In this article, we shall study the use of principle of homogeneity to derive the form of the physical equation.

Example – 01:

The hydrostatic pressure P of a liquid column depends upon density (d) and height of liquid (h) and acceleration due to gravity (g), using dimensional analysis derive the formula for pressure.

Solution:

let  P  ∝  d^x,   P  ∝  h^y,   P  ∝  g^z,

Combining above relations we have  P  ∝  d^x h^y g^z,

P  = k  d^x h^y g^z …………..  (1)

By principle of homogeneity of dimensions we have

[P]  = [d]^x  [h]^y [g]^z 

∴  [L^-1M¹T^-2]  = [L^-3M¹T⁰]^x  [L¹M⁰T⁰]^y [L¹M⁰T^-2]^z 

∴  [L^-1M¹T^-2]  = [L^-3xM^xT⁰][L^yM⁰T⁰][L^zM⁰T^-2z]

∴  [L^-1M¹T^-2]  = [L^{-3x + y + z} M^xT^-2z]

Considering equality of two sides we have

3x +y +z = -1 , x = 1 , -2z = -2 i.e. z = 1

∴ 3x +1+1 = -1

∴ 3x  = – 3

∴ x  = 1

Substituting x =1, y = 1 and z = 1 in equation (1) we get

P  = k  dh g

The value of constant in this case is 1, i.e. k = 1

P  = h d g

This is the formula for the pressure

Example – 02:

Derive an expression for kinetic energy of a body of mass ‘m’ moving with a velocity ‘v’, using dimensional analysis. Or Using dimensional analysis, show that the kinetic energy of a body of mass m moving with a velocity v varies as mv².

Solution:

Let us represent kinetic energy by a symbol E

let  E  ∝  m^x,   E  ∝  v^y

Combining above relations we have  E  ∝  m^x v^y 

E  =  k m^x v^y   …………..  (1)

By the principle of homogeneity of dimensions we have

[E]  = [m]^x  [v]^y 

∴  [L²M¹T^-2]  = [L⁰M¹T⁰]^x  [L¹M⁰T^-1]^y

∴  [L²M¹T^-2]  = [L⁰M^xT⁰][L^yM⁰T^-y]

∴  [L²M¹T^-2]  = [L^yM^xT^-y]

Considering the equality of two sides we have

y = 2 and x = 1

Substituting x =1 and y = 2 in equation (1) we get

E  =  k m¹ v²

This is the form of the equation for the kinetic energy of a body.

E ∝  m v²

Ans: Thus the kinetic energy of a body varies as mv².

Example – 03: 

The frequency ‘n’ of vibration of a wire under tension depends on tension ‘T’, mass per unit length ‘m’ and vibrating length ‘l’ of wire. Using dimensional analysis, obtain dependence of frequency on these quantities. OR The frequency n of vibration of a string of length l under tension T depends upon l, T and m where m is the mass per unit length of the string. Show that

Solution:

Given  n  ∝  T^x,   n  ∝  m^y and n  ∝  l^z

Combining above relations we have  n  ∝  T^x m^y l^z

n  =  k T^x m^y l^z   …………..  (1)

By the principle of homogeneity of dimensions we have

[n]  =  [T]^x [m]^y [l]^z

∴  [L⁰M⁰T^-1]  = [L¹M¹T^-2]^x [L^-1M¹T⁰]^y  [L¹M⁰T⁰]^z

∴  [L⁰M⁰T^-1]  = [L^xM^xT^-2x][L^-yM^yT⁰][L^zM⁰T⁰]

∴  [L⁰M⁰T^-1]  = [L^{x – y+ z}M^{x + y}T^-2x]

Considering the equality of two sides we have

x – y + z = 0, x + y = 0 and – 2x = -1

∴ x = 1/2

Substituting in x + y = 0

1/2 + y = 0

∴ y = – 1/2

Substituting x = 1/2 and y = -1/2 in x – y + z = 0

∴  1/2 – (-1/2) + z = 0

∴ 1 + z = 0

∴ z = -1

Substituting x =1/2, y = -1/2 and z = – 1 in equation (1) we get

n  =  k T^1/2 m^-1/2 l^{– 1}

n  ∝    T^1/2 m^-1/2 l^{– 1}

n  ∝   (l^{– 1}) (T^1/2 /m^1/2)

n  ∝   (1/l) (T/m)^1/2

Proved as required

Example – 04:

For a small sphere falling through a viscous medium of infinite extent, the force(F) opposing the motion depends directly upon the velocity (v) coefficient of viscosity (η) of the medium and the radius (r) of the sphere. From the dimensional analysis, show that F = 6πηrv. where the constant k = 6π.

