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The time dependence of a physical quanti...

The time dependence of a physical quantity P is given by `P=P_(0) exp (-alpha t^(2))`, where `alpha` is a constant and t is time. The constant `alpha`

A

is dimensionless

B

has dimensions `[T^(-2)]`

C

has dimensions `[T^(2)]`

D

has dimensions of `p`

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To find the dimension of the constant \( \alpha \) in the equation \( P = P_0 \exp(-\alpha t^2) \), we can follow these steps: ### Step 1: Understand the Equation The equation given is \( P = P_0 \exp(-\alpha t^2) \). Here, \( P \) is a physical quantity that depends on time \( t \), and \( P_0 \) is a constant. ### Step 2: Analyze the Exponential Function In the exponential function \( \exp(-\alpha t^2) \), the argument of the exponential must be dimensionless. This means that the term \( -\alpha t^2 \) must also be dimensionless. ### Step 3: Identify the Dimensions Let’s denote the dimension of \( t \) (time) as \( [T] \). Therefore, the dimension of \( t^2 \) is: \[ [t^2] = [T^2] \] ### Step 4: Set Up the Equation for Dimensions Since \( -\alpha t^2 \) is dimensionless, we can express this as: \[ [\alpha] \cdot [T^2] = 1 \] where \( 1 \) represents the dimensionless quantity. ### Step 5: Solve for the Dimension of \( \alpha \) From the equation above, we can rearrange it to find the dimension of \( \alpha \): \[ [\alpha] = \frac{1}{[T^2]} = [T^{-2}] \] ### Step 6: Conclusion Thus, the dimension of the constant \( \alpha \) is: \[ [\alpha] = [T^{-2}] \] ### Final Answer The dimension of \( \alpha \) is \( T^{-2} \). ---
To find the dimension of the constant \( \alpha \) in the equation \( P = P_0 \exp(-\alpha t^2) \), we can follow these steps: ### Step 1: Understand the Equation The equation given is \( P = P_0 \exp(-\alpha t^2) \). Here, \( P \) is a physical quantity that depends on time \( t \), and \( P_0 \) is a constant. ### Step 2: Analyze the Exponential Function In the exponential function \( \exp(-\alpha t^2) \), the argument of the exponential must be dimensionless. This means that the term \( -\alpha t^2 \) must also be dimensionless. ...
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