Question

The lower end of a capillary tube is immersed in mercury. The level of mercury in the tube is found to be 2 cm below the outer level. If the same tube is immersed in water, up to what height will the water rise in the capillary?

Answer 1

$h_{Hg}=\frac{2T_{Hg}cos\theta Hg}{r{\rho }_{Hg}g}$

$h_w=\frac{2T_{w}cos{\theta }_w}{r{\rho }_{w}g}$

Where, the symbols have their usual meanings.

$\frac{h_w}{h_{Hg}}=\frac{h_w}{h_{Hg}}\times \frac{\rho Hg}{\rho w}\times \frac{cos\theta }{cos{\theta }_{Hg}}$

$=\left(\frac{0.075}{0.465}\right)\times \left(13.6\right)\times \left(\frac{cos{0}^{\circ }}{cos{40}^{\circ }}\right)$

= 2.9

$\Rightarrow h_w=2\times \left(2.9\right)=5.8$cm