Solution:

Given  F  ∝  v^x,   F  ∝ η^y and F  ∝  r^z

Combining above relations we have  F  ∝  v^x η^y r^z

F  =  k v^x η^y r^z   …………..  (1)

By the principle of homogeneity of dimensions we have

[F]  =  [v]^x [η]^y [r]^z

∴  [L¹M¹T^-2]  = [L¹M⁰T^-1]^x [L^-1M¹T^-1]^y  [L¹M⁰T⁰]^z

∴  [L¹M¹T^-2]  = [L^xM⁰T^-x][L^-yM^yT^-y][L^zM⁰T⁰]

∴  [L¹M¹T^-2]  = [L^{x – y + z}M^yT^{-x – y}]

Considering the equality of two sides we have

x – y + z = 1,  y = 1 and x – y = – 2

Substituting y = 1 in x – y = 0

x – 1 = 0

∴ x = 1

Substituting x = 1 and y =1 in x – y + z = 1

∴ 1 – 1 + z = 1

∴ z = 1

Substituting x =1, y = 1 and z = 1 in equation (1) we get

F  =  k v¹ η¹ r ¹

Given k = 6π

∴  F = 6πηrv

Example – 05:

Using dimensional analysis, show that the velocity ‘v’ of a body falling freely under gravity varies with √gh . where ‘g’ is the acceleration due to gravity and ‘h’ is the distance through which the body falls.

Solution:

Given  v  ∝  g^x and   v ∝ h^y 

Combining above relations we have  v  ∝  g^x h^y

v  =  k g^x h^y   …………..  (1)

By the principle of homogeneity of dimensions we have

[v]  =  [g]^x [h]^y

∴  [L¹M⁰T^-1]  = [L¹M⁰T^-2]^x [L¹M⁰T⁰]^y

∴  [L¹M⁰T^-1]  = [L^xM⁰T^-2x][L^yM⁰T⁰]

∴  [L¹M⁰T^-1]  = [L^{x + y} M⁰T^-2x ]

Considering the equality of two sides we have

x + y = 1 and  – 2x = -1 i.e. x = 1/2

Substituting  in x + y = 1

1/2 + y = 1

∴ y =  1/2

Substituting x = 1/2 and y = 1/2  in equation (1) we get

v  =  k g^1/2 h^1/2 

∴ v  =  k (g h)^1/2 

∴ v  =  k √gh 

∴ v  ∝ √gh

Ans: The velocity ‘v’ of a body falling freely under gravity varies with √gh .

Example – 06:

The period T of revolution of the earth around the sun depends upon the orbital radius ‘R’, the mass ‘M’ of the Sun and gravitational constant ‘G’. Show by dimensional analysis that T² ∝ R³.

Solution:

Given  T  ∝  R^x,   T  ∝ M^y and T ∝  G^z

Combining above relations we have  T  ∝  R^x M^y G^z

T  =  k R^x M^y G^z   …………..  (1)

By the principle of homogeneity of dimensions we have

[T]  =  [R]^x [M]^y [G]^z

∴  [L⁰M⁰T¹]  = [L¹M⁰T⁰]^x [L⁰M¹T⁰]^y  [L³M^-1T^-2]^z

∴  [L⁰M⁰T¹]  = [L^xM⁰T⁰][L⁰M^yT⁰][L^3zM^-zT^-2z]

∴  [L⁰M⁰T¹]  = [L^{x + 3z} M^{y – z}T^-2z]

Considering the equality of two sides we have

x + 3z = 0, y – z = 0, -2z = 1 i.e. z = – 1/2,

Substituting z = – 1/2 in y – z = 0

∴ y – (-1/2) = 0

∴ y = -1/2

Substituting z = -1/2 in x + 3z = 0

∴ x + 3(-1/2) = 0

∴ x + – 3/2 = 0

∴ x = 3/2

Substituting x = 3/2, y = -1/2 and z = -1/2 in equation (1) we get

T  =  k R^3/2 M^-1/2 G ^-1/2

Squaring both the sides we get

T²  =  k² R³ M^-1 G ^-1

T²  =  k² (R³/GM)

Now the mass of the Sun M and G are constant

∴ T² ∝ R³

Example – 07:

The velocity v of sound in a gas depends upon the pressure P and density ρ, show by dimensional analysis that v ∝ √P/ρ.

Solution:

Given  v  ∝  P^x and   v ∝ ρ^y 

Combining above relations we have  v  ∝  P^x ρ^y

v  =  k P^x ρ^y   …………..  (1)

By the principle of homogeneity of dimensions we have

[v]  =  [P]^x [ρ]^y

∴  [L¹M⁰T^-1]  = [L^-1M¹T^-2]^x [L^-3M¹T⁰]^y

∴  [L¹M⁰T^-1]  = [L^-xM^xT^-2x][L^-3yM^yT⁰]

∴  [L¹M⁰T^-1]  = [L^{– x – 3y} M^{x + y}T^-2x ]

Considering the equality of two sides we have

x – 3y = 1, x + y = 0 and  – 2x = -1 i.e. x = 1/2

Substituting  in x + y = 0

1/2 + y = 0

∴ y =  -1/2

Substituting x = 1/2 and y = 1/2  in equation (1) we get

v  =  k P^1/2 ρ^-1/2 

∴ v  =  k (P/ρ)^1/2 

∴ v  =  k √P/ρ 

∴ v  ∝ √P/ρ 

Example – 08:

Reynold’s number R which determines the condition for streamline flow of a liquid through a tube depends upon the velocity ‘v’ of the liquid flow, the density ‘ρ’ of liquid, coefficient of viscosity of liquid ‘η’ and diameter ‘d’ of the tube. Using dimensional analysis show that R ∝ vρd/η.

