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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