Physics

In a test-tube, there is a liquid filled with surface tension T. The surface tension of the liquid depends upon the radius of the test-tube (r), height of the liquid column in tube (h), density of the liquid filled (d) and the acceleration due to gravity (g). Establish a relation for surface tension using the method of dimensions. 


SOLUTION
In the Question dependence of T is given on four quantities. But using dimension analysis we can determine the value of dimensions of three unknown quantities only. Therefore, we will consider one quantity independent. For our purpose, we are considering r as constant. Now applying the given dependence, we have, 
$$T \propto r^1$$
$$ T\propto h^a$$
$$T \propto d^b$$
and $$T \propto g^c$$ or $$ T \propto r^1h^ad^bg^c$$
or $$T=k r^1 /h^q d^bg^c$$.............(1)
The dimensional formula of L.H.S. 
$$=[M^1L^0T^{-2}]$$ 
The dimension of a formula of R.H.S 
$$[L^1][L]^a[M^1L^{-3}]^b[L^1T^{-2}]^c$$
$$=[L^1L^aM^bL^{-3b}L^cT^{-2c}]=[m^bL^{1+1-3b+c}T^{-2c}]$$
Since, for the correctness of formula (1), the dimensions of both the sides should be equal. 
Therefore, on comparing the dimensions of both the sides, we have,
 $$b = 1 $$...(2) 
$$1+a-3b+c=0$$.....(3) 
$$-2c = -2 $$...(4) 
From Eq. (4), c = 1 
Substituting the values of b and cis EQuestion (3), we get, 
$$1+a-3+1=0$$ 
or$$ a - 1 = 0\Rightarrow a = 1$$ 
Hence from Eq. (1), we have 
$$T = kr^1h^1d^1g^1$$
or $$T=k.rhdg$$
On the basic of practical result.
$$k=\dfrac{1}{2}$$
$$\therefore T=\dfrac{rhdg}{2}$$
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