If the dimensions of a physical quantity are | vedclass

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Question

Pressure, $p = \frac{{force}}{{area}} = \frac{{mass\, 	imes \,acceleration}}{{area}}$
$	herefore \,\left[ P ight] = \frac{{{M^1}L{T^{ - 2}}}}{{{L

Vedclass · Accepted Answer

pressure if $a=1,b=-1,c=-2$

Vedclass · Answer

Pressure, $p = \frac{{force}}{{area}} = \frac{{mass\, 	imes \,acceleration}}{{area}}$
$	herefore \,\left[ P ight] = \frac{{{M^1}L{T^{ - 2}}}}{{{L^2}}} = \left[ {{M^1}{L^{ - 1}}{T^{ - 2}}} ight] = \,{M^a}{L^b}{T^c}$
$	herefore \,a = 1,\,b =  - 1,\,c =  - 2.$

If the dimensions of a physical quantity are given by $M^aL^bT^c$ ,then physical quantity will be

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