The dimensions $[M L^{- 1} T^{- 2}]$ may be of

pressure

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energy per unit volume

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force

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linear momentum

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Solution

The correct options are
A pressure
D energy per unit volume
Pressure $P =$ force/ area $= F / A$
$[P] = [M L T^{- 2}] / [L^{2}] = [M L^{- 1} T^{- 2}]$

Energy per unit volume , $E =$ work done/volume $=$ (force $\times$ displement)/volume
$[E] = \frac{[M L T^{- 2}] [L]}{[L^{3}]} = [M L^{- 1} T^{- 2}]$

The dimension of force, $[F] = [M L T^{- 2}]$ and linear momentum, $[p] = [m v] = [M L T^{- 1}]$

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M L^{- 1} T^{- 2}

Three physical quantity having dimensions [ M L^-1 T^-2] are.

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