Question

1. Find the dimensional formulae of the following quantities :
(a) the universal constant of gravitation G,
(b) the surface tension S,
(c) the thermal conductivity k and
(d) the coefficient of viscosity η

Concept of Physics - 1 , HC VERMA , Chapter "Introduction to Physics"

Answer 1

Hello Dear.

(a) By using the newton law of gravitation,
We know,

F = G $\frac{m_{1} }{m_{2} }$ / r² 
where G is the gravitational constant.

So by this formula , we get
G = $\frac{Fr ^{2} }{m_{1} m_{2} }$

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
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(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]= $\frac{M}{L^3}$ × $\frac{L}{T^2}$ × L²

[S] = MT⁻²

Hence,we get the dimensional formula of surface tension =MT⁻² .

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(c) Dimension of Thermal conductivity 

∵ Q = [k A(θ₂ - θ₁)t] ÷ d
∴ k = Qd ÷ [A(θ₂ - θ₁)t]
Here Q is the heat energy , so the dimension of Q = [ML²T⁻²] ,
where A is the area t is the time & d is the thickness.

Now , K = [ML²T⁻² . L] / L².KT
K = MLT⁻³K⁻¹

Hence the dimension of thermal conductivity is MLT⁻³K⁻¹ . 
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(d) Dimension of coefficient of Viscosity .

F = η.A × $\frac{v_{2} - v_{1} }{ x_{2} - x_{1} }$ 

where F is the force , now changing into dimension.
MLT⁻² = (η) L² × $\frac{L/T}{L}$
MLT⁻² = (η) $\frac{L^2}{T}$
now , η = ML⁻¹T⁻¹
Hence dimensional of coefficient viscosity is ML⁻¹T⁻¹
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Hope it Helps.