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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

This question was previously asked in
WBJEE 2026 Physics and Chemistry Question Paper (24-May-2026)
The correct answer is
$LT^{-2}, L, T$

Determining Dimensions of Constants in Velocity Equation

We are given the velocity $v$ of a particle at time $t$ as: $v = at + \frac{b}{t+c}$ We need to find the dimensions of the constants $a$, $b$, and $c$. We will use the principle of dimensional homogeneity, which states that dimensions must be consistent throughout an equation.

Dimensional Analysis Steps

  1. Dimension of Velocity ($v$): Velocity is defined as displacement over time. Therefore, its dimensions are length per time. $[v] = LT^{-1}$
  2. Dimension of Constant $a$: The term $at$ must have the same dimensions as the velocity $v$, since it's added to other terms contributing to $v$. $[at] = [v]$ $[a][t] = LT^{-1}$ Since the dimension of time $t$ is $[t] = T$, we substitute this: $[a]T = LT^{-1}$ Solving for $[a]$: $[a] = \frac{LT^{-1}}{T} = LT^{-2}$
  3. Dimension of Constant $c$: In the term $t+c$, both $t$ and $c$ must have the same dimensions because quantities can only be added or subtracted if they have identical dimensions. $[c] = [t]$ $[c] = T$
  4. Dimension of Constant $b$: The term $\frac{b}{t+c}$ must also have the same dimensions as velocity $v$. $\left[\frac{b}{t+c}\right] = [v]$ $\frac{[b]}{[t+c]} = LT^{-1}$ From step 3, we know $[t+c] = [t] = T$. Substituting this: $\frac{[b]}{T} = LT^{-1}$ Solving for $[b]$: $[b] = LT^{-1} \times T = L$

Resulting Dimensions

The dimensions are found to be:

  • Dimension of $a$: $LT^{-2}$
  • Dimension of $b$: $L$
  • Dimension of $c$: $T$

This matches option 3.

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