MCQ
If pressure $P$, velocity $V$ and time $T$ are taken as fundamental physical quantities, the dimensional formula of force is
- ✓$P{V^2}{T^2}$
- B${P^{ - 1}}{V^2}{T^{ - 2}}$
- C$PV{T^2}$
- D${P^{ - 1}}V{T^2}$
✓
Answer
Correct option: A.
$P{V^2}{T^2}$
a
$F\,\, = \,\,{P^\alpha }{v^\beta }{T^\gamma }$
$F\,\, = \,\,{P^\alpha }{v^\beta }{T^\gamma }$
$\left[ {{M^1}{L^1}{T^{ - 2}}} \right]\,\, = \,\,{\left[ {{M^1}{L^{ - 1}}{T^2}} \right]^\alpha }\,{\left[ {L{T^{ - 1}}} \right]^\beta }{\left[ T \right]^\gamma }$
$\,\,\left[ {{M^1}{L^1}{R^{ - 2}}} \right]\,\, = \,\,\left[ {{M^\alpha }{L^{ - \alpha \, + \;\beta }}{T^{ - 2\alpha - \beta + \gamma }}} \right]$
$\alpha \,\, = \,\,1\,;\,\, - \alpha \,\, + \;\,\beta \,\, = \,\,1\,$
$\beta \,\, = \,\,\,2,\,\,2 - 2\alpha \,\, - \,\,\beta \,\, + \;\,\gamma \,\, = - 2$
$\therefore \,\, - 2\,\, - \,\,2\,\, + \;\,\gamma \,\, = - 2\,\,$
$\,\gamma \,\, = \,\,2$
$ \Rightarrow \,\,F\,\, = \,\,P{v^2}{T^2}$
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