The dimensions $M L^{- 1} T^{- 2}$ may correspond to

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linear momentum

Energy per unit volume

pressure

work done by a force

answer is C, D.

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Detailed Solution

Pressure $\frac{F}{A} = \frac{M L T^{- 2}}{L^{2}} = M L^{- 1} T^{- 2}$ and energy per unit volume $\frac{E}{V} = \frac{M L^{2} T^{- 2}}{L^{3}} = M L^{- 1} T^{- 2}$ have same dimensions.

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