The correct answer is A because:

Since k is a constant, we can omit it from the equation, leaving
- Q ∝ $\frac{r^{3} (p_{1} - p_{2})}{L} {(\frac{M}{R T})}^{\frac{1}{2}}$
The units of each quantity:
- Radius, r = m
- Change in pressure, (p₁ - p₂) = Δp = N m^–2 = (kg m s^–2) m^–2
- Length, L = m
- Molar mass, M = kg mol^–1
- Temperature, T = K
- Molar gas constant, R = J K^–1 mol^–1 = (kg m² s^–2) K^–1 mol^–1

The units on the RHS of the equation are
[Q] = $[r]^{3} [∆ p] {[L]}^{- 1} [M]^{\frac{1}{2}} {[R T]}^{- \frac{1}{2}}$
[Q] = (m)³ (kg m s^–2 m^–2) (m)^–1 $(k g m o l^{- 1})^{\frac{1}{2}} ((k g m^{2} s^{- 2} K^{- 1} m o l^{- 1}) (K))^{- \frac{1}{2}}$
- [Q] = kg (m³ m m^–2 m^–1) s^–2 ${(m^{2} s^{- 2})}^{- \frac{1}{2}}$
- [Q] = kg m s^–2 ${(m^{2} s^{- 2})}^{- \frac{1}{2}}$
- [Q] = kg m s^–2 m^–1 s
- [Q] = kg s^–1
Therefore, option A is correct

If you find it easier to visualise, you can calculate the unit using fractions

$[Q] = \frac{[r]^{3} [∆ p]}{[L]} {[\frac{M}{R T}]}^{\frac{1}{2}} = \frac{(m^{3}) (N m^{- 2})}{(m)} {(\frac{(k g m o l^{- 1})}{(J K^{- 1} m o l^{- 1}) (K)})}^{\frac{1}{2}}$

$[Q] = \frac{(m^{3}) (k g m s^{- 2} m^{- 2})}{(m)} {(\frac{(k g m o l^{- 1})}{(k g m^{2} s^{- 2} K^{- 1} m o l^{- 1}) (K)})}^{\frac{1}{2}}$

$[Q] = k g m s^{- 2} {(\frac{1}{m^{2} s^{- 2}})}^{\frac{1}{2}} = k g s^{- 1}$

	Physical Quantity	Measurable or estimated
A.	length of a football field	estimated
B.	age of the universe	measurable
C.	potential difference of a mains supply	estimated
D.	speed of sound	measurable

Measurements in Physics (DP IB Physics: SL)

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