
Dimensions of velocity gradient are same as that of:
$
{\text{A}}{\text{. Time period}} \\
{\text{B}}{\text{. Frequency}} \\
{\text{C}}{\text{. acceleration}} \\
{\text{D}}{\text{. length}} \\
$
Answer
598.5k+ views
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. =$\dfrac{{dv}}{{dx}}$
Now the dimensions of velocity are: $[L{T^{ - 1}}]$
Dimensions of distance: $[L]$
So, V.G. will have dimension:
$\dfrac{{[L{T^{ - 1}}]}}{{[L]}} = [{T^{ - 1}}]$
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,
$[\dfrac{1}{T}] = [{T^{ - 1}}]$
Acceleration- It is defined as the rate of change of velocity.
$
\dfrac{{dv}}{{dt}} = a \\
\dfrac{{[L{T^{ - 1}}]}}{{[T]}} = [L{T^{ - 2}}] \\
$
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.
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. =$\dfrac{{dv}}{{dx}}$
Now the dimensions of velocity are: $[L{T^{ - 1}}]$
Dimensions of distance: $[L]$
So, V.G. will have dimension:
$\dfrac{{[L{T^{ - 1}}]}}{{[L]}} = [{T^{ - 1}}]$
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,
$[\dfrac{1}{T}] = [{T^{ - 1}}]$
Acceleration- It is defined as the rate of change of velocity.
$
\dfrac{{dv}}{{dt}} = a \\
\dfrac{{[L{T^{ - 1}}]}}{{[T]}} = [L{T^{ - 2}}] \\
$
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.
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