Question

The time dependence of a physical quantity P is given by $P = {P_0}e( - \alpha {t^2})$ where $\alpha$ is a constant and t is time. The constant $\alpha$
$(a)$ is a dimensionless
$(b)$Has a dimension of P
$(c)$ Has a dimension of ${T^{ - 2}}$
$(d)$ Has a dimension of T

Answer 1

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.

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.