Match List-I with List-II.

Choose the correct answer from the options below:

We first write one standard defining relation for each physical quantity and then deduce its dimensional formula.

Case A: Coefficient of viscosity ($$\eta$$)
Shear stress $$\tau = \eta \dfrac{dv}{dy}\quad -(1)$$
  • Dimension of shear stress $$\tau$$ is the same as pressure: $$[M\,L^{-1}\,T^{-2}]$$.
  • Velocity gradient $$\dfrac{dv}{dy}$$ has dimension $$T^{-1}$$.
Therefore,
$$\eta = \dfrac{\tau}{dv/dy} \; \Rightarrow \; [M\,L^{-1}\,T^{-2}] \; / \; [T^{-1}] = [M\,L^{-1}\,T^{-1}]$$.
Hence $$\eta$$ corresponds to List-II (IV).

Case B: Intensity of wave ($$I$$)
Intensity $$I = \dfrac{\text{Power}}{\text{Area}} \quad -(2)$$
  • Power $$P = \dfrac{\text{Energy}}{\text{time}}$$. Energy has dimension $$[M\,L^{2}\,T^{-2}]$$, so
$$P = [M\,L^{2}\,T^{-3}]$$.
  • Area has dimension $$[L^{2}]$$.
Therefore,
$$I = \dfrac{[M\,L^{2}\,T^{-3}]}{[L^{2}]} = [M\,L^{0}\,T^{-3}]$$.
Hence $$I$$ corresponds to List-II (I).

Case C: Pressure gradient ($$\nabla P$$)
Pressure gradient means pressure per unit length:
$$\nabla P = \dfrac{P}{L} \quad -(3)$$
  • Pressure $$P$$ has dimension $$[M\,L^{-1}\,T^{-2}]$$.
  • Length $$L$$ has dimension $$[L]$$.
Therefore,
$$\nabla P = \dfrac{[M\,L^{-1}\,T^{-2}]}{[L]} = [M\,L^{-2}\,T^{-2}]$$.
Hence $$\nabla P$$ corresponds to List-II (II).

Case D: Compressibility ($$\beta$$)
Compressibility is the reciprocal of bulk modulus (pressure):
$$\beta = \dfrac{1}{P} \quad -(4)$$
Since the dimension of pressure is $$[M\,L^{-1}\,T^{-2}]$$, we get
$$\beta = [M^{-1}\,L\,T^{2}]$$.
Hence $$\beta$$ corresponds to List-II (III).

Final matching
(A) $$\eta$$  →  (IV) [ML$$^{-1}$$T$$^{-1}$$]
(B) $$I$$  →  (I) [ML$$^{0}$$T$$^{-3}$$]
(C) $$\nabla P$$  →  (II) [ML$$^{-2}$$T$$^{-2}$$]
(D) $$\beta$$  →  (III) [M$$^{-1}$$LT$$^{2}$$]

Thus the correct set is: (A)-(IV), (B)-(I), (C)-(II), (D)-(III), which is given in Option B.

Answer : Option B