Physics

# $V\propto { g }^{ x }{ h }^{ x }$, where $V$ is velocity, $g$ is  acceleration due to gravity and $h$ is height. Then $x$ and $y$ are:

$\dfrac { 1 }{ 2 } ,\dfrac { 1 }{ 2 }$

##### SOLUTION

$V \propto g^{x} h^{y}$

$LT^{-1}=(LT^{-2})^{x}(L)^{y}$

$L^{1}T^{-1}=L^{x+y}T^{-2x}$

$x+y=1$

$-2x=-1$
$\Rightarrow x=\dfrac{1}{2}, y=\dfrac{1}{2}$

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Single Correct Medium Published on 18th 08, 2020
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