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