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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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