Dimensions

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DIMENSIONS

 

 

Dimensions of a physical quantity are the powers (or exponents) to which the base quantities are raised to represent that quantity.

1. Dimensional formula :-

Dimensional formula of any physical quantity is that expression which represents how and which of the base quantities are included in that quantity.
It is written by enclosing the symbols for base quantities with appropriate powers in square brackets i.e. [ ]
Example :
Write dimensional formula of mass and speed.
Solution :
Dimensional formula of mass is [M1L0 T0 ] and that of speed (= distance/time) is [M0L1T–1]

2. Dimensional equation :-

The equation obtained by equating a physical quantity with its dimensional formula is called a dimensional equation.
e.g. [v] = [M0L1T–1]
For example [F] = [MLT–2] is a dimensional equation, [MLT–2] is the dimensional formula of the force and the dimensions of force are 1 in mass, 1 in length and –2 in time.

3. Dimensional Formula of Some Physical Quantities

S. No. Physical Quantity Dimensional Formula SI Unit Formula Used to find Dimensions
1 Area [L2] metre2 Area= length×length
 2 Volume [L3] metre3 Volume=length×length×length
3 Velocity or Speed [LT-1] ms-1 Velocity=displacement/time

Speed = distance/time

 4 Acceleration [LT-2] ms-2 a = Δv/t
 5 Force [MLT-2] newton (N) F=ma
6 Work or energy or Heat [ML2T-2] joule (J) W = F.s

U = mgh

 7 Density [ML-3] kg m-3 ρ = mass/volume
 8 Linear momentum or Impulse [MLT-1] kg ms-1 p = mv

I = Favg t

9 Power [ML2T-3] J s-1 or watt P = W/t
 10 Pressure or stress [ML-1T-2] Nm-2 P = F/A
 11 Strain Dimensionless Unitless Strain=change in dimension/original dimension
12 Modulus of elasticity [ML-1T-2] Nm-2 Modulus of elasticity = Stress/Strain
 13 Surface tension [MT-2] Nm-1 S = Force/length
14 Angular velocity or

Frequency

[T-1] rad/s ω = Δθ/t

ν = 1/T

15 Wavelength [L] m It is simply length
16 Torque [ML2T-2] N-m τ = r × F
 17 Angular acceleration [T-2] rad/s2 α = Δω/t
18 Angular momentum [ML2T-1] kg m2S-1 L = mvr = mωr2
 19 Moment of inertia [ML2] kg m2 I = mr2
20 Angle or Angular displacement [M0L0T0] radian Angle = Length of arc/radius
21 Charge [AT] Coulomb Q = I t
22 Gravitational constant [M-1L3T-2] Nm2/kg2 F = Gm1m2/r2
23 Planck’s constant [ML2T-1] J-s E = hν
24 Universal gas constant [ML2T-2θ-1μ-1] J/mol-K PV = nRT
25 Potential Difference [ML2T-3A-1] Volt V = W/q
26 Resistance [ML2T-3A-2] Ω(V/A) R = V/I
27 Capacitance [M-1L-2T4A2] farad(C/V) C = q/V
28 Inductance [ML2T-2A-2] Henry(Weber/ampere) U = LI2/2

⇒ L = 2U/I2

29 Intensity [ML0T-3] Watt/m2 I = Energy/Area-time
30 Electric Field Intensity [MLT-3A-1] N/C E = F/q
31 Magnetic Field Intensity [ML0T-2A-1] Tesla(N/A-m) F = BIL sinθ

⇒ B = F/ILsinθ

32 Electric permittivity [M-1L-3T4A2] C2N-1m-2 \displaystyle F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}
33 Magnetic Permeability [MLT-2A-2] H/m

or

Wb/Am

or

Nm/A2

\displaystyle B=\frac{{{\mu }_{0}}I}{2\pi R}

(Magnetic field due to a current carrying conductor at a distance d)

4. Applications of dimensional analysis :-

(i)  Principle of Homogeneity :-

For a physical equation to be correct dimensionally, the dimensions of all of its terms should be the same. In other words only those physical quantities can be added or subtracted, whose dimensions are same.

For example in \displaystyle S=ut+\frac{1}{2}a{{t}^{2}}

dimensions of \displaystyle ut = dimensions of  \displaystyle \frac{1}{2}a{{t}^{2}}.

