Question

Dimension of velocity gradient is:

$\begin{array}{rl}& A.[M^0{L}^0{T}^{-1}]\\ & B.[M{L}^{-1}{T}^{-1}]\\ & C.[M^{0}L{T}^{-1}]\\ & D.[M{L}^0{T}^{-1}]\end{array}$

Answer 1

Hint: First we have to figure out the conventional formula for velocity gradient. After that we have to break that formula into its individual constituents. Then relate all the individual constituents to their dimensional form, then bring all those dimensional forms into the equation of velocity gradient to represent the velocity gradient formula in dimensional form.

Complete step by step answer:
[M] = MASS.
[L] = LENGTH.
[T] = TIME.
[ I ] = CURRENT
We know that velocity Gradient is:
Velocity gradient=Velocity $\times$ [distance]${}^{-1}$ …… eq.1
Now we know that the dimensional formula for velocity is:
$v=[M^0{L}^1{T}^{-1}]$ ………..eq.2
And the dimensional formula for distance is:
$S=[M^0{L}^1{T}^0]$ ……….eq.3
Now if we substitute the dimensional formula for velocity that is eq.2 and the dimensional formula for distance that is eq.3, with eq.1 that is the equation of velocity gradient then,
We get,
$$V={[M^0{L}^1{T}^{-1}]\times [M^0{L}^1{T}^0]}^{-1}$$
$V=[M^0{L}^0{T}^{-1}]$
Therefore option A is the correct option.

Additional Information:
The equation that we get, when we equate a physical quantity with its dimensional formula is known as Dimensional analysis.
While doing dimensional analysis one of the most important things to remember is that the unit you wish to cancel out must be on the opposite side of the fraction.

Note: We know that the formula for velocity gradient is force divided by distance, now we also know that velocity is distance travelled divided by time, Now in dimensional analysis velocity gradient is length (L) divided by (T), and time is T. In this way we have to figure out the answer, break all the terms individually at first because there is a slight chance of mistake there while writing.