Solution:

Example – 09:

Assuming a period of vibration of a tuning fork depends upon a length l of its prongs. density rho of the material and elastic constant E of the material. Find the formula for the period of vibrations using dimensional analysis?

Solution:

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Science > Physics > Units and Measurements > Form of Physical Equation

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Example – 01:

Check the correctness of physical equation, s = ut +1/2at², Where u is the initial velocity, v is the final velocity, a is the acceleration, s is the displacement and t is the time in which the change occurs.

Solution:

Given equation is s = ut +1/2at²

L.H.S. = v, hence [L.H.S.] = [s] = [L¹M⁰T⁰]  ………………. (1)

R.H.S = ut +1/2at² , hence [R.H.S] = [u][t]+ [t]²

= [L¹M⁰T^-1][L⁰M⁰T¹] + [L¹M⁰T^-2][L⁰M⁰T¹]²

[R.H.S] = [L¹M⁰T⁰] + [L¹M⁰T^-2][L⁰M⁰T²]

[R.H.S] = [L¹M⁰T⁰] +  [L¹M⁰T⁰]

[R.H.S] = [L¹M⁰T⁰]  ……………. (2)

From (1) and (2) we have [L.H.S.] = [R.H.S.]

Hence by the principle of homogeneity, the given equation is dimensionally correct.

Example – 02:

Check the correctness of physical equation, F = mv²/r, Where F is the centripetal force acting on a body of mass m performing uniform circular motion along a circle of radius r with linear speed v.

Solution:

Given equation is F = mv²/r

L.H.S. = F, hence [L.H.S.] = [F] = [L¹M¹T^-2]  ………………. (1)

R.H.S = mv²/r , hence [R.H.S] = [m][v]²/[r]

[R.H.S] = [L⁰M¹T⁰][L¹M⁰T^-1]² / [L¹M⁰T⁰]

[R.H.S] = [L⁰M¹T⁰][L²M⁰T^-2] [L^-1M⁰T⁰]

[R.H.S] = [L¹M¹T^-2]     ……………. (2)

From (1) and (2) we have [L.H.S.] = [R.H.S.]

Hence by the principle of homogeneity, the given equation is dimensionally correct.

Example – 03:

Check the correctness of physical equation, v² = u² + 2as²

Solution:

Given equation is v² = u² + 2as²

L.H.S. = v², hence [L.H.S.] = [v]² = [L¹M⁰T^-1]² = [L²M⁰T^-2] ………………. (1)

R.H.S = u² + 2as²

hence [R.H.S] = [u]² + 2[a][s]²

[R.H.S] = [L¹M⁰T^-1]² + [L¹M⁰T^-2][L¹M⁰T⁰]²

[R.H.S] = [L²M⁰T^-2] + [L¹M⁰T^-2][L²M⁰T⁰]

[R.H.S] = [L²M⁰T^-2] + [L³M⁰T^-2]  ……………. (2)

From (1) and (2) we have [L.H.S.] ≠ [R.H.S.]

Hence by the principle of homogeneity, the given equation is dimensionally not correct.

Example – 04:

Check the correctness of physical equation, when the frequency of vibration ‘n’ of a string of length ‘l’ having mass per unit length ‘m’ kept under tension ‘F’ is given by

Solution:

Given equation is

L.H.S. = n, hence [L.H.S.] = [n] = [L⁰M⁰T^-1]  ………………. (1)

R.H.S = (1/2l) (F/m)^1/2 , hence [R.H.S] = [1/l](F/m)^1/2

[R.H.S] = [l]^-1[F]^1/2[M]^-1/2

[R.H.S] = [L¹M⁰T⁰]^-1[L¹M¹T^-2]^1/2 [L^-1M¹T⁰]^-1/2

[R.H.S] =  [L^-1M⁰T⁰][L^1/2M^1/2T^-1] [L^1/2M^-1/2T⁰]

[R.H.S] = [L^-1+1/2+1/2M^1/2-1/2T^-1]

[R.H.S] = [L⁰M⁰T^-1]     ……………. (2)

From (1) and (2) we have [L.H.S.] = [R.H.S.]

Hence by the principle of homogeneity, the given equation is dimensionally correct.

Example – 05:

Check the homogeneity (correctness of physical equation) of the equation, when the rate of flow of a liquid having a coefficient of viscosity’η’ through a capillary tube of length ‘l’ and radius ‘a’ under pressure head ‘p’ is given by

Solution:

Given equation is

From (1) and (2) we have [L.H.S.] = [R.H.S.]

Hence by the principle of homogeneity, the given equation is dimensionally homogeneous.