Illustration 1.

In x + αt, if x = distance and t is time then find dimensions of α.

Solution.

\displaystyle [x]=[\alpha t]=[\alpha ][t]

\displaystyle \Rightarrow [L]=[\alpha ][T]

\displaystyle \Rightarrow [\alpha ]=\frac{[L]}{[T]}=[L{{T}^{-1}}]

Illustration 2.

If in x = a + bt + ct2, if x = distance and t is time then find dimension of a, b and c.

Solution.

\displaystyle [x]=[a]=[bt]=[c{{t}^{2}}]

\displaystyle [x]=[a]=[b][t]=[c][{{t}^{2}}]

\displaystyle \Rightarrow [a]=[x]=[L]

Again

\displaystyle [b][t]=[x]=[L]

\displaystyle [b][T]=[L]

\displaystyle [b]=\frac{[L]}{[T]}=[L{{T}^{-1}}]

Finally

\displaystyle [c][{{t}^{2}}]=[x]=[L]

\displaystyle \Rightarrow [c]=\frac{[L]}{[{{t}^{2}}]}=\frac{[L]}{[{{T}^{2}}]}=[L{{T}^{-2}}]

Illustration 3.

In  \displaystyle v=at+\frac{b}{c-{{t}^{2}}}, v is velocity and t is time then find dimensions of a, b and c.

Solution.

\displaystyle [v]=[at]=\left[ \frac{b}{c-{{t}^{2}}} \right]

Using \displaystyle [v]=[at]

we get \displaystyle [L{{T}^{-1}}]=[a][T]

\displaystyle \Rightarrow [a]=\frac{[L{{T}^{-1}}]}{[T]}=[L{{T}^{-2}}]

Now we can also write that \displaystyle [c]=[{{t}^{2}}]

\displaystyle \Rightarrow [c]=[{{T}^{2}}]

Finally using \displaystyle [v]=\left[ \frac{b}{c-{{t}^{2}}} \right]

We get, \displaystyle [L{{T}^{-1}}]=\left[ \frac{b}{{{T}^{2}}} \right]

\displaystyle \Rightarrow [b]=[L{{T}^{-1}}][{{T}^{2}}]=[LT]

Illustration 4.

In \displaystyle \left( p+\frac{a}{{{V}^{2}}} \right)\left( V-b \right)=RT, p = pressure, V = volume, R = Universal gas constant and T = temperature. Find dimensions of a and b.

Solution.

We can write \displaystyle [p]=\left[ \frac{a}{{{V}^{2}}} \right]

\displaystyle \Rightarrow [a]=[p][{{V}^{2}}]

\displaystyle [a]=[M{{L}^{-1}}{{T}^{2}}][{{L}^{6}}]

\displaystyle [a]=[M{{L}^{5}}{{T}^{-2}}]

And \displaystyle [b]=[V]

So

\displaystyle [b]=[{{L}^{3}}]

Illustration 5.

In x = a sin(bt – c), x represents position of a particle at any time t. Find dimensions of a, b and c.

Solution.

Here (bt – c) is angle, and angle is dimensionless, hence \displaystyle [bt-c]=[{{M}^{0}}{{L}^{0}}{{T}^{0}}]

\displaystyle \Rightarrow [c]=[{{M}^{0}}{{L}^{0}}{{T}^{0}}]

And \displaystyle [bt]=[{{M}^{0}}{{L}^{0}}{{T}^{0}}]

\displaystyle \Rightarrow [b]=\frac{[{{M}^{0}}{{L}^{0}}{{T}^{0}}]}{[t]}=\frac{[{{M}^{0}}{{L}^{0}}{{T}^{0}}]}{[T]}

\displaystyle [b]=[{{M}^{0}}{{L}^{0}}{{T}^{-1}}]

Finally sin(bt – c) is dimensionless hence \displaystyle [x]=[a]

\displaystyle \Rightarrow [a]=[L]

Illustration 6.

In y = a (ebt),  y represents position of a particle at any time t. Find dimensions of a and b.

Solution.

