Question

Check the correctness of relation dimensionally H = rρg/2Scosθ Where His height, r is radius, ρ is density, S is surface tension and g is acceleration due to gravity.

Answer 1

Given Equation:

• $\sf{H = \dfrac{r \rho g}{2 S Cos\theta}}$

To Check:

• Whether the given equation is dimensionally correct or not.

Solution:

Given that:

• $\sf{H = \dfrac{r \rho g}{2 S Cos\theta}}$

Where:

• H = Height
• r = radius
• ρ = density
• g = acceleration due to gravity
• S = Surface Tension

Now let us check whether the given equation is dimensionally correct or not.

LHS,

Height is measured in length

Therefore,

Dimension of H = [M⁰L¹T⁰]

RHS,

$\sf{\dfrac{r \rho g}{2 S Cos\theta}}$

We know,

r is measured in length.

Therefore,

Dimension of r = [M⁰L¹T⁰]

______

$\sf{Density = \dfrac{kg}{m^3}}$

$\implies \sf{Density = \dfrac{M}{L^3}}$

$\implies \sf{Dimension\:of\:Density (\rho) = [M^1 L^{-3} T^0]}$

______

$\sf{Acceleration = \dfrac{m}{s^2}}$

$\sf{Acceleration = \dfrac{L}{T^2}}$

$\implies \sf{Dimension\:of\: g = [M^0 L^1 T^{-2}]}$

______

$\sf{Surface\:Tension = \dfrac{N}{m}}$

$\implies \sf{Surface\:Tension = \dfrac{MLT^{-2}}{L}}$

$\implies \sf{Dimension\:of\:S = [M^1 L^0 T^{-2}]}$

_______

Constants and angles are dimensionless. Therefore, there is no dimension of 2 and Cosθ.

_______

Now, putting the Dimensions in the RHS of the formula:

$\sf{\dfrac{[M^0 L^1 T^0] [M^1 L^{-3} T^0] [M^0 L^1 T^{-2}]}{M^1 L^0 T^{-2}]}}$

$= \sf{\dfrac{[M^{0 + 1 + 0} L^{1 + (-3) + 1} T^{0 + 0 + (-2)}]}{[M^1 L^0 T^{-2}]}}$

$= \sf{\dfrac{[M^1 L^{2 - 3} T^{-2}]}{M^1 L^0 T^{-2}]}}$

$= \sf{\dfrac{[M^1 L^{-1} T^{-2}]}{M^1 L^0 T^{-2}]}}$

$= \sf{\dfrac{[\cancel{M^1} L^{-1} \cancel{T^{-2}}]}{\cancel{M^1} \cancel{T^{-2}}]}}$

$= \sf{[M^0 L^{-1} T^0]}$

Here, LHS ≠ RHS.

∴ The given equation is dimensionally incorrect.

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