Physics

# The position of a particle at time $t$ is given by the relation $x = \dfrac { v _ { 0 } } { \alpha } \left( 1 - e ^ { - \alpha t } \right)$. The dimensional formula for $\alpha ^ { 2 } \ v _ { 0 } ^ { 3 }$ will be :

##### SOLUTION
$x=\cfrac { { v }_{ 0 } }{ \alpha } \left( 1-{ e }^{ -\alpha t } \right) ....(A)$

From equation (A) since exponential is dimensionless

$\alpha t$ is dimensionless quantity

$\Rightarrow \alpha t=\left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 } \right]$

$\quad \therefore \left[ \alpha \right] =\cfrac { \left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 } \right] }{ \left[ T \right] }$

$\Rightarrow \left[ \alpha \right] =\left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ -1 } \right] ...(1)$

Also,

$\left[ \cfrac { { v }_{ 0 } }{ \alpha } \right] =\left[ x \right]$

$\left[ { v }_{ 0 } \right] =\left[ x \right] \left[ \alpha \right] ...(2)$

from (2) and (1)

$\left[ { v }_{ 0 } \right] =\left[ x \right] \left[ \alpha \right] =\left[ L \right] \left[ { M }^{ 0 }{ L }^{ 0}{ T }^{ -1 } \right]$

$=\left[ { M }^{ 0 }{ L }^{ 1 }{ T }^{ -1 } \right] ....(3)$

Dimensional formula of $\alpha ^2\ {v_0}^3$ can be obtained from from (1) and (3)

= $[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ -1 }]^2$ $\left[ { M }^{ 0 }{ L }^{ 1 }{ T }^{ -1 } \right]^3$

$=\left[ { M }^{ 0 }{ L }^{ 3 }{ T }^{ -5 } \right]$

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Subjective Medium Published on 18th 08, 2020
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