Question

If the equation for the velocity of a particle at time 't' is \(v = at + \frac{b}{t+c}\), then the dimensions of a, b, c are respectively

• A. \(LT^{-2}, L, T \)
• B. \(L^2, L, T \)
• C. \(LT^{-2}, LT, L \)
• D. \(L, LT, L^2 \)

Hint:

The principle of homogeneity of dimensions is fundamental in dimensional analysis. It states two key rules: 1. Quantities that are added or subtracted must have the same dimensions. 2. The dimensions of the terms on both sides of an equation must be identical. This principle helps verify the correctness of physical equations and derive dimensions of unknown physical quantities.

Answer 1

The Correct Option is A

Step 1: Apply the principle of homogeneity of dimensions.
The principle of homogeneity of dimensions states that in a valid physical equation, the dimensions of each term on both sides of the equation must be the same. Also, quantities that are added or subtracted must have the same dimensions. The given equation for velocity \(v\) is: \[ v = at + \frac{b}{t+c} \] The dimension of velocity \([v]\) is \(LT^{-1}\) (Length per unit Time). The dimension of time \([t]\) is \(T\). 
Step 2: Determine the dimension of \(c\).
In the term \((t+c)\) in the denominator, \(t\) and \(c\) are added. For quantities to be added, they must have the same dimensions.
Therefore, the dimension of \(c\) must be the same as the dimension of \(t\). \[ [c] = [t] = T \] 
Step 3: Determine the dimension of \(a\).
Consider the term \(at\). According to the principle of homogeneity, the dimension of \(at\) must be equal to the dimension of \(v\). \[ [at] = [v] \] \[ [a][t] = LT^{-1} \] Substitute the dimension of \(t\): \[ [a]T = LT^{-1} \] Solve for \([a]\): \[ [a] = \frac{LT^{-1}}{T} = LT^{-1}T^{-1} = LT^{-2} \] 
Step 4: Determine the dimension of \(b\).
Consider the term \(\frac{b}{t+c}\). Its dimension must also be equal to the dimension of \(v\). \[ \left[\frac{b}{t+c}\right] = [v] \] We already found that \([t+c] = T\) (since \(t\) and \(c\) have dimension \(T\)). \[ \frac{[b]}{T} = LT^{-1} \] Solve for \([b]\): \[ [b] = LT^{-1} \cdot T = L \] 
Step 5: Summarize the dimensions.
The dimensions are:
Dimension of a: \(LT^{-2}\)
Dimension of b: \(L\)
Dimension of c: \(T\)
Thus, the dimensions of a, b, c are respectively \(LT^{-2}, L, T\).
The final answer is $\boxed{LT^{-2}, L, T}$.