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Question: What is the dimension of \[\dfrac{1}{2}a{{t}^{2}}\] ?...

What is the dimension of 12at2\dfrac{1}{2}a{{t}^{2}} ?

Explanation

Solution

To find the dimension of this quantity, just put the dimensions of a and t2{{t}^{2}} in the formula, where ‘a’ is the acceleration and ‘t’ is time. Try thinking of the units of ‘a’ and ‘t’ to know the dimensions of these quantities.

Formula used:
The dimensional formula of acceleration(a) is M0L1T−2{{M}^{0}}{{L}^{1}}{{T}^{-2}} and time(t) is T1{{T}^{1}}.
where, M=M= Mass, L=L= Length and T=T= Time.

Complete step by step answer:
Before getting into the solution, Let’s understand the importance and a few important points about dimensional analysis. It can be used to check the correctness of any equation.For example, you don’t remember the formula of time and are confused between the two options whether
Time=SpeedDistance\text{Time}=\dfrac{\text{Speed}}{\text{Distance}}
⇒Time=DistanceSpeed\Rightarrow\text{Time}=\dfrac{\text{Distance}}{\text{Speed}}

We can verify this using dimensional analysis and making sure that the dimensions on both sides of the equation are the same. Let’s see how.We know the dimensional formula of speed(m/s) is LT−1L{{T}^{-1}} and dimensional formula of distance(m) is L1{{L}^{1}}. So, let’s put these formulas into the two formulas above and check.

If we put this in first formula we get,
[T]=[L][T]L−1=[T]−1\left[ T \right]={{\dfrac{\left[ L \right]\left[ T \right]}{L}}^{-1}}{{=}^{{}}}{{\left[ T \right]}^{-1}}
Clearly, LHS≠\ne RHS
Putting it in the second formula, we get,
[T]=L[L][T]−1=[T]\left[ T \right]=\dfrac{L}{\left[ L \right]{{\left[ T \right]}^{-1}}}={{\left[ T \right]}^{{}}}
Here, LHS==RHS and this is the correct formula.

It can help in checking the consistency of a dimensional equation. All constants (numbers,angle,π\pi ) are dimensionless. There are constants like planck’s constant, gravitational constant which do have dimensions. Now let’s solve the question, dimensional formula of 12at2\dfrac{1}{2}a{{t}^{2}} is

\Rightarrow \dfrac{1}{2}\times \dfrac{{{L}^{1}}}{{{T}^{2}}}\times {{T}^{2}} \\\ \therefore \dfrac{1}{2}{{\left[ L \right]}^{1}}$$ **Hence, the dimension of $$\dfrac{1}{2}a{{t}^{2}}$$ is $$\dfrac{1}{2}{{\left[ L \right]}^{1}}$$.** **Note:** Some rules for dimensional analysis are: -Two physical quantities can only be equated if they have the same dimensions. -Two physical quantities can only be added/subtracted if they have the same dimensions. -Angles are dimensionless but have units (degrees or radians). Dimensional analysis gives no information about whether a physical quantity is a scalar or vector.