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The velocity $v$ of a particle at time $t$ is given by $v = at + \dfrac{b}{t + c}$, where $a$, $b$ and $c$ are constants. The dimensions of $a$, $b$ and $c$ are respectively :-
A
$LT^{-2}$, $L$ and $T$
B
$L^2$, $T$ and $LT^2$
C
$LT^2$, $LT$ and $L$
D
$L$, $LT$ and $T^2$
Detailed Solution
By the principle of homogeneity, only quantities of the same dimensions can be added, and every term of an equation must have the same dimensions.
In $(t + c)$, $c$ is added to time, so $[c] = [t] = T$.
$[v] = [at] \Rightarrow [a] = \dfrac{[v]}{[t]} = \dfrac{LT^{-1}}{T} = LT^{-2}$
$[v] = \dfrac{[b]}{[t + c]} \Rightarrow [b] = [v][t] = (LT^{-1})(T) = L$
So the dimensions of $a$, $b$ and $c$ are $LT^{-2}$, $L$ and $T$ respectively.
In $(t + c)$, $c$ is added to time, so $[c] = [t] = T$.
$[v] = [at] \Rightarrow [a] = \dfrac{[v]}{[t]} = \dfrac{LT^{-1}}{T} = LT^{-2}$
$[v] = \dfrac{[b]}{[t + c]} \Rightarrow [b] = [v][t] = (LT^{-1})(T) = L$
So the dimensions of $a$, $b$ and $c$ are $LT^{-2}$, $L$ and $T$ respectively.