Question

The position of a particle is given by x = at + bt$^{3/2$ where a and b are constants and t is the time. Then dimension of $\frac{b}{a}$ is?}

Hint:

To save time, you don't always need to find the absolute dimensions. Notice that $x = a(t) + b(t^{3/2})$. For terms to add, $[at] = [bt^{3/2}] \implies [b/a] = [t] / [t^{3/2}] = [T] / [T^{3/2}] = [T^{-1/2}]$.

Answer 1

Step 1: Understanding the Concept:
This problem uses the Principle of Dimensional Homogeneity. This principle states that in any physically correct mathematical equation, the dimensions of all the terms added or subtracted on either side must be identical.
Step 2: Key Formula or Approach:
For the equation \(x = at + bt^{3/2}\):
\[ [\text{Dimension of } x] = [\text{Dimension of } at] = [\text{Dimension of } bt^{3/2}] \]
We will find the individual dimensions of constants '\(a\)' and '\(b\)' and then divide them to find the dimension of \(b/a\).
Step 3: Detailed Explanation:
Let's denote dimensions with square brackets \([\ ]\).
Dimension of position \(x\), \([x] = [L]\)
Dimension of time \(t\), \([t] = [T]\)
From the principle of homogeneity:
1. \([at] = [x]\)
\([a][T] = [L] \implies [a] = [L][T]^{-1} = [L T^{-1}]\)
2. \([bt^{3/2}] = [x]\)
\([b][T^{3/2}] = [L] \implies [b] = [L][T]^{-3/2} = [L T^{-3/2}]\)
Now we need to find the dimension of the ratio \(b/a\):
\[ \left[\frac{b}{a}\right] = \frac{[b]}{[a]} \]
\[ \left[\frac{b}{a}\right] = \frac{[L T^{-3/2}]}{[L T^{-1}]} \]
\[ \left[\frac{b}{a}\right] = [L]^{1-1} [T]^{-3/2 - (-1)} \]
\[ \left[\frac{b}{a}\right] = [L]^0 [T]^{-3/2 + 1} \]
\[ \left[\frac{b}{a}\right] = [T]^{-1/2} \]
Step 4: Final Answer:
The dimension of \(b/a\) is \([T^{-1/2}]\).