We read your photo as:

Show that the equation pi a v = (4)/(3) pi a3 ( - ') g is dimensionally homogeneous.

Physics

ScanSolve AI Answer

Step-by-step solution

4 steps

i. An equation is considered homogeneous if the dimensions of all terms on both sides of the equation are identical. This means that if you express each physical quantity in terms of its fundamental dimensions (mass [M][M], length [L][L], and time [T][T]), the resulting dimensional formula must be the same for every term in the equation.

ii. To show that the equation ηπav=43πa3(ρ−ρ′)g\eta \pi a v = \frac{4}{3} \pi a^3 (\rho - \rho') g is homogeneous, we need to determine the dimensions of each term on both sides.

  1. Identify the dimensions of each variable.

    • η\eta (coefficient of viscosity): [ML−1T−1][M L^{-1} T^{-1}] (given as kg m−1s−1\text{kg m}^{-1} \text{s}^{-1})
    • aa (radius): [L][L]
    • vv (speed): [LT−1][L T^{-1}]
    • ρ\rho (density): [ML−3][M L^{-3}]
    • ρ′\rho' (density): [ML−3][M L^{-3}]
    • gg (acceleration due to gravity): [LT−2][L T^{-2}]
    • π\pi and 43\frac{4}{3} are dimensionless constants.

  2. Calculate the dimension of the Left Hand Side (LHS).

    The LHS is ηπav\eta \pi a v.

    Dimension(LHS)=Dimension(η)×Dimension(a)×Dimension(v)\text{Dimension(LHS)} = \text{Dimension}(\eta) \times \text{Dimension}(a) \times \text{Dimension}(v) Dimension(LHS)=[ML−1T−1]×[L]×[LT−1]\text{Dimension(LHS)} = [M L^{-1} T^{-1}] \times [L] \times [L T^{-1}] Dimension(LHS)=[ML−1+1+1T−1−1]\text{Dimension(LHS)} = [M L^{-1+1+1} T^{-1-1}] Dimension(LHS)=[MLT−2]\text{Dimension(LHS)} = [M L T^{-2}]

  3. Calculate the dimension of the Right Hand Side (RHS).

    The RHS is 43πa3(ρ−ρ′)g\frac{4}{3} \pi a^3 (\rho - \rho') g.
    First, for terms to be subtracted, they must have the same dimensions. Since ρ\rho and ρ′\rho' are both densities, their dimensions are the same:

    Dimension(ρ−ρ′)=[ML−3]\text{Dimension}(\rho - \rho') = [M L^{-3}]

    Now, substitute all dimensions into the RHS expression:

    Dimension(RHS)=Dimension(a3)×Dimension(ρ−ρ′)×Dimension(g)\text{Dimension(RHS)} = \text{Dimension}(a^3) \times \text{Dimension}(\rho - \rho') \times \text{Dimension}(g) Dimension(RHS)=[L3]×[ML−3]×[LT−2]\text{Dimension(RHS)} = [L^3] \times [M L^{-3}] \times [L T^{-2}] Dimension(RHS)=[ML3−3+1T−2]\text{Dimension(RHS)} = [M L^{3-3+1} T^{-2}] Dimension(RHS)=[MLT−2]\text{Dimension(RHS)} = [M L T^{-2}]

  4. Compare the dimensions of the LHS and RHS.

    We found that:

    Dimension(LHS)=[MLT−2]\text{Dimension(LHS)} = [M L T^{-2}] Dimension(RHS)=[MLT−2]\text{Dimension(RHS)} = [M L T^{-2}]

    Since the dimensions of both sides of the equation are identical, the equation is dimensionally homogeneous.

    The equation ηπav=43πa3(ρ−ρ′)g\eta \pi a v = \frac{4}{3} \pi a^3 (\rho - \rho') g is indeed homogeneous.

Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Physics Questions

QExplain why dimensional homogeneity does not guarantee the correctness of a physical equation, and…QUsing the principle of homogeneity, determine the dimensions of the constant 'a' in the van der…QExplain the concept of dimensional homogeneity in physical equations and determine the base units of…QGiven formula for volume (V) = (pi d² h)/(4)QGiven that V = (1)/(3)pi r2 h , make h the subject.
See all Physics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp