Question

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

• A. a- $[{{L}^{2}}]$ b- $[T]$ c- $[L{{T}^{2}}]$
• B. a- $[L{{T}^{2}}]$ b- $[LT]$ c- $[L]$
• C. a- $[L{{T}^{-2}}]$ b- $[L]$ c- $[T]$
• D. a- $[L]$ b- $[LT]$ c- $[{{T}^{2}}]$

Answer 1

Answer: (C)

a- $[L{{T}^{-2}}]$ b- $[L]$ c- $[T]$

Answer 2

$v=at+\frac{b}{t+c}$ $[at]=[v]=[L{{T}^{-1}}]$ $\therefore$ $[a]=\frac{[L{{T}^{-1}}]}{[T]}=[L{{T}^{-2}}]$ Dimensions of $c=[t]=[T]$ (we can add quantities of same dimensions only). $\left[ \frac{b}{t+c} \right]=[v]=[L{{T}^{-1}}]$ $[b]=[L{{T}^{-1}}][T]=[L]$