Question

Dimensions of velocity gradient are same as that of:
$$\begin{gathered} \text{A}\text{. Time period} \\ \text{B}\text{. Frequency} \\ \text{C}\text{. acceleration} \\ \text{D}\text{. length} \end{gathered}$$

Answer 1

Hint: Find the dimension of the velocity gradient and then compare it with the dimensions of the options.

Complete step-by-step solution -
The velocity gradient can be defined as the rate of change of velocity along with the distance.
So, V.G. =${\displaystyle \frac{dv}{dx}}$
Now the dimensions of velocity are: $[L{T}^{-1}]$
Dimensions of distance: $[L]$
So, V.G. will have dimension:
${\displaystyle \frac{[L{T}^{-1}]}{[L]}}=[T^{-1}]$
Dimensions of:
Time period- It is the time taken for one cycle of any periodic quantity to complete. So it will have simple units of time [T].
Frequency- it is the number of times a cycle occurs per unit time. It is the inverse of the time period, so it has dimensions,
$[{\displaystyle \frac{1}{T}}]=[T^{-1}]$
Acceleration- It is defined as the rate of change of velocity.
$$\begin{gathered} {\displaystyle \frac{dv}{dt}}=a \\ {\displaystyle \frac{[L{T}^{-1}]}{[T]}}=[L{T}^{-2}] \end{gathered}$$
Length – It is a measure of the distance between two points. It has dimensions [L].
So, we observe that the dimensions of the velocity gradient are similar to that of frequency.
The correct option is (B).

Note: The gradient of a quantity represents the change along the distance. So any quantity’s gradient will be w.r.t. distance. Having similar dimensions does not imply that the physical quantities are physically similar.