The velocity $v$ of a particle at time $t$ is given by $v=a t+\frac{b}{t+c}$, where $a, b$ and $c$ are…

The velocity $v$ of a particle at time $t$ is given by $v=a t+\frac{b}{t+c}$, where $a, b$ and $c$ are constants. The dimensions of $a, b$ and $c$ are:
  1. L, LT and $\mathrm{LT}^2$
  2. $\mathrm{LT}^{-2}, \mathrm{~L}$ and $\mathrm{T}$
  3. $\mathrm{L}^2, \mathrm{~T}$ and $\mathrm{LT}^2$
  4. $\mathrm{LT}^2$, LT and L.

Solution

Using principle of homogeniety.
Dimension of $v=$ Dimension of $a t=$
Dimenstion of $\frac{b}{t+c}$
$\begin{array}{l}
\therefore a t =\left[\mathrm{LT}^{-1}\right] \\
\Rightarrow |a| =\frac{\left[\mathrm{LT}^{-1}\right]}{[\mathrm{T}]}=\left[\mathrm{LT}^{-2}\right] \\
\text {And } \frac{[b]}{[t]} =\left[\mathrm{LT}^{-1}\right] \\
{[b]} =\left[\mathrm{LT}^{-1}\right][\mathrm{T}]=[\mathrm{L}] \\
\text {And } c =\mathrm{T} \\
\Rightarrow c =[\mathrm{T}]
\end{array}$

Asked in: NEET 2006

Practice more Units and Dimensions questions on Aicharya