Question

Force F and density D are related as F=(alpha)/(beta+sqrtd), Then find the dimensions of alpha and beta

Answer 1

$\huge\fcolorbox{red}{pink}{Answer}$

✏ Relation:

$\dag \: \boxed{ \rm\pink{F = \frac{ \alpha}{ \beta + \sqrt{D} } }} \: \dag$

✏ Calculation:

$\implies \rm \: \beta = \sqrt{D } = {( {M}^{1} {L}^{ - 3} )}^{ \frac{1}{2} } \\ \\ \therefore \: \underline{ \boxed{ \bold{ \rm{ \blue{ \beta = {M}^{ \frac{1}{2} }{L}^{ \frac{ - 3}{2} } } }}}} \\ \\ \implies \rm \: \alpha = F \times \beta = ( {M}^{1} {L}^{1} {T}^{ - 2} ) \times ( {M}^{ \frac{1}{2} } {L}^{ \frac{ - 3}{2} } ) \\ \\ \therefore \: \underline{ \boxed{ \bold{ \rm{ \orange{ \alpha = {M}^{ \frac{3}{2} } {L}^{ \frac{ - 1}{2} } {T}^{ - 2} }}}}}$

Answer 2

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{ \alpha ={ M}^{ \frac{3}{2} }{ L}^{ - \frac{1}{2} } {T}^{ - 2}. }$