MHT CET · Physics · Units and Dimensions
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
- 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 \(\left[\mathrm{L}^{1} \mathrm{M}^{1} \mathrm{~T}^{2}\right]\)
Answer & Solution
Correct Answer
(A) \(\left[\mathrm{L}^{0} \mathrm{M}^{-1} \mathrm{~T}^{2}\right]\)
Step-by-step Solution
Detailed explanation
\(\left[S]=[L] \quad, [t]=[T] \quad,[P]=\left[M L^{-1} T^{-2}\right]\right.\)
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 \((i i / i) \Rightarrow \frac{[D]}{[C]}=\frac{\left[M^{-1}L^{0} T^{4}\right]}{\left[T^{2}\right]}\)
\(=\left[L^{0} M^{-1} T^{2}\right]\)
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 \((i i / i) \Rightarrow \frac{[D]}{[C]}=\frac{\left[M^{-1}L^{0} T^{4}\right]}{\left[T^{2}\right]}\)
\(=\left[L^{0} M^{-1} T^{2}\right]\)
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