Question

The velocity $v$ of a particle at time $t$ is given by $v=at+\frac{b}{t+c}$ , where $a$, $b$ and $c$ are constants. The dimensions of $a$, $b$ and $c$ are, respectively,

• A. $\left[LT^{-2}\right]$, $\left[L\right]$ and $\left[T\right]$
• B. $\left[L^{2}\right]$, $\left[T\right]$ and $\left[LT^{2}\right]$
• C. $\left[LT^{2}\right]$, $\left[LT\right]$ and $\left[L\right]$
• D. $\left[L\right]$, $\left[LT\right]$ and $\left[T^{2}\right]$

Answer 1

The Correct Option is A

To determine the dimensions of the constants \(a\), \(b\), and \(c\) in the given velocity equation \(v = at + \frac{b}{t+c}\), we need to analyze each component's contribution to the velocity dimension. The given formula can be broken down as follows:

1. Understand the dimension of velocity v:
   Velocity \(v\) has the dimension of \(\left[LT^{-1}\right]\) where \(L\) stands for length and \(T\) stands for time.
2. Analyze the term \(at\):
   The term \(at\) must have the same dimension as velocity. Therefore, $[a][t] = [LT^{-1}]$.
   Since time \(t\) has the dimension \(\left[T\right]\), we find:
   $[a] = [LT^{-1}][T^{-1}] = [LT^{-2}]$
3. Examine the term \(\frac{b}{t+c}\):
   This term also needs to match the dimension of velocity \(\left[LT^{-1}\right]\).
   Since \(t\) and \(c\) are both terms with dimensions that simplify to time (i.e., \([T]\)), the denominator \(t+c\) has a dimension \([T]\). Thus, we derive:
   $\frac{[b]}{[T]} = [LT^{-1}]$
   Therefore, multiplying both sides by \([T]\):
   [b] = [LT^{-1}][T] = [L]
4. Evaluate the dimension of \(c\):
   As previously discussed, \(t\) and \(c\) have dimensions that combine to form the denominator in \(\frac{b}{t+c}\). Thus, both \(t\) and \(c\) must share the same dimension, leading to:
   [c] = [T]

Based on this analysis, the dimensions are:

• For \(a\): [LT^{-2}]
• For \(b\): [L]
• For \(c\): [T]

Thus, the correct answer is: \left[LT^{-2}\right], \left[L\right], and \left[T\right].