Physics

# If $y$ represents pressure and $x$ represents velocity  gradient, then the dimensions of $\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } }$ are

$\left[ ML^{ -1 }T^{ 0 } \right]$

##### SOLUTION
Dimensions of $\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } }$  = dimension of $\dfrac { y }{ { x }^{ 2 } }$
Dimension of y $M{ L }^{ -1 }{ T }^{ -2 }$
Dimension of x ${ T }^{ -1 }$
$\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \quad =\dfrac { M{ L }^{ -1 }{ T }^{ -2 } }{ { \left( { T }^{ -1 } \right) }^{ 2 } } \\ =M{ L }^{ -1 }{ T }^{ 0 }$

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Single Correct Medium Published on 18th 08, 2020
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