Question

calculate dimensions of Linear Momentum and surface tension in terms of velocity density and frequency as fundamental units

Answer 1

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



Hope it helps.

Answer 2

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____Answer is in Attachment !!______

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Hope it helps...

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