Question

Find the dimensions of a/b in the relation P =ax+bt2

Answer 1

According to the rules of dimensions, all quantities should have the same dimensions. Thus , dimensions of P = dimension of ax = dimension of bt2.

We know that P = [ML^-1T^-2]

So , ax= [ML^-1T^-2]

a= [ML^-2T^-2]

And, bt^2 = [ML^-1T^-2]

b = [ML^-1T^-4]

Now, just find a/b = [L^-1 T^2]

Answer 2

Hope u like my process
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There is no relation between pressure and velocity or speed.

But it can be said that..

Change in velocity per unit time is directly proportional to the pressure applied.

So,

$= > p = \frac{v}{t} = ( {l } \: {t}^{ - 2} )$
Now..
$= > p = ax + b {t}^{2} \\ \\ or. \: \: (l \: \: {t}^{ - 2} ) = a \: (l) + b( {t}^{2} ) \\ \\ or. \: \: (l \: \: {t}^{ - 2} ) = (l \: \: {t}^{ - 2} )(a( {t}^{2} ) + b( {l}^{ - 1} {t}^{4} ) \\ \\ thus \: \: a \: = {t}^{ - 2} \\ \\ and \: \: \: b = l \: {t}^{ - 4} \\ \\ (since \: \:(a( { t }^{ 2} ) + b( {l}^{ - 1} \: {t}^{4} )) \: \: is \: dimensionless) \\ \\ so \\ \\ \frac{a}{b} = \frac{ {t}^{ - 2} }{l \: {t}^{ - 4} } = ( {l}^{ - 1} \: \: {t}^{2} )$
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