Question

If the dimensions of a physical quantity are given by $M^{a}L^{b}T^{c}$, then the physical quantity will be
(a) velocity if a = 1, b = 0, c = – 1
(b) acceleration if a = 1, b = 1, c = – 2
(c) force if a = 0, b = – 1, c = – 2
(d) pressure if a = 1, b = – 1, c = – 2

Answer 1

c is the correct option

Answer 2

Answer:

D) a = 1, b = – 1, c = – 2

Explanation:

Dimensions of physical quantity = M^{a}L^{b}T^{c}  ( Given)

Pressure [P] = [M1L–1T–2] F

Thus,

Pressure [P} = Force/Area = Mass× acceleration/ area

[P] = M1 LT-² / L²

    = [ M1L-1T-²]

    = M^{a}L^{b}T^{c}

Therefore, a = 1, b = – 1, c = – 2

The physical quantity will be pressure.