Q125 / 150 · MHT CET 2020 (20 Oct Shift 1) · Physics

In the equation, pressure $\mathrm{P}=\frac{\mathrm{c}-\mathrm{t}^{2}}{\mathrm{DS}}, \mathrm{S}$ and $\mathrm{t}$ represent the distance and time respectively. The dimensions of $\left(\frac{D}{c}\right)$ are

  • Correct answer
    A
    $\left[\mathrm{L}^{0} \mathrm{M}^{-1} \mathrm{~T}^{2}\right]$
  • B
    $\left[\mathrm{L}^{0} \mathrm{M}^{1} \mathrm{~T}^{1}\right]$
  • C
    $\left[\mathrm{L}^{1} \mathrm{M}^{-1} \mathrm{~T}^{-2}\right]$
  • D
    $\left[\mathrm{L}^{1} \mathrm{M}^{1} \mathrm{~T}^{2}\right]$

Solution

$\left[S]=[L] \quad, [t]=[T] \quad,[P]=\left[M L^{-1} T^{-2}\right]\right.$ Also. $[C]=\left[T^{2}\right] \dots(i)$ Now, $\left[M L^{-1} T^{-2}\right]=\frac{\left[T^{2}\right]}{[D][L]}$ $\Rightarrow[D]=\left[\begin{array}{lll}M^{-1} L^{0} T^{4}\end{array}\right] \dots(ii)$ Now $(i i / i) \Rightarrow \frac{[D]}{[C]}=\frac{\left[M^{-1}L^{0} T^{4}\right]}{\left[T^{2}\right]}$ $=\left[L^{0} M^{-1} T^{2}\right]$

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

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