1.Units, Dimensions and Measurement
hard

The mass of a liquid flowing per second per unit area of cross section of a 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

A

$x = y$

B

$x = -y$

C

$x = -y^2$

D

$y = x^2$

Solution

$\frac{M}{A t} \propto P^{x} v^{y}$

$\Rightarrow \mathrm{M} \mathrm{L}^{-2} \mathrm{T}^{-1}=\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{T}^{-2}\right]^{\mathrm{x}}\left[\mathrm{L}^{1} \mathrm{T}^{-1}\right]^{\mathrm{y}}$

$=\mathrm{M}^{\mathrm{x}} \mathrm{L}^{-\mathrm{x}+\mathrm{y}} \mathrm{T}^{-2 \mathrm{x}-\mathrm{y}}$

$\mathrm{x}=1,-\mathrm{x}+\mathrm{y}=-2$ and $-2 \mathrm{x}-\mathrm{y}=-1$

From here, we get $y=-1 .$ Thus, $x=-y$

Std 11
Physics

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