If momentum (P), area (A) and time (T) are ta | vedclass

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Question

$\mathrm{E}=\mathrm{KP}^{\mathrm{a}} \mathrm{A}^{\mathrm{b}} \mathrm{T}^{\mathrm{c}}$

$\mathrm{ML}^{2} \mathrm{T}^{-2}=\left[\mathrm{MLT}^{-1}\righ

Vedclass · Accepted Answer

$\left[ {P{A^{1/2}}{T^{ - 1}}} \right]$

Vedclass · Answer

$\mathrm{E}=\mathrm{KP}^{\mathrm{a}} \mathrm{A}^{\mathrm{b}} \mathrm{T}^{\mathrm{c}}$

$\mathrm{ML}^{2} \mathrm{T}^{-2}=\left[\mathrm{MLT}^{-1}\right]^{\mathrm{a}}\left[\mathrm{L}^{2}\right]^{\mathrm{b}}[\mathrm{T}]^{\mathrm{c}}$

$\mathrm{ML}^{2} \mathrm{T}^{-2}=\left[\mathrm{M}^{3} \mathrm{L}^{\mathrm{a}+2 \mathrm{b}} \mathrm{T}^{-\mathrm{a}+\mathrm{c}}\right]$

$a=1 \quad a+2 b=2 \quad-2=-a+c$

                 $\mathrm{b}=1 / 2 \quad \mathrm{c}=-1$

$\mathrm{E}=\left[\mathrm{PA}^{1 / 2} \mathrm{T}^{-1}\right]$

If momentum $(P)$, area $(A)$ and time $(T)$ are taken to be fundamental quantities then energy has dimensional formula

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