In order to measure physical quantities in the sub-atomic world, the quantum theory often employs energy [E], angular momentum [J] and velocity [c] as fundamental dimensions instead of the usual mass, length and time. Then, the dimension of pressure in this theory is

Question

Aicharya · Accepted Answer

$[\mathrm{E}]=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]$
$[\mathrm{J}]=\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]$
(ii) $\ldots . .$ (i
$[\mathrm{C}]=\left[\mathrm{LT}^{-1}\right]$
(iii) Solving (i), (ii) and (iii) we get,
$\left[\frac{\mathrm{E}}{\mathrm{C}^{2}}\right]=[\mathrm{M}], \quad\left[\frac{\mathrm{JC}}{\mathrm{E}}\right]=[\mathrm{L}]$ and $\left[\frac{\mathrm{J}}{\mathrm{E}}\right]=[\mathrm{T}]$
Now, [Pressure] $=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$
$=\left[\frac{\mathrm{E}}{\mathrm{C}^{2}}\right] \times\left[\frac{\mathrm{E}}{\mathrm{JC}}\right] \times\left[\frac{\mathrm{E}^{2}}{\mathrm{~J}^{2}}\right]=\frac{[\mathrm{E}]^{4}}{\left[\mathrm{~J}^{3}\right]\left[\mathrm{C}^{3}\right]}$ .

In order to measure physical quantities in the sub-atomic world, the quantum theory often employs energy [E]…

Solution