Example – 06:

Check the homogeneity (correctness of physical equation)of the equation, when the periodic time ‘T’ of vibration of the magnet of the moment of inertia ‘I’, magnetic moment, ‘M’ vibrating in magnetic induction ‘B’ is given by

Solution:

Given equation is

From (1) and (2) we have [L.H.S.] = [R.H.S.]

Hence by the principle of homogeneity, the given equation is dimensionally homogeneous.

Example – 07:

Check the homogeneity of the equation, when the terminal velocity ‘v’ of a small sphere of radius ‘a’ and density ‘ρ’ falling through a liquid of density ‘σ’ and coefficient of viscosity ‘η’ is given by

Solution:

Given equation is

From (1) and (2) we have [L.H.S.] = [R.H.S.]

Hence by the principle of homogeneity, the given equation is dimensionally homogeneous.

Example – 08:

A force F is given by F = at + bt², where t is the time. Find the dimensional formula of ‘a’ and ‘b’.

Solution:

By the principle of homogeneity the dimensions on either side of the physical equation must be the same. Now two physical quantities can be added or subtracted if and only if their dimensions are the same.

Dimensions of F = Dimensions of at

[L¹M¹T^-2] = [a] [t]

[a] =[L¹M¹T^-2] / [T]

[a]=[L¹M¹T^-3]

Similarly, dimensions of F  = Dimensions of bt²

[L¹M¹T^-2] = [b] [T] ²

[b]= [L¹M¹T^-2] / [T²]

[b]= [L¹M¹T^-4]

Ans: The dimensional formula of ‘a’ and ‘b’ are [M¹L¹T^-3 ] and [M¹L¹T^-4].

Example – 09:

The distance covered by a body in time t is given by the relation S = a + bt + ct². What are the dimension of a,b,c ? Also write the quantities they represent.

Solution: 

By the principle of homogeneity the dimensions on either side of the physical equation must be the same. Now two physical quantities can be added or subtracted if and only if their dimensions are the same.

Dimensions of S = Dimensions of a

[L¹M⁰T⁰] = [a]

Dimensions of ‘a’ are that of displacement. Hence ‘a’ represent the displacement.

Dimensions of S = Dimensions of bt

[L¹M⁰T⁰] = [b] [t]

[b] =[L¹M⁰T⁰] / [T]

[b]=[L¹M⁰T^-1]

Dimensions of ‘b’ are that of velocity. Hence ‘b’ represent the velocity.

Dimensions of S  = Dimensions of ct²

[L¹M⁰T⁰] = [c] [T] ²

[c]= [L¹M⁰T⁰] / [T²]

[b]= [L¹M⁰T^-2]

Dimensions of ‘c’ are that of acceleration. Hence ‘c’ represent the acceleration.

Ans: The dimensional formula of ‘a’, ‘b’ and ‘c’ are [L¹M⁰T⁰],  [L¹M⁰T^-1], and [L¹M⁰T^-2] and the quantities represented by a, b and c are displacement, velocity and acceleration.

Example – 10:

According to Laplace’s formula, the velocity ( V) of sound in a gas is given by v = (γP / d)^1/2, where P is the pressure, d is the density of a gas. What is the dimensional formula for γ?

Solution:

Ans: The dimensional formula for γ is [ M⁰ L⁰ T⁰ ]

Example – 11:

In an equation { P + a / V² } (V – b) = RT, where P is the pressure, V is the volume, T is the temperature and a, b, R are constants. What is the dimensional formula of a /b?

Solution:

Two physical quantities can be added or subtracted if and only if their dimensions are the same.

Thus the dimensions of P and a / V² are same

Dimensions of b = Dimensions of V

[b] = [L³M⁰T⁰]

Now [a]/[b] = [L⁵M¹T^-2]/ [L³M⁰T⁰] = [L²M¹T^-2]

Ans: The dimensional formula of a /b  is [ M¹ L² T^-2 ]

Example – 12:

A force is given in terms of distance x and time t by F = A sin Ct + B cos Dx. Then what are the dimensions of A/B and C/D?

Solution:

By the principle of homogeneity the dimensions on either side of a physical equation must be the same. The trigonometric ratios sin and cos are pure ratios of length, hence are dimensionless quantity.

Thus [A] = [B] =[F] = [ M¹ L¹ T^-2 ]

Thus dimensions of A/B are [ M⁰ L⁰ T⁰ ]

Now the angle is dimensionless quantity

[C][t] = [ M⁰ L⁰ T⁰ ]

[C] = [ M⁰ L⁰ T⁰ ]/ [T] = [ M⁰ L⁰ T^-1 ]

[D][x] = [ M⁰ L⁰ T⁰ ]

[D] = [ M⁰ L⁰ T⁰ ]/ [L] = [ M⁰ L^-1 T⁰ ]

[C]/[D] = [ M⁰ L⁰ T^-1 ]/ [ M⁰ L^-1 T⁰ ] = [ M⁰ L¹ T^-1 ]

Ans: The dimensions of A/B and C/D are [M⁰L⁰T⁰] [M⁰L¹ T^-1]

Example – 13:

In a equation { P + a / V² } ( V – b) = constant , where P is the pressure, V is the volume and a, b are constants. Find the unit of ‘a’.

