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Coefficient of viscosity:
From Newton’s law of viscosity:
$$F=\eta A\frac{dv}{dx}$$
$$\eta=\frac{F}{A}\cdot\frac{dx}{dv}$$
$$[\eta]=\frac{MLT^{-2}}{L^2}\cdot\frac{L}{LT^{-1}}=[ML^{-1}T^{-1}]$$
So, (A) matches with (III)
Surface tension:
$$\text{Surface tension}=\frac{\text{Force}}{\text{length}}$$
$$\frac{MLT^{-2}}{L}=[MT^{-2}]$$
So, (B) matches with (IV)
Angular momentum:
$$L=r\times p=r\cdot mv$$
$$[L]=[L]\cdot[MLT^{-1}]=[ML^2T^{-1}]$$
So, (C) matches with (II)
Rotational kinetic energy:
$$E=\frac{1}{2}I\omega^2$$
$$[I]=ML^2,\quad[\omega]=T^{-1}$$
$$[E]=ML^2\cdot T^{-2}=[ML^2T^{-2}]$$
So, (D) matches with (I)
Final matching:
(A)−(III), (B)−(IV), (C)−(II), (D)−(I)
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