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Hint: In this question use the concept that to expand the exponential power the power of the exponential should be dimensionless. Make $\alpha {t^2}$ dimensionless using the fact that the multiplication of the respective dimensions should be one that is $\left[ \alpha \right]\left[ {{T^2}} \right] = 1$. This will help approaching the problem.

Given expression:

$P = {P_o}{e_\alpha }\left( { - \alpha {t^2}} \right)$

Here ${e_\alpha }$ is nothing but the exponential term, so write the equation in standard form we have,

$ \Rightarrow P = {P_o}{e^{ - \alpha {t^2}}}$

Now it is given that $\alpha $is a constant and t is a time.

As we all know P is the symbol of pressure.

And ${P_o}$ is also the representation of the pressure so P and ${P_o}$ have the same dimensions.

Now as we know to expand the exponential power the power of the exponential should be dimensionless.

Therefore the dimension of $\alpha {t^2}$ should have nothing i.e. it must be dimensionless.

As we know (t) is time so the dimension of the t is [T].

So the dimension of the square of the (t) is, ${t^2}$ = [${T^2}$].

Now $\alpha {t^2}$is dimensionless, so the multiplication of the respective dimensions is one.

Therefore,

$\left[ \alpha \right]\left[ {{t^2}} \right] = 1$

Now substitute the dimension of ${t^2}$ we have,

$ \Rightarrow \left[ \alpha \right]\left[ {{T^2}} \right] = 1$

$ \Rightarrow \left[ \alpha \right] = \dfrac{1}{{\left[ {{T^2}} \right]}} = \left[ {{T^{ - 2}}} \right]$

So this is the required dimension of the $\alpha $.

Note – Dimension formula is the expression for the unit of a physical quantity in terms of the fundamental quantities. The fundamental quantities are mass (M), Length (L) and time (T). A dimensional formula is expressed in terms of power of M, L and T. The trick point here was that the exponential \[P = {P_0}e( - \alpha {t^2})\] resembles exactly the same as $P = {P_o}{e^{ - \alpha {t^2}}}$, since the dimensions of P has to be similar to that of \[{P_0}\], thus exponential terms has to be dimensionless.

__Complete step-by-step solution__-Given expression:

$P = {P_o}{e_\alpha }\left( { - \alpha {t^2}} \right)$

Here ${e_\alpha }$ is nothing but the exponential term, so write the equation in standard form we have,

$ \Rightarrow P = {P_o}{e^{ - \alpha {t^2}}}$

Now it is given that $\alpha $is a constant and t is a time.

As we all know P is the symbol of pressure.

And ${P_o}$ is also the representation of the pressure so P and ${P_o}$ have the same dimensions.

Now as we know to expand the exponential power the power of the exponential should be dimensionless.

Therefore the dimension of $\alpha {t^2}$ should have nothing i.e. it must be dimensionless.

As we know (t) is time so the dimension of the t is [T].

So the dimension of the square of the (t) is, ${t^2}$ = [${T^2}$].

Now $\alpha {t^2}$is dimensionless, so the multiplication of the respective dimensions is one.

Therefore,

$\left[ \alpha \right]\left[ {{t^2}} \right] = 1$

Now substitute the dimension of ${t^2}$ we have,

$ \Rightarrow \left[ \alpha \right]\left[ {{T^2}} \right] = 1$

$ \Rightarrow \left[ \alpha \right] = \dfrac{1}{{\left[ {{T^2}} \right]}} = \left[ {{T^{ - 2}}} \right]$

So this is the required dimension of the $\alpha $.

__Hence option (C) is the correct answer.__Note – Dimension formula is the expression for the unit of a physical quantity in terms of the fundamental quantities. The fundamental quantities are mass (M), Length (L) and time (T). A dimensional formula is expressed in terms of power of M, L and T. The trick point here was that the exponential \[P = {P_0}e( - \alpha {t^2})\] resembles exactly the same as $P = {P_o}{e^{ - \alpha {t^2}}}$, since the dimensions of P has to be similar to that of \[{P_0}\], thus exponential terms has to be dimensionless.

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