Solution:

Two physical quantities can be added or subtracted if and only if their units are the same.

Thus the units of P and a / V² are same

Unit of P = Unit of a /(unit of V)²

dyne cm^-2 = unit of a / (cm³) ²

dyne cm^-2 = unit of a / cm⁶

unit of a = dyne cm^-2 x cm⁶ = dyne cm⁴

Ans: The unit of a is dyne cm⁴

Example – 14:

What are the dimensions of constant C in the equation P + ½ρv² + hρg = C? Where P is pressure, ρ is a density, v is the velocity of liquid flow, h is the height and g is the acceleration due to gravity.

Solution:

Two physical quantities can be added or subtracted if and only if their units are the same. Thus the dimension of P, ½ρv², hρg, and C should be that of pressure.

Dimensions of C = Dimensions of P = [M¹L^-1 T^-2]

Ans: The dimensions of C are[M¹L^-1 T^-2]

Example – 15:

The velocity v, the acceleration a and displacement s of a body in motion are related by the expression v² ∝ a^xs^y. Find x and y.

Solution:

Given v² ∝ a^xs^y

∴  v² = k a^xs^y

∴  [M⁰L¹ T^-1]² = [M⁰L¹ T^-2]^x[M⁰L¹ T⁰]^y

∴  [M⁰L² T^-2]² = [M⁰L^x T^-2x] [M⁰L^y T⁰]

∴  [M⁰L² T^-2]² = [M⁰L^{x + y} T^-2x]

∴ x + y = 2 and -2x = -2 i.e. x = 1 and y = 1

Ans: x = 1 and y = 1

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Science > Physics > Units and Measurements > More Problems on Dimensional Analysis

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In this article, we shall study the applications of dimensional analysis.

Applications of Dimensional Analysis:

• To check the correctness of physical equation: 
• To Find Dimensions of New Physical Quantity:
• To derive the form of a physical equation:
• To derive the relation between different units of different systems of a physical quantity:

To check the correctness of physical equation: 

We can check the correctness of the physical equation using the principle of homogeneity. By the principle of homogeneity of dimensions, the dimensions of all the terms on the two sides of an equation must be the same.

Example – 01:

To check the correctness of physical equation, v = u + at, Where ‘u’ is the initial velocity, ‘v’ is the final velocity, ‘a’ is the acceleration and ‘t’ is the time in which the change occurs.

Solution:

Given equation is v = u + at

L.H.S. = v, hence [L.H.S.] = [v] = [L¹M⁰T^-1]  ………………. (1)

R.H.S = u + at, hence [R.H.S] = [u] + [a][t]

[R.H.S] = [L¹M⁰T^-1] + [L¹M⁰T^-2][L⁰M⁰T¹]

[R.H.S] = [L¹M⁰T^-1] +  [L¹M⁰T^-1]

[R.H.S] = [L¹M⁰T^-1]  ……………. (2)

From (1) and (2) we have [L.H.S.] = [R.H.S.]

Hence by principle of homogeneity the given equation is dimensionally correct.

Example – 02:

To check the correctness of physical equation, v² = u² + 2as, Where ‘u’ is the initial velocity, ‘v’ is the final velocity, ‘a’ is the acceleration and s is the displacement.

Solution:

Given equation is v² = u² + 2as

L.H.S. = v, hence [L.H.S.] = [v]² = [L¹M⁰T^-1]² = [L²M⁰T^-2]  ………………. (1)

R.H.S = u² + 2as, hence [R.H.S] = [u]² + [a][s]

[R.H.S] = [L¹M⁰T^-1]² + [L¹M⁰T^-2][L¹M⁰T⁰]

[R.H.S] = [L²M⁰T^-2] + [L²M⁰T^-2]

[R.H.S] = [L²M⁰T^-2]  ……………. (2)

From (1) and (2) we have [L.H.S.] = [R.H.S.]

Hence by the principle of homogeneity the given equation is dimensionally correct.

Example – 03:

To check the correctness of physical equation, a = v²/r², Where ‘a’ is the centripetal acceleration of a body performing uniform circular motion along a circle of radius ‘r’ with linear speed ‘v’.

Solution:

Given equation is a = v²/r²

L.H.S. = a, hence [L.H.S.] = [a] = [L¹M⁰T^-2]  ………………. (1)

R.H.S = v²/r² , hence [R.H.S] = [v]²/[r]²

[R.H.S] = [L¹M⁰T^-1]² / [L¹M⁰T⁰]²

[R.H.S] = [L²M⁰T^-2] / [L²M⁰T⁰]

[R.H.S] = [L⁰M⁰T^-2]     ……………. (2)

From (1) and (2) we have [L.H.S.] ≠ [R.H.S.]

Hence by the principle of homogeneity, the given equation is dimensionally not correct.

To Find Dimensions of New Physical Quantity:

To find the form of a physical equation, we use a physical equation which contains the quantity, whose physical dimensions are to be found. Then the equation is rearranged to find the quantity whose dimensions are to be found is written. Knowing dimensions on the right hand side, dimensions of a physical quantity can be obtained.

