Question 23: Units , Dimensions and Error Analysis

Question

The time dependence of a physical quantity P is give by $\mathrm{P=P}_0 \exp \left(-\alpha t^2\right)$ where $\alpha$ is a constant and $t$ is time. The constant $\alpha$ will

Accepted Answer

The expression given is $\mathrm{P=P}_0 \exp \left(-\alpha t^2\right)$, where $\mathrm{P}$ is a physical quantity, $\alpha$ is a constant, and $t$ is time. Since the argument of the exponential function must be dimensionless (as the exponential function can only operate on dimensionless quantities), the term $-\alpha t^2$ must be dimensionless. This means that $\alpha$ must be chosen such that when it is multiplied by $t^2$ (where $t$ has dimensions of time, $[T]$), the result is a dimensionless quantity.
Let's consider the dimensions of each component:
The dimensions of $\mathrm{P}$ and $\mathrm{P}_0$ are the same, as they are both representations of the physical quantity $\mathrm{P}$.
The dimension of time, $t$, is $[T]$.
The exponential function does not add or change dimensions; it requires its argument to be dimensionless.
To ensure the argument of the exponential $-\alpha t^2$ is dimensionless, $\alpha$ must have dimensions that cancel out the dimensions of $t^2$. Since the dimensions of $t^2$ are $[T]^2$, $\alpha$ must have dimensions that, when multiplied by $[T]^2$, result in a dimensionless quantity. Therefore, $\alpha$ must have the dimensions of $[T]^{-2}$.
So, the correct option is:
Option D: Have dimensions of $t^{-2}$.

Answer

Have dimensions as that of $\mathrm{P}$

Answer

Have dimensions equal to that of $\mathrm{Pt}^2$

Answer

Have no dimensions

Answer

Have dimensions of $t^{-2}$

Units , Dimensions and Error Analysis

The time dependence of a physical quantity P is give by \mathrm{P=P}_0 \exp \left(-\alpha t^2\right) where \alpha is a constant and t is time. The constant \alpha will

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