In a capillary tube having the cross-sectional area 'A', the water rises to a height ′ h 1 ′ . If the cross-sectional area is reduced by 'A/25', the rise of water in capillary is

Question

In a capillary tube having the cross-sectional area 'A', the water rises to a height ′ h 1 ′ ​ . If the cross-sectional area is reduced by 'A/25', the rise of water in capillary is ′ h 2 ′ ​ . If h 2 ​ = x h 1 ​ , the value of x is

TestHub · Accepted Answer

The height of capillary rise is given by $h = \frac{2 T cos θ}{ρ g r}$

Since the cross-sectional area is $A = π r^{2}$ , we have $r \propto A$

Therefore, $h \propto \frac{1}{A}$

This gives $h_{1} A_{1} = h_{2} A_{2}$

Given $A_{1} = A$ and the area is reduced by $\frac{A}{25}$ , the new area is $A_{2} = A - \frac{A}{25} = \frac{24 A}{25}$

Substituting these values:

$h_{1} A = h_{2} \frac{24 A}{25}$

$h_{2} = h_{1} \frac{25}{24} = h_{1} \frac{5}{2 6}$

Comparing with $h_{2} = x h_{1}$ , we get $x = \frac{5}{2 6}$

Answer: $\frac{5}{2 6}$

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