking of a vector into two or more vectors is called resolution of vectors. The vectors obtained after resolution are called components of vectors. Generally, a vector is resolved into two components, which are at right angle to each other such resolution of vectors is called rectangular resolution and components are called rectangular components.

__Rectangular Resolution of Vector:__

Let a vector OP−→−��→=A →A →is inclined at an angle θθ with x-axis. Draw PN⊥⊥ON.

In Δ∆ONP, using triangle law a vector, we have,

OP−→−��→ = ON−→−��→ + NP−→−��→

OP−→−��→ = Ax−→��→ + Ay−→��→

Where, ON−→−��→ = Ax−→��→ and NP−→−��→ = Ay−→��→ are the resolved parts of vector A →A → along x-axis and y-axis respectively.

So they are called x-component and y-component of vector A →A → respectively.

Magnitude of Vector A →A →

In right angled Δ∆ONP.

OP^{2} = ON^{2} + NP^{2}

A = ∣∣∣A →∣∣∣A → = A2x+A2y−−−−−−−−−√�2�+�2�

Magnitude of component vectors Ax−→��→ and Ay−→��→ fromΔ∆ONP

cosθθ = ONOP����

cosθθ = AxA���

Ax = A.cosθθ (along x-axis)

and, sinθθ = NPOP����

sinθθ = AyA���

Ay = A.sinθθ (along y-axis)

Rectangular representation of vector

Here, Ax−→��→ = Ax iˆ�^

Ax−→��→= A.cosθθ iˆ�^

Similarly, Ay−→��→ = Ay iˆ�^

Ay−→��→ = A.sinθθ iˆ�^

Now, ∣∣∣A →∣∣∣A → = A2x+A2y−−−−−−−−−√�2�+�2�

or, ∣∣∣A →∣∣∣A → = A2cos2θ+A2sin2θ−−−−−−−−−−−−−−−−√�2cos2�+�2sin2�

or, ∣∣∣A →∣∣∣A → = A2.(cos2θ+sin2θ)−−−−−−−−−−−−−−−√�2.(cos2�+sin2�)

or, ∣∣∣A →∣∣∣A → = A2.(1)−−−−−−√�2.(1)

or, ∣∣∣A →∣∣∣A → = A2−−−√�2

or, ∣∣∣A →∣∣∣A → = A

Thus, vector A →A →can be represented as

A →A →=Ax−→��→+Ay−→��→

A →A →= Axiˆ�^ + Ayjˆ�^

A →A → = A.cosθθiˆ�^ + A.sinθθjˆ�^

In three dimension

A →A →= Ax−→��→+Ay−→��→+Az−→��→

A →A →= iˆ�^Ax + jˆ�^Ay + kˆ�^Az

and, ∣∣∣A →∣∣∣A → = A2x+A2y+A2z−−−−−−−−−−−−−−√�2�+�2�+�2