Physics

If $L$ is the inductance, $'i'$ is current in the circuit, $\dfrac { 1 }{ 2 } { Li }^{ 2 }$ has the dimensions of:

Work

SOLUTION
$L \rightarrow M{ L }^{ 2 }{ I }^{ -2 }{ T }^{ -2 }$
$I\rightarrow I$
${ LI }^{ 2 } = (M{ L }^{ 2 }{ I }^{ -2 }{ T }^{ -2 }){ I }^{ 2 }= M{ L }^{ 2 }{ T }^{ -2 } \rightarrow (unit \ of \ work)$
$Power = f\times v = ML{ T }^{ -2 } \times L{ T }^{ -1 }\quad =\quad M{ L }^{ 2 }{ T }^{ -3 }$
$Pressure = f/A = \dfrac { ML{ T }^{ -2 } }{ { L }^{ 2 } } = { ML }^{ -1 }{ T }^{ -2 }$
$force = { M }^{ 1 }{ L }^{ 1 }{ T }^{ -2 }$

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Single Correct Medium Published on 18th 08, 2020
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Subjects 8
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