Question

The velocity $v$ (in $cm/s$ ) of a particle is given in terms of time $t$ (in seconds) 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)

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