Q93 / 125 · BITSAT 2012 · Physics

If the force is given by $\mathrm{F}=\mathrm{at}+\mathrm{bt}^{2}$ with $\mathrm{t}$ as time. The dimensions of a and b are

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

Solution

Dimension of at = Dimension of F $[a t]=[F] \Rightarrow[a]=\left[\frac{F}{t}\right]$ $[b]=\left[\frac{M L T^{-2}}{T}\right] \Rightarrow[a]=\left[M L T^{-3}\right]$ Dimension of $\mathrm{bt}^{2}=$ Dimension of $\mathrm{F}$ $\left[\mathrm{bt}^{2}\right]=[\mathrm{F}] \Rightarrow[\mathrm{b}]=\left[\frac{\mathrm{F}}{\mathrm{t}^{2}}\right]$ $[\mathrm{b}]=\left[\frac{\mathrm{ML} \mathrm{T}^{-4}}{\mathrm{~T}^{2}}\right] \Rightarrow[\mathrm{b}]=\left[\mathrm{MLT}^{-4}\right]$

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

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