Question

⭐❤〰Hey mates〰❤⭐

Find the dimensional formula of the following quantities :-

a) the universal constant of gravitation G

b) the surface tension S ,

c) the coefficient of viscosity.

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Answer 1

Hey mate,..

a) By using the newton law of gravitation,

We know,

F = G  / r² 

where G is the gravitational constant.

So by this formula , we get

G = 

so now changing into dimensions.

∴ [G] = [F]L² /M² = MLT⁻².L² / M² 

   [G] =  M⁻¹L³T⁻²

Hence the dimensional formula of universal constant of gravitation is M⁻¹L³T⁻²

b) Dimension of surface tension→

   ∵ S = ρgrh / 2

   where

S = Surface Tension, ρ = Density, g = Acceleration due to gravity, h = Height

∴ changing into dimension , [S] = (ρ)(g) L²

[S]=  ×  × L²

[S] = MT⁻²

c) Dimension of coefficient of Viscosity .

F = η.A ×  v2-v1/x2-x1

where F is the force , now changing into dimension.

MLT⁻² = (η) L² × L/T /L

MLT⁻² = (η) 

now , η = ML⁻¹T⁻¹

Hence dimensional of coefficient viscosity is ML⁻¹T⁻¹.

Hope it will help you

Answer 2

$\sf{\large {\underline {DIMENSIONAL\:FORMULAE}}}$ :

a ). Universal Gravitation Constant 'G' :

$\sf{\underline {FORMULA}}$ :$\tt{\frac{Force*{distance} ^{2}}{Mass*Mass}}$

$\sf{\underline {Unit}}$ : $\tt{N{m} ^{2}{kg}^{-2}}$

➡️ $[\tt{{M}^{-1} {L} ^{3}{T}^{-2}} ]$

b ). Surface Tension :

$\sf{\underline {FORMULA}}$ : $\tt{\frac{Force} {Length}}$

$\sf{\underline {Unit}}$ : $\tt{N{m} ^{-1}}$

➡️$[\tt{M {L} ^{0}{T}^{-2}}]$

c ). Coefficient of Viscosity :

$\sf{\underline {FORMULA}}$ : $\tt{\frac{Force*distance} {Area*velocity}}$

$\sf{\underline {Unit}}$ : $\tt{N{m} ^{-2}s}$

➡️ $[\tt{M {L} ^{-1}{T}^{-1}}]$

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$\tt{\fbox{NOTE \:-\: Dimensional \:Formulae \:are\:always \:written\:within\:a\:square\:bracket.}}$