The exponent “bt” is dimensionless, hence \displaystyle [bt]=[{{M}^{0}}{{L}^{0}}{{T}^{0}}]

\displaystyle \Rightarrow [b]=\left[ \frac{{{M}^{0}}{{L}^{0}}{{T}^{0}}}{T} \right]=[{{M}^{0}}{{L}^{0}}{{T}^{-1}}]

Also ebt, is just a numerical value, so \displaystyle [a]=[y]

\displaystyle \Rightarrow [a]=[{{M}^{0}}L{{T}^{0}}]

Illustration 7.

In \displaystyle p=\frac{\alpha }{{{t}^{2}}}\exp \left( \frac{-\beta x}{t} \right), p is pressure and x is position. Find the dimensions of α and β.

Solution.

Here \displaystyle p=\frac{\alpha }{{{t}^{2}}}\exp \left( \frac{-\beta x}{t} \right), can be written as \displaystyle p=\frac{\alpha }{{{t}^{2}}}{{e}^{\left( \frac{-\beta x}{t} \right)}}

Now \displaystyle {{e}^{\left( \frac{-\beta x}{t} \right)}}, is dimensionless, so \displaystyle [p]=\left[ \frac{\alpha }{{{t}^{2}}} \right]

\displaystyle \Rightarrow [\alpha ]=[p][{{t}^{2}}]=[M{{L}^{-1}}{{T}^{-2}}][{{T}^{2}}]

\displaystyle \Rightarrow [\alpha ]=[M{{L}^{-1}}{{T}^{0}}]

Now, \displaystyle \left[ \frac{-\beta x}{t} \right]=[{{M}^{0}}{{L}^{0}}{{T}^{0}}]

\displaystyle \Rightarrow [\beta ]=\frac{[{{M}^{0}}{{L}^{0}}{{T}^{0}}][T]}{[L]}=[{{M}^{0}}{{L}^{-1}}T]

Illustration 8.

In \displaystyle I=\alpha {{e}^{\left( \frac{-\beta {{t}^{2}}}{RC} \right)}}, I = electric current, R = resistance, C = capacitance and t = time. Find dimensions of α and β.

Solution.

Here \displaystyle \left[ \frac{-\beta {{t}^{2}}}{RC} \right]=[{{M}^{0}}{{L}^{0}}{{T}^{0}}]

Hence

\displaystyle [\alpha ]=[I]=[A]

Again as \displaystyle \left[ \frac{-\beta {{t}^{2}}}{RC} \right]=[{{M}^{0}}{{L}^{0}}{{T}^{0}}]  \displaystyle \Rightarrow [\beta {{t}^{2}}]=[RC]

\displaystyle \Rightarrow [\beta ][{{T}^{2}}]=[M{{L}^{2}}{{T}^{-3}}{{A}^{-2}}][{{M}^{-1}}{{L}^{-2}}{{T}^{4}}{{A}^{2}}]

\displaystyle [\beta ]=[{{M}^{0}}{{L}^{0}}{{T}^{-1}}]

 

(ii) To check the correctness of a dimensional equation :-

If in a given physical equation if dimensions of L.H.S. = dimensions of R.H.S. , then this type of equation is called dimensionally correct.

Note :- A dimensionally correct equation may or may not be numerically correct. For example in the equation \displaystyle T=\sqrt{\frac{l}{g}}, the dimensions of L.H.S. = [T] and dimensions of R.H.S. \displaystyle =\left[ \sqrt{\frac{l}{g}} \right]=\left[ \sqrt{\frac{L}{L{{T}^{-2}}}} \right]=\left[ \sqrt{{{T}^{2}}} \right]=[T], so here dimensions of L.H.S. = dimensions of R.H.S. but still the equation is wrong, and the correct equation is Time period of a simple pendulum(T) = \displaystyle 2\pi \sqrt{\frac{l}{g}}

Illustration 9.

Check the dimensional correctness of the equation \displaystyle F=\frac{m{{v}^{2}}}{r}, where m = mass, v = velocity and r is radius.

Solution.

Here,  dimensions of L.H.S. \displaystyle =[F]=[ML{{T}^{-2}}]

and dimensions of R.H.S. \displaystyle =\left[ \frac{m{{v}^{2}}}{r} \right]=\frac{[M]{{[L{{T}^{-1}}]}^{2}}}{[L]}=[ML{{T}^{-2}}]

dimensions of L.H.S. = dimensions of R.H.S.

