Question

The time dependence of physical quantity P is given by P = P₀$e^{- \alpha t^{2} + \beta t + \gamma}$, where α, β, γ are constants and their dimensions are given by (where t is time) –

• A. $M^{0}\text{ }\text{L}^{0}\text{ T}\text{ }^{\text{-2}}\text{, }\text{M}^{0}\text{ }\text{L}^{0}\text{ }\text{T}^{\text{ -1}}\text{, }\text{M}^{0}\text{ }\text{L}^{0}\text{ }\text{T}^{0}$
• B. $M^{0}\text{ }\text{L}^{\text{-1}}\text{,T}\text{ }^{\text{-2}}\text{, }\text{M}^{0}\text{ }\text{L}^{0}\text{ T}\text{ }^{\text{-1}}\text{, }\text{M}^{0}\text{ }\text{L}^{0}\text{ T}$
• C. $M^{0}\text{ }\text{L}^{0}\text{ T}\text{ }^{\text{-1}}\text{, M L T}\text{ }^{\text{-2}}\text{, }\text{M}^{0}\text{ }\text{L}^{0}\text{ T}\text{ }^{\text{-1}}$
• D. $\text{M, L, T, M L }\text{T}^{0}\text{, }\text{M}^{0}\text{ }\text{L}^{0}\text{ }\text{T}^{0}$

Answer 1

Answer: (A)

$M^{0}\text{ }\text{L}^{0}\text{ T}\text{ }^{\text{-2}}\text{, }\text{M}^{0}\text{ }\text{L}^{0}\text{ }\text{T}^{\text{ -1}}\text{, }\text{M}^{0}\text{ }\text{L}^{0}\text{ }\text{T}^{0}$

Answer 2

[– α t² + βt + γ] = 1

[α] = T ^–2, [β] = T ^–1, [γ] = 1