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If the dimensions of a physical quantity are given by $M^{a} L^{b} T^{c}$ , then the physical quantity will be

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force if a = 0, b = – 1, c = – 2

velocity if a = 1, b = 0, c = – 1

acceleration if a = 1, b = 1, c = – 2

pressure if a = 1, b = –1, c = –2

answer is D.

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Detailed Solution

Pressure,

$\begin{aligned} P = \frac{force}{area} = \frac{mass \times acceleration}{area} \\ ∴ [P] = \frac{M^{1} {LT}^{- 2}}{L^{2}} = [M^{1} L^{- 1} T^{- 2}] = M^{a} L^{b} T^{c} . \\ ∴ a = 1, b = - 1, c = - 2 . \end{aligned}$

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