Question

If the dimensions of a physical quantity are given by $\left[M^a{L}^b{T}^c\right]$, 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

The correct option is D pressure if $a=1,b=-1,c=-2$

The dimensions of the mentioned physical quantities are
velocity $=\left[M^0{L}^1{T}^{-1}\right]$
acceleration $=\left[M^0{L}^1{T}^{-2}\right]$
force $=\left[M^1{L}^1{T}^{-2}\right]$
pressure $=\left[M^1{L}^{-1}{T}^{-2}\right]$
Hence, on comparing the values, the right answer is $\left(d\right)$