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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Units and Measurement

Dimensional analysis of equation (velocity)$^x$=(pressure difference)$^{\Large\frac{3}{2}}$(density)$^{\Large\frac{-3}{2}}$ give values of $x$ to be

\[(a)\;6 \quad (b)\;1 \quad (c)\;-3 \quad (d)\;3\]

1 Answer

 
$[LT^{-1}]^x=[MLT^{-2}.L^{-2}]^{\large\frac{3}{2}}$$[ML^{-3}]^{\large\frac{-3}{2}}$
$\qquad\quad=L^3T^{-3}=[LT^{-1}]^3$
$\qquad\quad=>x=3$
Hence d is the correct answer.

 

answered Jun 18, 2013 by meena.p
edited Jan 9, 2014 by meena.p
 

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