Question

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

• A. $\left[L{T}^{-2}\right],\left[L\right]$ and $\left[T\right]$
• B. $\left[L^2\right],\left[T\right]$ and $\left[L{T}^2\right]$
• C. $\left[L{T}^2\right],\left[LT\right]$ and $\left[L\right]$
• D. $\left[L\right],\left[LT\right]$ and $\left[T^2\right]$

Answer 1

Answer: (A)

$\left[L{T}^{-2}\right],\left[L\right]$ and $\left[T\right]$

Given,

$v=at+\frac{b}{t+c}$

Two entities can only be added if their dimensions are same

$⟹$ $\left[c\right]=\left[T\right]$

$\left[at\right]=\left[v\right]=\left[L{T}^{-1}\right]$

$⟹\left[a\right]=\frac{\left[L{T}^{-1}\right]}{\left[T\right]}=\left[L{T}^{-2}\right]$

$\frac{\left[b\right]}{\left[T\right]}=\left[L{T}^{-1}\right]$

$⟹\left[b\right]=\left[L\right]$

So correct answer is option (A).