Question

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Answer 1

$\large {\bold{ \color{yellow}{ \underline{ \color{yellow}{Question : - }}}}}$

Find the dimensions of a and b in the relation $p \: = \frac{a - {x}^{2} }{bt}$, where x is distance , t is Time and P is Pressure .

$\large{ \bold{ \color{yellow}{ \underline{ \color{yellow}{Solution : - }}}}}$

$p \: = \frac{a - {x}^{2} }{bt} \\ \: \: a \: = {x}^{2} = [ {l}^{2} ]$

Again, $b \: = \frac{a - {x}^{2} }{P\times t} \: = \frac{ {l}^{2} }{( {ML}^{ - 1} {T}^{2}) \: \times t }$

$= [ {M}^{ - 1} {L}^{3} {T}^{1} ]$

$\frac{a}{b} = \frac{ {L}^{2} }{ {M}^{ - 1} {L}^{3} {T}^{1} } = [ {M}^{1} {L}^{ - 1 } {T}^{ - 1} ]$

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Answer 2

$\large {\bold{ \color{blue}{ \underline{ \color{blue}{Question : - }}}}}$

Find the dimensions of a and b in the relation

$\begin{gathered}p \: = \frac{a - {x}^{2} }{bt} \\ \end{gathered}$

where x is distance , t is Time and P is Pressure .

$\large{ \bold{ \color{violet}{ \underline{ \color{violet}{Ãñswër : - }}}}}$

$\begin{gathered}p \: = \frac{a - {x}^{2} }{bt} \\ \: \: a \: = {x}^{2} = [ {l}^{2} ] \end{gathered}$

Again, $\begin{gathered}b \: = \frac{a - {x}^{2} }{P\times t} \: = \frac{ {l}^{2} }{( {ML}^{ - 1} {T}^{2}) \: \times t } \\ \end{gathered}$

$\begin{gathered} = [ {M}^{ - 1} {L}^{3} {T}^{1} ] \\ \end{gathered}$

$\begin{gathered} \frac{a}{b} = \frac{ {L}^{2} }{ {M}^{ - 1} {L}^{3} {T}^{1} } = [ {M}^{1} {L}^{ - 1 } {T}^{ - 1} ] \\ \end{gathered}$

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