Question
In order to measure physical quantities in the sub-atomic world, the quantum theory often employs energy $$\left[ E \right],$$ angular momentum $$\left[ J \right]$$ and velocity $$\left[ c \right]$$ as fundamental dimensions instead of the usual mass, length and time. Then, the dimension of pressure in this theory is
A.
$$\frac{{{{\left[ E \right]}^4}}}{{{{\left[ J \right]}^3}{{\left[ c \right]}^3}}}$$
B.
$$\frac{{{{\left[ E \right]}^2}}}{{\left[ J \right]\left[ c \right]}}$$
C.
$$\frac{{\left[ E \right]}}{{{{\left[ J \right]}^2}{{\left[ c \right]}^2}}}$$
D.
$$\frac{{{{\left[ E \right]}^3}}}{{{{\left[ J \right]}^2}{{\left[ c \right]}^2}}}$$
Answer :
$$\frac{{{{\left[ E \right]}^4}}}{{{{\left[ J \right]}^3}{{\left[ c \right]}^3}}}$$
Solution :
$$\eqalign{
& \left[ E \right] = \left[ {M{L^2}{T^{ - 2}}} \right]\,......\left( {\text{i}} \right) \cr
& \left[ J \right] = \left[ {M{L^2}{T^{ - 1}}} \right]\,......\left( {{\text{ii}}} \right) \cr
& \left[ C \right] = \left[ {L{T^{ - 1}}} \right]\,......\left( {{\text{iii}}} \right) \cr} $$
Solving (i), (ii) and (iii) we get,
$$\left[ {\frac{E}{{{C^2}}}} \right] = \left[ M \right],\,\left[ {\frac{{JC}}{E}} \right] = \left[ L \right]\,\,{\text{and}}\,\,\left[ {\frac{J}{E}} \right] = \left[ T \right]$$
Now, $$\left[ {{\text{Pressure}}} \right] = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$$
$$ = \left[ {\frac{E}{{{C^2}}}} \right] \times \left[ {\frac{E}{{JC}}} \right] \times \left[ {\frac{{{E^2}}}{{{J^2}}}} \right] = \frac{{{{\left[ E \right]}^4}}}{{\left[ {{J^3}} \right]\left[ {{C^3}} \right]}}$$