Question Number 94782 by I want to learn more last updated on 21/May/20
$$\mathrm{The}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{physical}\:\mathrm{quantities}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$$\:\:\mathrm{v}\:\:=\:\:\sqrt{\frac{\mathrm{P}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{n}}}{\mathrm{x}}}\:,\:\:\mathrm{where}\:\:\mathrm{P}\:\mathrm{is}\:\mathrm{the}\:\mathrm{pressure}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{dimention}\:\mathrm{of}\:\:\:\mathrm{n}\:\:\mathrm{and}\:\:\mathrm{x}. \\ $$
Answered by mr W last updated on 21/May/20
![v=[(m/s)] p=[(N/m^2 )]=(1/n) ⇒n=[(m^2 /N)] (([(N/m^2 )])/x)=([(m/s)])^2 =[(m^2 /s^2 )] ⇒x=[(N/m^2 )×(s^2 /m^2 )]=[((kgm)/s^2 )×(s^2 /m^4 )]=[((kg)/m^3 )] i.e. x is the density.](Q94823.png)
$${v}=\left[\frac{{m}}{{s}}\right] \\ $$$${p}=\left[\frac{{N}}{{m}^{\mathrm{2}} }\right]=\frac{\mathrm{1}}{{n}} \\ $$$$\Rightarrow{n}=\left[\frac{{m}^{\mathrm{2}} }{{N}}\right] \\ $$$$\frac{\left[\frac{{N}}{{m}^{\mathrm{2}} }\right]}{{x}}=\left(\left[\frac{{m}}{{s}}\right]\right)^{\mathrm{2}} =\left[\frac{{m}^{\mathrm{2}} }{{s}^{\mathrm{2}} }\right] \\ $$$$\Rightarrow{x}=\left[\frac{{N}}{{m}^{\mathrm{2}} }×\frac{{s}^{\mathrm{2}} }{{m}^{\mathrm{2}} }\right]=\left[\frac{{kgm}}{{s}^{\mathrm{2}} }×\frac{{s}^{\mathrm{2}} }{{m}^{\mathrm{4}} }\right]=\left[\frac{{kg}}{{m}^{\mathrm{3}} }\right] \\ $$$${i}.{e}.\:{x}\:{is}\:{the}\:{density}. \\ $$
Commented by I want to learn more last updated on 21/May/20
$$\mathrm{I}\:\mathrm{appeciate}\:\mathrm{sir}. \\ $$