Example – 04:

Find dimensions of universal gravitation constant (G).

Solution:

If m₁ and m₂ are two masses separated by a distance r from each other then the force of gravitation acting between them is given by Newton’s law of gravitation

Hence dimensions of universal gravitation constant are [L³M^-1T^-2]

Example – 05:

Find dimensions of the coefficient of viscosity (η).

Solution:

Le F be the viscous force acting between two layers of liquid area A having velocity difference of dv between them. Let dx be the separation between the two layers  and η is coefficient of viscosity, then by Newton’s law of viscosity

Hence dimensions of the coefficient of viscosity are [L^-1M¹T^-1]

Example – 06:

Find dimensions of electrical permittivity (ε_o).

Solution:

If q₁ and q₂ are two masses separated by a distance r from each other then the magnitude of electrostatic force acting between them is given by Coulomb’s law

Hence dimensions of electrical permittivity are [L^-3M^-1T⁴I²]

To derive the form of a physical equation:

To find the form of a physical equation, we first consider all the physical quantities on which a given physical quantity is likely to depend. Then, by the application of the principle of homogeneity of dimensions, we eliminate those quantities on which the physical quantity does not depend.

Example – 07:

The period (T) of a simple pendulum is assumed to depend on length (l) of the pendulum, acceleration due to gravity (g) and mass (m) of the bob of the pendulum. If the constant of proportionality is 2π, then find the equation for the time period of the simple pendulum.

Solution:

let  T  ∝ l^x,   T  ∝  g^y,   T  ∝  m^z,

Combining above relations we have  T ∝ l^x g^y m^z,

T  = k l^x g^y m^z …………..  (1)

By principle of homogeneity of dimensions we have

[T]  = [l]^x  [g]^y [m]^z 

∴  [L⁰M⁰T¹]  = [L¹M⁰T⁰]^x  [L¹M⁰T^-2]^y [L⁰M¹T⁰]^z 

∴  [L⁰M⁰T¹]= [L^xM⁰T⁰][L^yM⁰T^-2y][L⁰M^zT⁰]

∴   [L⁰M⁰T¹] = [L^x+y M^z T^-2y]

Considering equality of two sides we have

x + y = 0 , z = 0 , -2y = 1

∴ y = -1/2

x + y = 0

∴ x -1/2 = 0

∴ x  = 1/2

Substituting x =1/2, y = -1/2 and z = 0 in equation (1) we get

T  = k l^1/2 g^-1/2 m⁰ 

T  = k l^1/2 g^-1/2 m⁰ 

The value of constant in this case is 2π , i.e. k = 2π

This is the formula for time period of the simple pendulum.

Example – 08:

The centripetal force (F) acting on a body is assumed to depend on the mass (m) of the body, its linear velocity (v) and radius of circular path (r). Show that F ∝ mv²/r

Solution:

let  F  ∝ m^x,  F  ∝  v^y,  F  ∝  r^z,

Combining above relations we have  F ∝ m^x v^y r^z,

F  = k m^x v^y r^z …………..  (1)

By principle of homogeneity of dimensions we have

[F]  = [m]^x  [v]^y [r]^z 

∴  [L¹M¹T^-2]  = [L⁰M¹T⁰]^x  [L¹M⁰T^-1]^y [L¹M⁰T⁰]^z 

∴ [L¹M¹T^-2]= [L⁰M^xT⁰][L^yM⁰T^-y][L^zM⁰T⁰]

∴   [L¹M¹T^-2] = [L^y+z M^x T^-y]

Considering equality of two sides we have

y + z = 1 , x = 1 , – y = -2

∴ y = 2

y + z = 1

∴ 2 + z = 1

∴ z  = -1

Substituting x =1, y = 2 and z = -1 in equation (1) we get

F  = k m¹ v² r^-1 

∴  F  = k. m¹ v² / r

∴  F ∝ mv²/r (Proved as required)

To derive the relation between different units of different systems of a physical quantity:

The dimensional analysis is used to find the conversion factor when the system of units is changed from one type to other.

Example – 09:

Derive the relation between newton (N) and dyne.

Solution:

newton and dyne are units of force in S.I. and c.g.s. system respectively

Dimensions of force are [L¹M¹T^-2]

Example – 10:

Derive the relation between joule (J) and erg.

Solution:

joule and erg are units of energy in S.I. and c.g.s. system respectively

Dimensions of force are [L²M¹T^-2]

Limitations of Dimensional Analysis:

• The constant of the physical equation cannot be found using dimensional analysis. These constants are to be determined by experiments.
• A physical equation is dimensionally correct does not mean that the equation is scientifically correct.
• It is not useful when the trigonometric or exponential functions are involved.
• This method can be used only for the relations having a product or division relation. It is not used for addition or subtractive relation. Thus it is not used for deriving complex relations.
• This method does not give information about dimensional constant such as Universal gravitation constant G, Planck’s constant h, Rydberg’s constant R, etc.
• In this method, we compare the powers of the fundamental quantities to obtain a number of independent equations for finding the unknown powers. Since the total number of such equations cannot exceed the number of fundamental quantities, we cannot use this method to obtain the relation of the quantity of interest depends upon more parameters than the number of fundamental quantities used.
• In many problems, it is difficult to guess the parameters on which the quantity of interest depends. This requires a trained, subtle, and intuitive mind.

