The velocity $v$ (in m/s) of a particle is given in terms of time t (in seconds) by the equation, $v = a t + \frac{b}{(t + c)}$ . The dimensions of $a, b, c$ are:

L, T, {L T}^{2}

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{L T}^{2}, L T, L

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{L T}^{- 2}, L, T

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L^{2}, L T, T^{2}

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Solution

The correct option is B ${L T}^{- 2}, L, T$
$v = a t + \frac{b}{(t + c)}$
Since t is being added to c, hence c should have same dimension as t.
$c = T^{1}$
$a t \to M L T^{- 1} (m / s)$
$a = L T^{- 2}$
$\frac{b}{t + c} = L T^{- 1} (m / s)$
$b T^{- 1} = L T^{- 1} \Rightarrow b = L^{1}$
Hence, option C is correct.

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