Physics, asked by srishtiaggarwal68, 7 months ago

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

Answers

Answered by nirman95
2

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.

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