Question 24: 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

Choose the correct answer: