Q151 / 160 · AP EAMCET 2010 · Physics

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

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

Solution

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

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

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