Question

find dimensional formulq of a and b, in the relation
p=a exp (-bx)
where p= power
x=distance​

Answer 1

Given:

Relation as follows:

$\boxed{p = a {e}^{( - bx)} }$

To find:

Dimensions of a and b.

Calculation:

We know that power of an exponential is dimension-less.

$\therefore \: \bigg \{bx \bigg \} = \bigg \{ {M}^{0} {L}^{0} {T}^{0} \bigg\}$

$= > \: \bigg \{b \times L \bigg \} = \bigg \{ {M}^{0} {L}^{0} {T}^{0} \bigg\}$

$= > \: \bigg \{b \bigg \} = \bigg \{ \dfrac{{M}^{0} {L}^{0} {T}^{0}}{L} \bigg\}$

$\boxed{= > \: \bigg \{b \bigg \} = \bigg \{ {M}^{0} {L}^{ - 1} {T}^{0} \bigg\} }$

Again , we know that LHS of an equation have same dimensions as that of that of RHS:

$\therefore \: \bigg \{p \bigg \} = \bigg \{a \bigg \}$

$= > \: \bigg \{ \dfrac{force \times displacement}{time} \bigg \} = \bigg \{a \bigg \}$

$= > \: \bigg \{ \dfrac{ ML{T}^{ - 2} \times L}{T} \bigg \} = \bigg \{a \bigg \}$

$= > \: \bigg \{a \bigg \} = \bigg \{ \dfrac{ M{L}^{2} {T}^{ - 2} }{T} \bigg \}$

$\boxed{ = > \: \bigg \{a \bigg \} = \bigg \{ M{L}^{2} {T}^{ - 3} \bigg \} }$

Hope It Helps.