Q113 / 160 · JIPMER 2006 · Physics

A force $F$ is given by $F=a t+b t^2$, where, $t$ is time. What are the dimensions of $a$ and $b$ ?

  • A
    $\left[\mathrm{MLT}^{-1}\right]$ and $\left[\mathrm{MLT}^0\right]$
  • B
    $\left[\mathrm{MLT}^{-3}\right]$ and $\left[\mathrm{ML}^2 \mathrm{~T}^4\right]$
  • C
    $\left[\mathrm{MLT}^{-4}\right]$ and $\left[\mathrm{MLT}^1\right]$
  • Correct answer
    D
    $\left[\mathrm{MLT}^{-3}\right]$ and $\left[\mathrm{MLT}^{-4}\right]$

Solution

Force $F=a t+b t^2$ From principle of homogeneity, Dimensions of $a t=$ dimensions of $F$ $\therefore$ Dimensions of $a=\frac{[F]}{[t]}=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{T}]}$ $=\left[\mathrm{MLT}^{-3}\right]$ Similarly, dimensions of $b=\frac{[F]}{\left[t^2\right]}=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{T}^2\right]}=\left[\mathrm{MLT}^{-4}\right]$

Hence, the correct answer is option: 'D'.

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