Question

Group-I indicates different properties of fluid and Group-II defines their basic dimensions in terms of Force (F), Length (L), and Time (T).

Which one of the following options identifies the correct match between Group-I and Group-II?

• A. P–1, Q–3, R–2
• B. P–2, Q–1, R–3
• C. P–3, Q–2, R–1
• D. P–2, Q–3, R–1

Hint:

Always rewrite $M$ in terms of $F, L, T$ using $M = \tfrac{F T^2}{L}$. This makes dimensional conversion into $F,L,T$ form very systematic.

Answer 1

The Correct Option is C

Step 1: Recall fundamental dimensions.
We are asked to express properties in terms of Force ($F$), Length ($L$), and Time ($T$). We know: \[ F = MLT^{-2} \Rightarrow M = \frac{F T^2}{L}. \] So we can replace $M$ everywhere with $F T^2 / L$.

Step 2: Dynamic viscosity $\mu$.
Standard dimensional formula of viscosity: \[ [\mu] = M L^{-1} T^{-1}. \] Substitute $M = \tfrac{F T^2}{L}$: \[ [\mu] = \left(\frac{F T^2}{L}\right) L^{-1} T^{-1} = F L^{-2} T. \] This matches Group-II option (3). So, \[ P \rightarrow 3. \]

Step 3: Surface tension $\sigma$.
Definition: Surface tension = force per unit length. \[ [\sigma] = \frac{F}{L} = F L^{-1}. \] This matches Group-II option (2). So, \[ Q \rightarrow 2. \]

Step 4: Density $\rho$.
Standard dimensional formula of density: \[ [\rho] = M L^{-3}. \] Substitute $M = \tfrac{F T^2}{L}$: \[ [\rho] = \frac{F T^2}{L} \cdot L^{-3} = F L^{-4} T^2. \] This matches Group-II option (1). So, \[ R \rightarrow 1. \]



Step 5: Final matching.
\[ P \rightarrow 3, Q \rightarrow 2, R \rightarrow 1. \] Final Answer:
\[ \boxed{\text{Option (C): P–3, Q–2, R–1}} \]