Change of Fundamental Quantities:

Example – 11:

If length ‘L’, force ‘F’ and time ‘T’ are taken as fundamental quantities. What would be the dimensional equation of mass and density?

Solution:

We have Force =  Mass × Acceleration

∴ Mass = Force / Acceleration

∴ [Mass] = [Force] / [Acceleration] = [F¹]/ [L¹T^-2] = [F¹L^-1T²]

Density = Mass / Volume

[Density] = [Mass] / [Volume]= [F¹L^-1T²] /[L³]= [F¹L^-4T²]

Dimensions of mass are [F¹L^-1T²] and that of density are  [F¹L^-4T²]

Example – 13:

If velocity (V), time (T) and force (F) were chosen as fundamental quantities, find dimensions of Mass?

Solution:

We have Force =  Mass × Acceleration

∴ Mass = Force / Acceleration

∴ Mass = Force / (Velociy/ time)

∴ Mass = (Force × time) / Velociy

∴ [Mass] = ([Force] × [time]) / [Velociy]

∴ [Mass] = ([F] × [T]) / [V]

∴ [Mass] = [F T V^-1]

Dimensions of mass are [F T V^-1]

Example – 14:

If the force (F), acceleration (A), and time (T) are taken as fundamental Units, then find the dimensions of energy.

We have Force =  Mass × Acceleration

∴ Mass = Force / Acceleration

∴ [Mass] = [F¹] / [A¹] = [F¹A^-1]

We have Acceleration = Velociy/ time

∴ Acceleration = (distance/time)/ time

∴ Acceleration = distance/time²

∴ distance = acceleration x time²

∴ [Length] = [Acceleration][time]² = [AT²]

Now, energy  = mass x gravitational acceleration x height

∴ [Energy] = [mass] x [acceleration] x [length]

∴ [Energy] = [F¹A^-1] x [A^-1] x [AT²]

∴ [Energy] = [F¹ A^-1 T²]

Dimensions of energy are [F¹ A^-1 T²]

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In this article, we shall study the concept of dimensions of physical quantities and to find dimensions of given physical quantity.

Dimensions of Physical Quantity:

The power to which fundamental units are raised in order to obtain the unit of a physical quantity is called the dimensions of that physical quantity. Dimensions of physical quantity do not depend on the system of units.

If ‘A’ is any physical quantity, then the dimensions of A are represented by [A]. Mass, length and time are represented by M, L, T respectively. Therefore, the dimensions of fundamental quantities are as follows:

• [Mass] = [M]
• [Length] = [L]
• [Time] = [T]

Now electric current (I) and temperature (K) are also considered as fundamental quantities and using them the dimensions of electrical quantities are found. An expression, which gives the relation between the derived units and fundamental units in terms of dimensions is called a dimensional equation. Thus the dimensional equation of speed is[L¹M⁰T^-1] and 1, 0, -1 are called dimensions.

Dimensions of Some Physical Quantities:

Derivation of Dimensions of Some Physical Quantity:

We should know following facts before finding dimensions of other physical quantities.

The dimensions of length (l), breadth (b), height (h), depth (d), thickness (t), width (w), circumference (c), perimeter (p), distance (S), displacement (S), radius (r), diameter (D) are [L¹M⁰T⁰]

Dimensions of mass are [L⁰M¹T⁰] and that of time are [L⁰M⁰T¹].

Dimensions and Unit of Area (A):

Area = Length × Breadth

∴ A = l × b

∴ [A] = [l] × [b]

∴ [A] =[L¹M⁰T⁰][L¹M⁰T⁰]

∴  [A] =[L²M⁰T⁰]

Dimensions of area are [L²M⁰T⁰]

S.I. Unit of the area is square metre (m²). c.g.s. unit of area is square centimetre (cm²)

Dimensions and Unit of Volume (V):

Volume = Length × Breadth × Height

∴  V = l × b × h

∴  [V] = [l] × [b] × [h]

∴  [V] = [L¹M⁰T⁰][L¹M⁰T⁰][L¹M⁰T⁰]

∴  [V] = [L³M⁰T⁰]

Dimensions of volume are [L³M⁰T⁰]

S.I. Unit of volume is cubic metre (m³). c.g.s. unit of volume is cubic centimetre3 (cm³).

Dimensions and Unit of Density (ρ or d):

Dimensions of density are [L^-3M¹T⁰]

S.I. Unit of density is kilogram per cubic metre (kg m3 or kg m^-3). c.g.s. unit of density is gram per cubic centimetre (g cm^-3)

Dimensions and Unit of Velocity or Speed (v):

Dimensions of velocity or speed are [L¹M⁰T^-1]

S.I. Unit of velocity or speed is metre per second (m s^-1). c.g.s. unit of velocity or speed is centimetre per second (cm s^-1).

