Physics, asked by masterspiral72, 8 months ago

find the dimensions of linear momentum and surface tension in term of velocity v , density p and frequency v as fundamental quantity​

Answers

Answered by Thoroddinson
6

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

Please mark me as brainliest please if you have any doubt ask me in comment

Answered by bezetezhil26
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

Similar questions