Question

Suppose [V] = L3, [rho]=ML–3, and [t] = T.

(a) What is the dimension of ∫rhodV?
(b) What is the dimension of dV/dt?
(c) What is the dimension of rho(dV/dt)?

Answer 1

In dimensional analysis, the integral of density over volume gives mass (M), the rate of volume change over time is represented by L³T⁻¹, indicating volume flow rate, and the product of density and volume flow rate gives mass flow rate, MT⁻¹.

Step-by-step explanation:

When considering the dimensions of physical quantities, we need to ensure that the equations are dimensionally consistent. Let's analyze the dimensions for the given scenarios:

1. Dimension of ∫rhodV: Since the dimension of ρ (rho) is ML⁻³ and the dimension of V is L³, the product of these would give us M (mass).
2. Dimension of dV/dt: The dimension of V is L³ and [t] is T, so the dimension of dV/dt is L³T⁻¹, which represents a rate of volume change over time.
3. Dimension of ρ(dV/dt): Multiplying the density ρ (ML⁻³) with the rate of volume change over time (L³T⁻¹) gives us ML⁻³ × L³T⁻¹ = MT⁻¹, which signifies mass flow rate.

These analyses are crucial in physics to ensure that the equations used to describe physical phenomena are valid.