First slide

Dimensions of physical quantities

Question

The mass of the liquid flowing per second per unit area of cross section of the tube is proportional to P^x and v^y , where P is the pressure difference and v is the velocity, then the relation between x and y is

Moderate

A

$x = y$
B

$x = - y$
C

$y^{2} = x$
D

$y = - x^{2}$

Solution

$\begin{array}{l} \frac{M}{A t} \propto P^{x} v^{y} \\ \Rightarrow M L^{- 2} T^{- 1} = {[M L^{- 1} T^{- 2}]}^{x} {[L^{1} T^{- 1}]}^{y} \\ \Rightarrow M L^{- 2} T^{- 1} = = M^{x} L^{- x + y} T^{- 2 x - y} \\ o n c o m p a r i n g w e g e t \\ x = 1, - x + y = - 2, and - 2 x - y = - 1 \end{array}$
From here, we get y = -1.
Thus, x = -y.

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App