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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 = F_avgt
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 s^-1 or Hertz(Hz)	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 R)

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 $\displaystyle [b]=[V]$

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

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 (e^bt), 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 e^bt, 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

mass (m)
velocity (v)
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)m^a v^b R^c

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 $v$ in a stretched string depends on tension ( $T)$ 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 $v in terms of T, A & ρ using method of dimensions.$

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)T^a A^b ρ^c

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” power of velocity (V), “b” power of force (F) and “c” power of time (t)

Then m = (some constant) V^aF^bT^c
Equating dimensions of both the sides
[M¹L⁰T⁰]=[L¹T⁻¹]^a[M¹L¹T⁻²]^b[T]^c
[M¹L⁰T⁰] = [M^bL^a+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) V^aF^bT^c

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

[ML²T⁻²] = [M^bL^a+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 “p” power of velocity of light (c), “q” power of Planck’s constant (h) and “r” power of gravitational constant (G)

Then m = (some constant) c^ph^qG^r

Equating dimensions of both the sides
[M¹L⁰T⁰] = [LT⁻¹]^p[ML²T⁻¹]^q[M^-1L³T^-2]^r

[M¹L⁰T⁰] = [M^q-rL^p+2q+3rT^-p-q-2r]

⇒ q-r = 1, p+2q+3r = 0 and -p-q-2r = 0

On solving we get p = 1/2 ; q = 1/2 ; r = -1/2

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

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

Then l = (some constant) c^ph^qG^r

Equating dimensions of both the sides
[M⁰L¹T⁰] = [LT⁻¹]^p[ML²T⁻¹]^q[M^-1L³T^-2]^r

[M⁰L¹T⁰] = [M^q-rL^p+2q+3rT^-p-q-2r]

⇒ q-r = 0, p+2q+3r = 1 and -p-q-2r = 0

On solving we get p = -3/2 ; q = 1/2 ; r = 1/2

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

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

Then t = (some constant) c^ph^qG^r

Equating dimensions of both the sides
[M⁰L⁰T¹] = [LT⁻¹]^p[ML²T⁻¹]^q[M^-1L³T^-2]^r

[M⁰L⁰T¹] = [M^q-rL^p+2q+3rT^-p-q-2r]

⇒ q-r = 0, p+2q+3r = 0 and -p-q-2r = 1

On solving we get p = -5/2 ; q = 1/2 ; r = 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 [M^aL^bT^c]

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

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

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