Question

In a capillary tube having area of cross-section 'A', water rises to a height 'h'. If cross-sectional area is reduced to $\frac{A}{9}$. the rise of water in the capillary tube is

• A. $4h$
• B. $3h$
• C. $2h$
• D. h

Answer 1

Answer: (B)

$3h$

$h=\frac{2T\cos \theta }{r\rho g}$

$⟹h\propto \frac{1}{r}$

And $A=\pi r^2⟹r\propto \sqrt{A}$

$⟹h\propto \frac{1}{\sqrt{A}}$

Now, $\frac{h_2}{h_1}=\sqrt{\frac{A_1}{A_2}}$

$⟹\frac{h_2}{h}=\sqrt{\frac{A}{A/9}}=3$

$⟹h_2=3h$

Hence, answer is option-(B).