Hence, equation is dimensionally correct.

Illustration 10.

Check the dimensional correctness of the equation \displaystyle g=\frac{4}{3}\pi GR\rho , where g = acceleration due to gravity, G = Universal gravitational constant, R = Radius of Earth and ρ = average density of Earth.

Solution.

Here,  dimensions of L.H.S. = \displaystyle [\rho ]=[L{{T}^{-2}}] and dimensions of R.H.S. = \displaystyle \left[ \frac{4}{3}\pi GR\rho \right]=[G][R][\rho ]=[{{M}^{-1}}{{L}^{3}}{{T}^{-2}}][L][M{{L}^{-3}}]=[L{{T}^{-2}}]

dimensions of L.H.S. = dimensions of R.H.S.

Hence, equation is dimensionally correct.

Illustration 11.

Check the dimensional correctness of the equation \displaystyle \tau =I\times \alpha , where τ = torque, I = moment of inertia and α = angular acceleration.

Solution.

Here,  dimensions of L.H.S. = \displaystyle [\tau ]=[rFsin\theta ]=[r][F]=[L][ML{{T}^{-2}}]=[M{{L}^{2}}{{T}^{-2}}] and dimensions of R.H.S. = \displaystyle [I\times \alpha ]=[I][\alpha ]=[M{{L}^{2}}][{{T}^{-2}}]=[M{{L}^{2}}{{T}^{-2}}]

dimensions of L.H.S. = dimensions of R.H.S.

Hence, equation is dimensionally correct.

Illustration 12.

Check the dimensional correctness of the equation \displaystyle L=m{{\omega }^{2}}{{R}^{2}}, where L = angular momentum, m = mass, ω = angular velocity and R = radius.

Solution.

Here,  dimensions of L.H.S. = \displaystyle [L]=[M{{L}^{2}}{{T}^{-1}}] and dimensions of R.H.S. = \displaystyle [m{{\omega }^{2}}{{R}^{2}}]=[m][{{\omega }^{2}}][{{R}^{2}}]=[M]{{[{{T}^{-1}}]}^{2}}{{[L]}^{2}}=[M{{L}^{2}}{{T}^{-2}}]

Here dimensions of L.H.S. ≠ dimensions of R.H.S.

Hence, equation is dimensionally wrong.

Illustration 13.

Check the dimensional correctness of the equation \displaystyle v=x\sqrt{\frac{3k}{2m}}, where v = velocity, x = position, m = mass and k = spring constant.

Solution.

Here,  dimensions of L.H.S. = [v] = [LT-1] and dimensions of R.H.S. = \displaystyle \left[ x\sqrt{\frac{3k}{2m}} \right]=[x]\left[ \sqrt{\frac{3k}{2m}} \right]=[L][\left[ \sqrt{\frac{M{{T}^{-2}}}{M}} \right]]=[L][{{T}^{-1}}]=[L{{T}^{-1}}]

dimensions of L.H.S. = dimensions of R.H.S.

Hence, equation is dimensionally correct.

 

(iii) To derive the relation between physical quantities :-

Illustration 14.

If force(F) depends depends upon following physical quantities

  1. mass (m)
  2. velocity (v)
  3. radius (R)

Then express force in terms of m,v and R.

Solution.

Let force depends on “a” power of mass(m), “b” power of velocity(v) and “c” power of radius(R)

Then F = (some constant)ma vb Rc

Now let us put the dimensions of quantities on both sides…

\displaystyle [ML{{T}^{-2}}]={{[M]}^{a}}{{[L{{T}^{-1}}]}^{b}}{{[L]}^{c}}

\displaystyle \Rightarrow [ML{{T}^{-2}}]=[{{M}^{a}}{{L}^{b+c}}{{T}^{-b}}]

⇒ a = 1, b + c = 1 and -b = -2

On solving we get

a = 1, c = -1 and b = 2

So finally \displaystyle F=\frac{m{{v}^{2}}}{R}

Illustration 15.

The speed of transverse wave in a stretched string depends on tension ( in the string, mass density of the material of the string (ρ) and cross section area of the string(A). Find the formula of wave

Solution.

