Question

b) If velocity, v=(A/B)[1-evB] where t is time, A and B are constants, write the dimensions of A and B.​

Answer 1

According to the principle of dimensional homogenity

Dimension of v = dimension of at  =  dimension of t+cb

Dimension of a =  [t][v]=TLT−2=[LT−2]

Dimension of b = [v][t]=[LT−1T]=[L] 

Dimension of c = [t]=[T]

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Answer 2

Given:

Velocity has the following relationship with A and B as follows:

$\sf{v = \dfrac{A}{B} \bigg \{1 - {e}^{vB} \bigg \} }$

Here v is Velocity and A and B are constants.

To find:

Dimensions of A and B.

Calculation:

We know that an exponential is dimension-less as follows :

$\therefore \sf{ \bigg \{vB \bigg \} = \bigg \{{M}^{0}{L}^{0} {T}^{0} \bigg \} }$

$= > \sf{ \bigg \{{M}^{0}{L}^{1}{T}^{ - 1} \times B \bigg \} = \bigg \{{M}^{0}{L}^{0} {T}^{0} \bigg \} }$

$= > \sf{ \bigg \{ B \bigg \} = \bigg \{{M}^{0}{L}^{ - 1} {T}^{1} \bigg \} }$

Now , we know that :

$\therefore \: \sf{ \bigg \{ \dfrac{A}{B} \bigg \}= \bigg \{v \bigg \}}$

$= > \: \sf{ \bigg \{ A \bigg \}= \bigg \{v \times B\bigg \}}$

$= > \: \sf{ \bigg \{ A \bigg \}= \bigg \{{M}^{0} L{T}^{ - 1} \times {L}^{ - 1} T \bigg \}}$

$= > \: \sf{ \bigg \{ A \bigg \}= \bigg \{{M}^{0}{L}^{0} {T}^{ 0} \bigg \}}$

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