Physics

If $F$ is given by $F$ = (${\alpha / \beta}$) (${1 - e ^{{- \beta t^2} / m}}$), where $F$ is force and $m$ & $t$ are mass and time, then dimension of $\alpha$ will be :

SOLUTION
Since, $F=\cfrac { \alpha }{ \beta } \left( 1-{ e }^{ -\cfrac { \beta { t }^{ 2 } }{ m } } \right)$

Since exponential is an dimensionless quantity

Hence,
$-\cfrac { \beta { t }^{ 2 } }{ m }$ is a dimensionless quantity

$\therefore$  $\left[ \cfrac { \beta { t }^{ 2 } }{ m } \right] =\left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 } \right]$

$\Rightarrow \beta \left[ \cfrac { { T }^{ 2 } }{ M } \right] =\left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 } \right]$

Hence, $\quad \beta =\cfrac { \left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 } \right] \left[ M \right] }{ \left[ { T }^{ 2 } \right] }$

and the two quantities cannot subtracted added until they don't have same dimension

$\left[ F \right] =\cfrac { \alpha }{ \beta }$

since, $\left[ F \right] =\left[ { M }^{ 1 }{ L }^{ 1 }{ T }^{ -2 } \right]$

$\left[ { M }^{ 1 }{ L }^{ 1 }{ T }^{ -2 } \right] =\cfrac { \left[ \alpha \right] }{ \left[ { M }^{ 1 }{ L }^{ 0 }{ T }^{ -2 } \right] }$

$\therefore \left[ \alpha \right] =\left[ { M }^{ 1 }{ L }^{ 1 }{ T }^{ -2 } \right] \left[ { M }^{ 1 }{ L }^{ 0 }{ T }^{ -2 } \right]$

$\left[ \alpha \right] =\left[ { M }^{ 2 }{ L }^{ 1 }{ T }^{ -4 } \right]$

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Subjective Medium Published on 18th 08, 2020
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Realted Questions

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Match the physical quantities given in Column I with suitable dimensions expressed in Column II.
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