Question

find the dimensions of linear momentum and surface tension in terms of velocity frequency and density as a fundamental quantity ​

Answer 1

Answer:

Explanation:

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

Thus, The Relation will be,

 

  p = v⁴ d/f³

For the Surface Tension,

Let the Relation will be ⇒

γ = vᵃ dᵇ fⁿ

Dimension of the Surface tension is MLT⁻²

MLT⁻² = [LT⁻¹]ᵃ [ML⁻³]ᵇ [T⁻¹]ⁿ

∴ MLT⁻² = LᵃT⁻ᵃ Mᵇ L⁻³ᵇ T⁻ⁿ

MLT⁻² = Mᵇ Lᵃ⁻³ᵇ T⁻ⁿ⁻ᵃ

On comapring,

b = 1

a - 3b = 1

a - 3 = 1 ⇒ a = 4

-n -a = -2

-n - 4 = -2

n = -2

Hence, the Relation is  γ = v⁴d/f²