Question

A physical quantity $P$ is given by the relation, $P=P_0e^{(-\alpha t^2)}$. If $t$ denotes the time, the dimensions of constant $\alpha$ are

• A. $[T]$
• B. $[T^2]$
• C. $[T^{-1}]$
• D. $[T^{-2}]$

Answer 1

Answer: (D)

(D) $[T^{-2}]$

Answer 2

The given relation is $P=P_0e^{(-\alpha t^2)}$.
For the equation to be dimensionally consistent, the argument of the exponential function must be dimensionless.
The argument is $-\alpha t^2$. The negative sign and the constant $P_0$ are dimensionless.
So, the dimension of $\alpha t^2$ must be $[M^0L^0T^0]$.
Let the dimension of $\alpha$ be $[\alpha]$.
The dimension of $t$ (time) is $[T]$. The dimension of $t^2$ is $[T^2]$.
Therefore, $[\alpha][T^2] = [M^0L^0T^0]$.
$[\alpha] = \frac{[M^0L^0T^0]}{[T^2]} = [T^{-2}]$.