Let velocity (v) depends on “a” power of tension (T), “b” power of area (A) and “c” power of density (ρ)

Then v = (some constant)Ta Ab ρc

Now let us put the dimensions of quantities on both sides…

\displaystyle [L{{T}^{-1}}]={{[ML{{T}^{-2}}]}^{a}}{{[{{L}^{2}}]}^{b}}{{[M{{L}^{-3}}]}^{c}}

\displaystyle [L{{T}^{-1}}]=[{{M}^{a+c}}{{L}^{a+2b-3c}}{{T}^{-2a}}]

On solving we get

a = 1/2,  b = -1/2 and c = -1/2

So finally \displaystyle v=\sqrt{\frac{T}{A\rho }}

Illustration 16.

If velocity (V), force (F) and time (T) are chosen as fundamental quantities, express
(i) mass and
(ii) energy
in terms of V, F and T
Solution.
Let mass (m) depends on “a” velocity (V), “b” power force (F) and “c” power of time (t)
Then m = (some constant) VaFbTc
Equating dimensions of both the sides
[M1L0T0]=[L1T−1]a [M1L1T−2]b [T]c
[M1L0T0] = [MbLa+bT−a−2b+c]
On solving we get a = – 1, b = 1, c = 1

So m = (some constant) (V−1F1T1)

Similarly, we can also express energy in terms of V, F, T

Let E = (some constant) VaFbTc

Equating dimensions of both the sides
[ML2T−2] = [LT−1]a[MLT−2]b[T]c

[ML2T−2] = [MbLa+bT−a−2b+c]

⇒ 1 = b; 2 = a + b ; -2 = -a – 2b + c

On solving we get a = 1 ; b = 1 ; c = 1

∴   E = (some constant) V1F1T1

Illustration 17.

If velocity of light c, Planck’s constant h and gravitational constant G are taken as fundamental quantities, then express mass, length and time in terms of dimensions of these quantities.

Solution.

(i) Let mass (m) depends on “a” power of velocity of light (c), “b” power of Planck’s constant (h) and “c” power of gravitational constant (G)
Then m = (some constant) cahbGc
Equating dimensions of both the sides
[M1L0T0] = [LT−1]a [ML2T−1]b [M-1L3T-2]c
[M1L0T0] = [Mb-c La+2b+3c T-a-b-2c]

⇒ b-c = 1, a+2b+3c = 0 and -a-b-2c = 0

On solving we get a = 1/2 ; b = 1/2 ; c = -1/2

∴  \displaystyle m=(some\text{ constant})\sqrt{\frac{hc}{G}}

(ii) Let length (l) depends on “a” power of velocity of light (c), “b” power of Planck’s constant (h) and “c” power of gravitational constant (G)

Then l = (some constant) cahbGc
Equating dimensions of both the sides
[M0L1T0] = [LT−1]a [ML2T−1]b [M-1L3T-2]c
[M0L1T0] = [Mb-c La+2b+3c T-a-b-2c]

⇒ b-c = 0, a+2b+3c = 1 and -a-b-2c = 0

On solving we get a = -3/2 ; b = 1/2 ; c = 1/2

\displaystyle l=(some\text{ constant})\sqrt{\frac{hG}{{{c}^{3}}}}

(iii) Let time (t) depends on “a” power of velocity of light (c), “b” power of Planck’s constant (h) and “c” power of gravitational constant (G)

Then t = (some constant) cahbGc
Equating dimensions of both the sides
[M0L0T1] = [LT−1]a [ML2T−1]b [M-1L3T-2]c
[M0L0T1] = [Mb-c La+2b+3c T-a-b-2c]

⇒ b-c = 0, a+2b+3c = 0 and -a-b-2c = 1

On solving we get a = -5/2 ; b = 1/2 ; c = 1/2

\displaystyle t=(some\text{ constant})\sqrt{\frac{hG}{{{c}^{5}}}}

 

(iv) To convert a physical quantity from one system of units to the other :-

This is a fact that magnitude of a physical quantity remains same whatever system is used for measurement i.e. magnitude = numeric value (n) × unit (u) = constant or n1u1 = n2u2

Now let:-

n1 = numerical value in I system
n2 = numerical value in II system

M1 = unit of mass in I system
M2 = unit of mass in II system

L1 = unit of length in I system
L2 = unit of length in II system

T1 = unit of time in I system
T2 = unit of time in II system

So if a quantity is represented by [MaLbTc]

