If the dimensions of a physical quantity are given by $M^aL^bT^c$ , the physical quantity will be.
Answer & explanation
Correct answer: option 1
Dimension of P= $\frac{Force}{Area}$
[P] = $\frac{[MLT^{-2}]}{[L^2]}$ = $[ML^{-1}T^{-2}]$
i.e a = 1, b = -1, c= -2
Dimension of velocity = $[LT^{-1}]$ a = 0 , b = 1 , c = -1
Dimension of acceleration = $[LT^{-2}]$ a = 0 , b = 1 , c = -2