Physics

If $$y$$ represents the pressure and $$x$$ represents the position then dimensional expression of $$\int { \dfrac { dx }{ \sqrt { p^{ 2 }- y^2}  }  } $$


ANSWER

$$[M^{-1}L^2T^{2}]$$


SOLUTION
Dimension of pressure = $$ ML^{-1}T^{-2}$$
Dimensions of p and y has to be equal
Dimensions of $$ dx = L^{1}$$
$$ \therefore $$ Dimensions of $$\int { \dfrac { dx }{ \sqrt { p^2 - y^2 }  }  }$$ 
Dimension of $$ \cfrac{dx}{y} = \cfrac{L^{1}}{ML^{-1}T^{-2}}$$
$$ = M^{-1}L^2T^2$$
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