Then

n1u1 = n2u2

\displaystyle {{n}_{1}}[M_{1}^{a}L_{1}^{b}T_{1}^{c}]={{n}_{2}}[M_{2}^{a}L_{2}^{b}T_{2}^{c}]

\displaystyle \Rightarrow {{n}_{2}}={{n}_{1}}{{\left[ \frac{{{M}_{1}}}{{{M}_{2}}} \right]}^{a}}{{\left[ \frac{{{L}_{1}}}{{{L}_{2}}} \right]}^{b}}{{\left[ \frac{{{T}_{1}}}{{{T}_{2}}} \right]}^{c}}

Illustration 18.

Using dimensions find the value of acceleration due to gravity ‘g’ in MKS system. The numerical value of ‘g’ in CGS system is 980.

Solution.

Here

n1 = 980
n2 = ?

M1 = 1gram
M2 = 1Kg

L1 = 1cm
L2 = 1m

T1 = 1s
T2 = 1s

[g] = [M0LT-2]

⇒ a = 0, b = 1 and c = -2

Using \displaystyle \Rightarrow {{n}_{2}}={{n}_{1}}{{\left[ \frac{{{M}_{1}}}{{{M}_{2}}} \right]}^{a}}{{\left[ \frac{{{L}_{1}}}{{{L}_{2}}} \right]}^{b}}{{\left[ \frac{{{T}_{1}}}{{{T}_{2}}} \right]}^{c}}

\displaystyle \Rightarrow {{n}_{2}}=980{{\left[ \frac{1gram}{1Kg} \right]}^{0}}{{\left[ \frac{1cm}{1m} \right]}^{1}}{{\left[ \frac{1s}{1s} \right]}^{-2}}=980{{\left[ \frac{1gram}{1Kg} \right]}^{0}}{{\left[ \frac{1cm}{100cm} \right]}^{1}}{{\left[ \frac{1s}{1s} \right]}^{-2}}

\displaystyle {{n}_{2}}=980\left[ \frac{1}{100} \right]=9.8

Illustration 19.

Find by dimensional method the value of Y(Young’s Modulus) in SI system, given Y = 20 × 1011 dyne cm-2.

Solution.

Dimensions of Y,  [Y] = [ML-1T-2]

⇒ a = 1, b = -1 and c = -2

n1 = 20 × 1011
n2 = ?

M1 = 1gram
M2 = 1Kg

L1 = 1cm
L2 = 1m

T1 = 1s
T2 = 1s

Using \displaystyle \Rightarrow {{n}_{2}}={{n}_{1}}{{\left[ \frac{{{M}_{1}}}{{{M}_{2}}} \right]}^{a}}{{\left[ \frac{{{L}_{1}}}{{{L}_{2}}} \right]}^{b}}{{\left[ \frac{{{T}_{1}}}{{{T}_{2}}} \right]}^{c}}

We get, \displaystyle \Rightarrow {{n}_{2}}=20\text{ }\times \text{ }{{10}^{11}}{{\left[ \frac{1gram}{1Kg} \right]}^{1}}{{\left[ \frac{1cm}{1m} \right]}^{-1}}{{\left[ \frac{1s}{1s} \right]}^{-2}}=20\text{ }\times \text{ }{{10}^{11}}{{\left[ \frac{1gram}{1000gm} \right]}^{1}}{{\left[ \frac{1cm}{100cm} \right]}^{-1}}{{\left[ \frac{1s}{1s} \right]}^{-2}}

\displaystyle \Rightarrow {{n}_{2}}=20\text{ }\times \text{ }{{10}^{11}}{{\left[ \frac{1}{1000} \right]}^{1}}{{\left[ \frac{1}{100} \right]}^{-1}}=20\text{ }\times \text{ }{{10}^{11}}\left[ \frac{1}{1000} \right]\left[ \frac{100}{1} \right]

\displaystyle {{n}_{2}}=20\text{ }\times \text{ }{{10}^{10}}

So Y = 20 × 1011N/m2

 

 

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Manoj Kumar Verma

Hi, I'm Manoj Kumar Verma, a physics faculty having 7 years of teaching experience. I have done B.Tech (E.E.). I am also a YouTuber and Blogger. This blog is dedicated to help students learn the physics concepts easily.

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