Question

Calculate the dimensions of linear momentum and surface tension in terms of velocity (upsilon), density (rho) and frequency (V) as fundamental units.

Answer 1

$\huge{\underline{\sf{Solution:-}}}$

For Linear Momentum,

Dimensions of the Velocity[v] = Dimension of the Length/Dimension of the Time.

 = L/T

 = LT⁻¹

Dimension of the Frequency[f] = 1/Dimension of the Time

 = 1/T

 = T⁻¹

Dimension of the Density[d] = Dimension of Mass/Dimension of the Volume

= M/L³

= ML⁻³

Dimension of Linear Momentum[p] = Dimension of (Mass × Velocity]

 = MLT⁻¹ 

Now, Let the Relation between the Momentum, Velocity, Density and Frequency be 

 p = vᵃ dᵇ fⁿ

 Putting the Dimension of the Quantities.

 MLT⁻¹ = [LT⁻¹]ᵃ [ML⁻³]ᵇ [T⁻¹]ⁿ

MLT⁻¹ = LᵃT⁻ᵃ Mᵇ L⁻³ᵇ T⁻ⁿ

MLT⁻¹ = Mᵇ Lᵃ⁻³ᵇ T⁻ⁿ⁻ᵃ

On comparing,

b = 1, 

a - 3b = 1 

⇒ a - 3(1) = 1

⇒ a = 3 +1 = 4

Also, -n - a = -1

-n -4 = -1

 n = 1 - 4 

 n = -3

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Answer 2

$\huge{\underline{\boxed{\sf{\blue{Solution:-}}}}}$

For Linear Momentum,

Dimensions of the Velocity[v] = Dimension of the Length / Dimension of the Time.

• = L/T

• = LT⁻¹

Dimension of the Frequency[f] = 1/Dimension of the Time

• = 1/T

• = T⁻¹

Dimension of the Density[d] = Dimension of Mass/Dimension of the Volume

• = M/L³

• = ML⁻³

Dimension of Linear Momentum[p] = Dimension of (Mass × Velocity]

• = MLT⁻¹

Now, Let the Relation between the Momentum, Velocity, Density and Frequency be

• p = vᵃ dᵇ fⁿ

Putting the Dimension of the Quantities.

• MLT⁻¹ = [LT⁻¹]ᵃ [ML⁻³]ᵇ [T⁻¹]ⁿ

• MLT⁻¹ = LᵃT⁻ᵃ Mᵇ L⁻³ᵇ T⁻ⁿ

• MLT⁻¹ = Mᵇ Lᵃ⁻³ᵇ T⁻ⁿ⁻ᵃ

On comparing,

• b = 1,

• a - 3b = 1

• a - 3(1) = 1

• a = 3 +1 = 4

Also, -n - a = -1

• -n -4 = -1

• n = 1 - 4

• $\sf{\boxd{ n = -3}}$

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