Question

Check the correctness of the given equation h = 2Tcosθ / rρg, where h is the height, T is the g, where h is the height, T is the surface tension,r is the radius of the tube, ρg, where h is the height, T is the is the density of the liquid and g is acceleration due to gravity.

Answer 1

Hope it helps.............

Answer 2

To check:

Correctness of the following equation :

$\boxed{h = \dfrac{2T \cos( \theta) }{r \rho g} }$

Solution:

In this type of questions , it is best to check the LHS and RHS of the equation using the dimensional analysis.

LHS:

$\therefore \: \bigg \{h \bigg \} = \bigg \{L \bigg \}$

RHS:

$\therefore \: \bigg \{\dfrac{2T \cos( \theta) }{r \rho g} \bigg \} = \bigg \{ \dfrac{ \bigg(ML{T}^{ - 2} \bigg)}{ \bigg(L \bigg) \bigg(M{L}^{ - 3} \bigg) \bigg(L{T}^{ - 2} \bigg) } \bigg \}$

$= > \: \bigg \{\dfrac{2T \cos( \theta) }{r \rho g} \bigg \} = \bigg \{ \dfrac{ \bigg(L{T}^{ - 2} \bigg)}{ \bigg(L \bigg) \bigg({L}^{ - 3} \bigg) \bigg(L{T}^{ - 2} \bigg) } \bigg \}$

$= > \: \bigg \{\dfrac{2T \cos( \theta) }{r \rho g} \bigg \} = \bigg \{ \dfrac{ \bigg(L \bigg)}{ \bigg(L \bigg) \bigg({L}^{ - 3} \bigg) \bigg(L \bigg) } \bigg \}$

$= > \: \bigg \{\dfrac{2T \cos( \theta) }{r \rho g} \bigg \} = \bigg \{L \bigg \}$

Hence, LHS = RHS.

So, the equation is dimensionally correct ✔️