If pressure P, velocity V and time T are take | vedclass

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Question

$F\,\, = \,\,{P^\alpha }{v^\beta }{T^\gamma }$

$\left[ {{M^1}{L^1}{T^{ - 2}}} \right]\,\, = \,\,{\left[ {{M^1}{L^{ - 1}}{T^2}} \right]^\alpha }\,{\

Vedclass · Accepted Answer

$P{V^2}{T^2}$

Vedclass · Answer

$F\,\, = \,\,{P^\alpha }{v^\beta }{T^\gamma }$

$\left[ {{M^1}{L^1}{T^{ - 2}}} ight]\,\, = \,\,{\left[ {{M^1}{L^{ - 1}}{T^2}} ight]^\alpha }\,{\left[ {L{T^{ - 1}}} ight]^\beta }{\left[ T ight]^\gamma }$

$\,\,\left[ {{M^1}{L^1}{R^{ - 2}}} ight]\,\, = \,\,\left[ {{M^\alpha }{L^{ - \alpha \, + \;\beta }}{T^{ - 2\alpha  - \beta  + \gamma }}} ight]$

$\alpha \,\, = \,\,1\,;\,\, - \alpha \,\, + \;\,\beta \,\, = \,\,1\,$

$\beta \,\, = \,\,\,2,\,\,2 - 2\alpha \,\, - \,\,\beta \,\, + \;\,\gamma \,\, =  - 2$

$	herefore \,\, - 2\,\, - \,\,2\,\, + \;\,\gamma \,\, =  - 2\,\,$

$\,\gamma \,\, = \,\,2$

$ \Rightarrow \,\,F\,\, = \,\,P{v^2}{T^2}$

List$-I$	List$-II$
$(a)$ Torque	$(i)$ ${MLT}^{-1}$
$(b)$ Impulse	$(ii)$ ${MT}^{-2}$
$(c)$ Tension	$(iii)$ ${ML}^{2} {T}^{-2}$
$(d)$ Surface Tension	$(iv)$ ${ML} {T}^{-2}$

If pressure $P$, velocity $V$ and time $T$ are taken as fundamental physical quantities, the dimensional formula of force is

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Match List$-I$ with List$-II.$

List$-I$ List$-II$

$(a)$ Torque $(i)$ ${MLT}^{-1}$

$(b)$ Impulse $(ii)$ ${MT}^{-2}$

$(c)$ Tension $(iii)$ ${ML}^{2} {T}^{-2}$

$(d)$ Surface Tension $(iv)$ ${ML} {T}^{-2}$

Choose the most appropriate answer from the option given below :

A dimensionally consistent relation for the volume V of a liquid of coefficient of viscosity ' $\eta$ ' flowing per second, through a tube of radius $r$ and length / and having a pressure difference $P$ across its ends, is