Question
Match List-I with List-II.
| List-I | List-II | ||
|---|---|---|---|
| A. | Coefficient of viscosity | I. | $$\left\lbrack {ML}^{- 1}{\text{ }T}^{- 2} \right\rbrack$$ |
| B. | Surface tension | II. | $$\left\lbrack {ML}^{2}{\text{ }T}^{- 2} \right\rbrack$$ |
| C. | Pressure | III. | $$\left\lbrack {ML}^{0}{\text{ }T}^{- 2} \right\rbrack$$ |
| D. | Surface energy | IV. | $$\left\lbrack {ML}^{- 1}{\text{ }T}^{- 1} \right\rbrack$$ |
Choose the correct answer from the options given below:
Options
Solution
(A) $\eta = \frac{Fdr}{Adv} = \frac{\left\lbrack {MLT}^{- 2} \right\rbrack\lbrack L\rbrack}{\left\lbrack L^{2} \right\rbrack\left\lbrack {LT}^{- 1} \right\rbrack} = \left\lbrack {ML}^{- 1}{\text{ }T}^{- 1} \right\rbrack$
(B) $S = \frac{F}{L} = \frac{\left\lbrack {MLT}^{- 2} \right\rbrack}{\lbrack L\rbrack} = \left\lbrack {MT}^{- 2} \right\rbrack$
(C) $P = \frac{F}{A} = \frac{\left\lbrack {MLT}^{- 2} \right\rbrack}{\left\lbrack L^{2} \right\rbrack} = \left\lbrack {ML}^{- 1}{\text{ }T}^{- 2} \right\rbrack$
(D) $E = S \times A = \left\lbrack {MT}^{- 2} \right\rbrack\left\lbrack L^{2} \right\rbrack = \left\lbrack {ML}^{2}{\text{ }T}^{- 2} \right\rbrack$
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