Answer
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Hint- Dimensional formula is an articulation for the unit of a physical quantity as far as the essential amounts. The fundamental quantities are mass (M), length (L), and time (T). A dimensional formula is communicated regarding powers of M, L and T.
Complete step-by-step answer:
First, we will find the dimensional formula of velocity-
$ \Rightarrow \left[ {{M^0}L{T^{ - 1}}} \right]$ as the formula for velocity is $v = \dfrac{{dis\tan ce}}{{time}}$
Now we will find out the dimensional formula for radius-
$ \Rightarrow \left[ {{M^0}L{T^0}} \right]$
The formula given in the question is $\dfrac{{{v^2}}}{r}$. Putting their dimensional formulas, we get-
$ \Rightarrow \dfrac{{{{\left[ {{M^0}L{T^{ - 1}}} \right]}^2}}}{{\left[ {{M^0}L{T^0}} \right]}} = \left[ {L{T^{ - 2}}} \right]$
As we know that $\left[ {L{T^{ - 2}}} \right]$ is the dimensional formula for acceleration.
Thus, option D is the correct option.
Note: Acceleration is the name we provide for any procedure where the velocity changes. Since velocity is a speed and a direction, there are just two different ways for you to quicken: change your speed or alter your direction—or change both.
Complete step-by-step answer:
First, we will find the dimensional formula of velocity-
$ \Rightarrow \left[ {{M^0}L{T^{ - 1}}} \right]$ as the formula for velocity is $v = \dfrac{{dis\tan ce}}{{time}}$
Now we will find out the dimensional formula for radius-
$ \Rightarrow \left[ {{M^0}L{T^0}} \right]$
The formula given in the question is $\dfrac{{{v^2}}}{r}$. Putting their dimensional formulas, we get-
$ \Rightarrow \dfrac{{{{\left[ {{M^0}L{T^{ - 1}}} \right]}^2}}}{{\left[ {{M^0}L{T^0}} \right]}} = \left[ {L{T^{ - 2}}} \right]$
As we know that $\left[ {L{T^{ - 2}}} \right]$ is the dimensional formula for acceleration.
Thus, option D is the correct option.
Note: Acceleration is the name we provide for any procedure where the velocity changes. Since velocity is a speed and a direction, there are just two different ways for you to quicken: change your speed or alter your direction—or change both.
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