Question:medium

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 dimension of \( a \), \( b \) and \( c \) are, respectively

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When performing dimensional analysis, always start with terms that are added or subtracted (like \( t + c \)). This immediately gives the dimension of one of the variables (here, \( [c] = T \)) without needing to analyze the rest of the equation first.
Updated On: May 28, 2026
  • \( L T^{-2}, L T, L \)
  • \( L, L T, T^2 \)
  • \( L T^{-2}, L, T \)
  • \( L^2, T, L T^2 \)
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
The Principle of Homogeneity of Dimensions states that in a physical equation, the dimensions of each term on both sides must be identical.
Additionally, only quantities with the same dimensions can be added or subtracted.
We use these rules to find the dimensions of unknown constants in an equation where the dimensions of other variables are known.
Step 2: Key Formula or Approach:
Dimensions of velocity \([v] = [L][T]^{-1} = LT^{-1}\).
Dimensions of time \([t] = [T]\).
Rule 1: \([v] = [at]\).
Rule 2: \([v] = [\frac{b}{t+c}]\).
Rule 3: \([t] = [c]\) (since they are added).
Step 3: Detailed Explanation:
1. Finding dimensions of \(c\):
In the term \(\frac{b}{t+c}\), the constant \(c\) is added to time \(t\).
Therefore, \(c\) must have the same dimensions as time.
\[ [c] = [t] = T \]
2. Finding dimensions of \(a\):
The term \(at\) must have the same dimensions as velocity \(v\).
\[ [at] = [v] \]
\[ [a][t] = LT^{-1} \]
\[ [a]T = LT^{-1} \]
\[ [a] = LT^{-1} \cdot T^{-1} = LT^{-2} \]
(This is the dimension of acceleration).
3. Finding dimensions of \(b\):
The entire term \(\frac{b}{t+c}\) must have dimensions of velocity.
Since the denominator \((t+c)\) has dimensions of time \(T\):
\[ [\frac{b}{t+c}] = [v] \]
\[ \frac{[b]}{T} = LT^{-1} \]
\[ [b] = LT^{-1} \cdot T = L \]
(This is the dimension of length).
Comparing our results \([a] = LT^{-2}\), \([b] = L\), and \([c] = T\) with the options:
Option (C) gives \(LT^{-2}, L, T\).
Step 4: Final Answer:
Applying the principle of homogeneity, we find that \(a\) has dimensions of acceleration, \(b\) has dimensions of length, and \(c\) has dimensions of time.
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