Dimensions and Unit of Acceleration (a or f):

Dimensions of acceleration are [L¹M⁰T^-2]

S.I. Unit of acceleration is metre per square second (m s^-2).

c.g.s. unit of acceleration is centimetre per square second (cm s^-2).

Dimensions and Unit of Force (F):

Force = Mass × Acceleration

∴  F = m× a

∴  [F] = [m] × [a]

∴  [F] = [L⁰M¹T⁰][L¹M⁰T^-2]

∴  [F] = [L¹M¹T^-2]

Dimensions of force are [L¹M¹T^-2]

S.I. Unit of force is kilogram metre per square second (kg m s^-2), This unit is known as newton (N).

c.g.s. unit of acceleration is gram centimetre per square second (g cm s^-2). This unit is known as dyne.

Dimensions and Unit of Momentum (p):

Momentum = Mass × Velocity

∴  p = m× v

∴  [p] = [m] × [v]

∴  [p] = [L⁰M¹T⁰][L¹M⁰T^-1]

∴  [p] = [L¹M¹T^-1]

Dimensions of momentum are [L¹M¹T^-1]

S.I. Unit of momentum is kilogram meter per second (kg m s^-1).

c.g.s. unit of momentum is gram centimeter per second (g cm s^-1)

Dimensions and Unit of Impulse of Force (J):

Impulse of Force = Force × Time

∴  J =F× t

∴  [J] = [F] × [t]

∴  [J] = [L¹M¹T^-2][L⁰M⁰T¹]

∴  [J] = [L¹M¹T^-1]

Dimensions of impulse of force are [L¹M¹T^-1]

S.I. Unit of impulse of force is kilogram meter per second (kg m s^-1). Common S.I. unit used is newton second (N s).. c.g.s. unit of impulse of force is gram centimeter per second (g cm s^-1). Common c.g.s. unit is dyne second.

Dimensions and Unit of Work (W):

Work Done = Force × Displacement

∴ W =F× s

∴  [W] = [F] × [s]

∴  [W] = [L¹M¹T^-2][L¹M⁰T⁰]

∴  [W] = [L²M¹T^-2]

Dimensions of work are [L²M¹T^-2]

S.I. Unit of work is kg square metre per square second. Commonly this unit is known as joule (J).

c.g.s. Unit of work is gram square centimetre per square second. Commonly this unit is known as erg.

Dimensions and Unit of Potential Energy (E or U):

Potential Energy = Mass × Acceleration due to gravity × Height

∴ E =m× g × h

∴  [E] = [m] × [g] × [h]

∴  [E] = [L⁰M¹T⁰][L¹M⁰T^-2][L¹M⁰T⁰]

∴  [E] = [L²M¹T^-2]

Dimensions of potential energy are [L²M¹T^-2]

S.I. Unit of potential energy is kg square metre per square second. Commonly this unit is known as joule (J).

c.g.s. Unit of potential energy is gram square centimetre per square second. Commonly this unit is known as erg.

Dimensions and Unit of Kinetic Energy (E):

Kinetic Energy = 1/2 Mass × Velocity²

∴ E =1/2×m ×v²

∴  [E] = [m] × [v]²

∴  [E] = [L⁰M¹T⁰][L¹M⁰T^-1]²

∴  [E] = [L⁰M¹T⁰][L²M⁰T^-2]

∴  [E] = [L²M¹T^-2]

Dimensions of kinetic energy are [L²M¹T^-2]

S.I. Unit of kinetic energy is kg square metre per square second. Commonly this unit is known as joule (J).

c.g.s. Unit of kinetic energy is gram square centimetre per square second. Commonly this unit is known as erg.

Dimensions and Unit of Pressure (P):

Dimensions of Pressure are [L^-1M¹T^-2] S.I. Unit of pressure is

S.I. Unit of pressure is newton per square metre or pascal (N m^-2 or Pa).

c.g.s. unit of pressure  is dyne per square centimetre (dyne cm^-2)

Dimensions and Unit of Power (P):

Dimensions of Power are [L²M¹T^-3]

S.I. Unit of Power is watt (W). c.g.s. unit of power  is erg per second (erg s^-1)

 Dimensions and Unit of Electric Charge:

Electric Charge = Electric Current × Time

∴ Q =I× t

∴  [Q] = [I] × [t]

∴  [Q] = [L⁰M⁰T⁰I¹][L⁰M⁰T¹]

∴  [Q] = [L⁰M⁰T¹I¹]

Dimensions of electric charge are [L⁰M⁰T¹I¹]

S.I. Unit of electric charge is coulomb (C)

Dimensions and Unit of Electric Potential:

Dimensions of electric potential are [L²M¹T^-3I^-1]

S.I. Unit of electric potential is volt (V)

Dimensions and Unit of Electric Resistance:

Dimensions of electric resistance are [L²M¹T^-3I^-2]

S.I. Unit of electric resistance is ohm (Ω)

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